<div dir="ltr" style="text-align: left;" trbidi="on"><h2>Sell my dream</h2><br />One morning at nine o’clock, while we were having breakfast<br />on the terrace of the Havana Riviera Hotel under a bright<br />sun, a huge wave picked up several cars that were driving<br />down the avenue along the seawall or parked on the<br />pavement, and embedded one of them in the side of the<br />hotel. It was like an explosion of dynamite that sowed panic<br />on all twenty floors of the building and turned the great<br />entrance window to dust. The many tourists in the lobby<br />were thrown into the air along with the furniture, and<br />some were cut by the hailstorm of glass. The wave must<br />have been immense, because it leaped over the wide two-<br />way street between the seawall and the hotel and still had<br />enough force to shatter the window.<br />The cheerful Cuban volunteers, with the help of the<br />fire department, picked up the debris in less than six hours,<br />and sealed off the gate to the sea and installed another,<br />and everything returned to normal. During the morning<br />nobody worried about the car encrusted in the wall, for<br />people assumed it was one of those that had been parked<br />on the pavement. But when the crane lifted it out of its<br />setting, the body of a woman was found secured behind<br />the steering wheel by a seat belt. The blow had been so<br />brutal that not a single one of her bones was left whole.<br />Her face was destroyed, her boots had been ripped apart,<br />and her clothes were in shreds. She wore a gold ring shaped<br />like a serpent, with emerald eyes. The police established<br />that she was the housekeeper for the new Portuguese<br />ambassador and his wife. She had come to Havana with<br />them two weeks before and had left that morning for the<br />market, driving a new car. Her name meant nothing to me<br />when I read it in the newspaper, but I was intrigued by the<br />snake ring and its emerald eyes. I could not find out,<br />however, on which finger she wore it.<br />This was a crucial piece of information, because I feared<br />she was an unforgettable woman whose real name I never<br />knew, and who wore a similar ring on her right forefinger<br />which, in those days, was even more unusual than it is<br />now. I had met her thirty-four years earlier in Vienna,<br />eating sausage with boiled potatoes and drinking draft beer<br />in a tavern frequented by Latin American students. I had<br />come from Rome that morning, and I still remember my<br />immediate response to her splendid soprano’s bosom, the<br />languid foxtails on her coat collar, and that Egyptian ring<br />in the shape of a serpent. She spoke an elementary Spanish<br />in a metallic accent without pausing for breath, and I<br />thought she was the only Austrian at the long wooden<br />table. But no, she had been born in Colombia and had<br />come to Austria between the wars, when she was little<br />more than a child, to study music and voice. She was<br />about thirty, and did not carry her years well, for she had<br />never been pretty and had begun to age before her time.<br />But she was a charming human being. And one of the<br />most awe-inspiring.<br />Vienna was still an old imperial city, whose<br />geographical position between the two irreconcilable worlds<br />left behind by the Second World War had turned it into a paradise of black marketeering and international espionage.<br />I could not have imagined a more suitable spot for my<br />fugitive compatriot, who still ate in the students’ tavern<br />on the corner only out of loyalty to her origins, since she<br />had more than enough money to buy meals for all her<br />table companions. She never told her real name, and we<br />always knew her by the Germanic tongue twister that we<br />Latin American students in Vienna invented for her: Frau<br />Frieda. I had just been introduced to her when I committed<br />the happy impertinence of asking how she had come to be<br />in a world so distant and different from the windy cliffs of<br />Quindio, and she answered with a devastating:<br />‘I sell my dreams.’<br />In reality, that was her only trade. She had been the<br />third of eleven children born to a prosperous shopkeeper<br />in old Caldas, and as soon as she learned to speak she<br />instituted the fine custom in her family of telling dreams<br />before breakfast, the time when their oracular qualities<br />are preserved in their purest form. When she was seven<br />she dreamed that one of her brothers was carried off by a<br />flood. Her mother, out of sheer religious superstition,<br />forbade the boy to swim in the ravine, which was his<br />favourite pastime. But Frau Frieda already had her own<br />system of prophecy.<br />‘What that dream means,’ she said, ‘isn’t that he’s<br />going to drown, but that he shouldn’t eat sweets.’<br />Her interpretation seemed an infamy to a five-year-old<br />boy who could not live without his Sunday treats. Their<br />mother, convinced of her daughter’s oracular talents,<br />enforced the warning with an iron hand. But in her first<br />careless moment the boy choked on a piece of caramel that<br />he was eating in secret, and there was no way to save him.<br />Frau Frieda did not think she could earn a living with<br />her talent until life caught her by the throat during the<br />cruel Viennese winters. Then she looked for work at the<br />first house where she would have liked to live, and when<br />she was asked what she could do, she told only the truth:<br />‘I dream.’ A brief explanation to the lady of the house was<br />all she needed, and she was hired at a salary that just<br />ed her minor expenses, but she had a nice room and<br />three meals a day—breakfast in particular, when the family<br />sat down to learn the immediate future of each of its<br />members: the father, a refined financier; the mother, a<br />joyful woman passionate about Romantic chamber music;<br />and two children, eleven and nine years old. They were all<br />religious and therefore inclined to archaic superstitions,<br />and they were delighted to take in Frau Frieda, whose<br />only obligation was to decipher the family’s daily fate<br />through her dreams.<br />She did her job well, and for a long time, above all<br />during the war years, when reality was more sinister than<br />nightmares. Only she could decide at breakfast what each<br />should do that day, and how it should be done, until her<br />predictions became the sole authority in the house. Her<br />control over the family was absolute: even the faintest sigh<br />was breathed by her order. The master of the house died<br />at about the time I was in Vienna, and had the elegance to<br />leave her a part of his estate on the condition that she<br />continue dreaming for the family until her dreams came<br />to an end.<br />I stayed in Vienna for more than a month, sharing the<br />straitened circumstances of the other students while I<br />waited for money that never arrived. Frau Frieda’s<br />unexpected and generous visits to the tavern were like<br />fiestas in our poverty-stricken regime. One night, in a beery<br />euphoria, she whispered in my ear with a conviction that<br />permitted no delay.<br />‘I only came to tell you that I dreamed about you last<br />night,’ she said. ‘You must leave right away and not come<br />back to Vienna for five years.’<br />Her conviction was so real that I boarded the last train<br />to Rome that same night. As for me, I was so influenced by<br />what she said that from then on I considered myself a<br />survivor of some catastrophe I never experienced. I still<br />have not returned to Vienna.<br />Before the disaster in Havana, I had seen Frau Frieda<br />in Barcelona in so unexpected and fortuitous a way that it<br />seemed a mystery to me. It happened on the day Pablo<br />Neruda stepped on Spanish soil for the first time since the<br />Civil War, on a stopover during a long sea voyage to<br />Valparaiso. He spent a morning with us hunting big game<br />in the second-hand bookstores, and at Porter he bought<br />an old, dried-out volume with a torn binding for which he<br />paid what would have been his salary for two months at<br />the consulate in Rangoon. He moved through the crowd<br />like an invalid elephant, with a child’s curiosity in the<br />inner workings of each thing he saw, for the world appeared<br />to him as an immense wind-up toy with which life invented<br />itself.<br />I have never known anyone closer to the idea one has<br />of a Renaissance pope: He was gluttonous and refined.<br />Even against his will, he always presided at the table.<br />Matilde, his wife, would put a bib around his neck that<br />belonged in a barbershop rather than a dining room, but it<br />was the only way to keep him from taking a bath in sauce.<br />That day at Carvalleiras was typical. He ate three whole<br />lobsters, dissecting them with a surgeon’s skill, and at the<br />same time devoured everyone else’s plate with his eyes<br />and tasted a little from each with a delight that made the<br />desire to eat contagious: clams from Galicia, mussels from<br />Cantabria, prawns from Alicante, sea cucumbers from the<br />Costa Brava. In the meantime, like the French, he spoke<br />of nothing but other culinary delicacies, in particular the<br />prehistoric shellfish of Chile, which he carried in his heart.<br />All at once he stopped eating, tuned his lobster’s antennae,<br />and said to me in a very quiet voice:<br />‘There’s someone behind me who won’t stop looking at<br />me. glanced over his shoulder, and it was true. Three<br />tables away sat an intrepid woman in an old-fashioned<br />felt hat and a purple scarf, eating without haste and staring<br />at him. I recognised her right away. She had grown old<br />and fat, but it was Frau Frieda, with the snake ring on her<br />index finger.<br />She was travelling from Naples on the same ship as<br />Neruda and his wife, but they had not seen each other on<br />board. We invited her to have coffee at our table, and I<br />encouraged her to talk about her dreams in order to astound<br />the poet. He paid no attention, for from the very beginning<br />he had announced that he did not believe in prophetic<br />dreams.<br />‘Only poetry is clairvoyant,’ he said.<br />After lunch, during the inevitable stroll along the<br />Ramblas, I lagged behind with Frau Frieda so that we could<br />renew our memories with no other ears listening. She told<br />me she had sold her properties in Austria and retired to<br />Oporto, in Portugal, where she lived in a house that she<br />described as a fake castle on a hill, from which one could<br />see all the way across the ocean to the Americas. Although<br />she did not say so, her conversation made it clear that,<br />dream by dream, she had taken over the entire fortune of<br />her ineffable patrons in Vienna. That did not surprise me,<br />however, because I had always thought her dreams were<br />no more than a stratagem for surviving. And I told her so.<br />She laughed her irresistible laugh. ‘You’re as impudent<br />as ever,’ she said. And said no more, because the rest of<br />the group had stopped to wait for Neruda to finish talking<br />in Chilean slang to the parrots along the Rambla de los<br />Pájaros. When we resumed our conversation, Frau Frieda<br />changed the subject.<br />‘By the way,’ she said, ‘you can go back to Vienna<br />now.’<br />Only then did I realise that thirteen years had gone by<br />since our first meeting.<br />‘Even if your dreams are false, I’ll never go back,’ I told<br />her. ‘Just in case.’<br />At three o’clock we left her to accompany Neruda to his<br />sacred siesta, which he took in our house after solemn preparations that in some way recalled the Japanese tea<br />ceremony. Some windows had to be opened and others closed<br />to achieve the perfect degree of warmth, and there had to<br />be a certain kind of light from a certain direction, and<br />absolute silence. Neruda fell asleep right away, and woke<br />ten minutes later, as children do, when we least expected<br />it. He appeared in the living room refreshed, and with the<br />monogram of the pillowcase imprinted on his cheek.<br />‘I dreamed about that woman who dreams,’ he said.<br />Matilde wanted him to tell her his dream.<br />‘I dreamed she was dreaming about me,’ he said.<br />‘That’s right out of Borges,’ I said.<br />He looked at me in disappointment.<br />‘Has he written it already?’<br />‘If he hasn’t he’ll write it sometime,’ I said. ‘It will be<br />one of his labyrinths.’<br />As soon as he boarded the ship at six that evening, Neruda<br />took his leave of us, sat down at an isolated table, and began<br />to write fluid verses in the green ink he used for drawing<br />flowers and fish and birds when he dedicated his books. At<br />the first ‘All ashore’ we looked for Frau Frieda, and found her<br />at last on the tourist deck, just as we were about to leave<br />without saying good-bye. She too had taken a siesta.<br />‘I dreamed about the poet,’ she said.<br />In astonishment I asked her to tell me her dream.<br />‘I dreamed he was dreaming about me,’ she said, and<br />my look of amazement disconcerted her. ‘What did you<br />expect? Sometimes, with all my dreams, one slips in that<br />has nothing to do with real life.’<br />I never saw her again or even wondered about her<br />until I heard about the snake ring on the woman who<br />died in the Havana Riviera disaster. And I could not resist the temptation of questioning the Portuguese<br />ambassador when we happened to meet some months<br />later at a diplomatic reception. The ambassador spoke<br />about her with great enthusiasm and enormous<br />admiration. ‘You cannot imagine how extraordinary she<br />was,’ he said. ‘You would have been obliged to write a<br />story about her.’ And he went on in the same tone, with<br />surprising details, but without the clue that would have<br />allowed me to come to a final conclusion.<br />‘In concrete terms,’ I asked at last, ‘what did she do?’<br />‘Nothing,’ he said, with a certain disenchantment. ‘She<br />dreamed<br /><br /><a name='more'></a><h2 style="text-align: left;">Eveline</h2><div><div>She sat at the window watching the evening invade the</div><div>avenue. Her head was leaned against the window curtains</div><div>and in her nostrils was the odour of dusty cretonne. She</div><div>was tired.</div><div>Few people passed. The man out of the last house</div><div>passed on his way home; she heard his footsteps clacking</div><div>along the concrete pavement and afterwards crunching on</div><div>the cinder path before the new red houses. One time there</div><div>used to be a field there in which they used to play every</div><div>evening with other people’s children. Then a man from</div><div>Belfast bought the field and built houses in it—not like</div><div>their little brown houses but bright brick houses with</div><div>shining roofs. The children of the avenue used to play</div><div>together in that field—the Devines, the Waters, the Dunns,</div><div>little Keogh the cripple, she and her brothers and sisters.</div><div>Ernest, however, never played: he was too grown up. Her</div><div>father used often to hunt them in and out of the field with</div><div>his blackthorn stick; but usually little Keogh used to keep</div></div>nix* and call out when he saw her father coming. Still they<br />seemed to have been rather happy then. Her father was<br />not so bad then; and besides, her mother was alive. That<br />was a long time ago; she and her brothers and sisters<br />were all grown up; her mother was dead. Tizzie Dunn was<br />dead, too, and the Waters had gone back to England.<br />Everything changes. Now she was going to go away like<br />the others, to leave her home.<br />Home! She looked round the room, reviewing all its<br />familiar objects which she had dusted once a week for so<br />many years, wondering where on earth all the dust came<br />from. Perhaps she would never see again those familiar<br />objects from which she had never dreamed of being divided.<br />And yet during all those years she had never found out the<br />name of the priest whose yellowing photograph hung on<br />the wall above the broken harmonium beside the coloured<br />print of the promises made to Blessed Margaret Mary<br />Alacoque. He had been a school friend of her father.<br />Whenever he showed the photograph to a visitor her father<br />used to pass it with a casual word: ‘He is in Melbourne<br />now.’<br />She had consented to go away, to leave her home. Was<br />that wise? She tried to weigh each side of the question. In<br />her home anyway she had shelter and food; she had those<br />whom she had known all her life about her. Of course she<br />had to work hard, both in the house and at business. What<br />would they say of her in the Stores when they found out<br />that she had run away with a fellow? Say she was a fool,<br />perhaps; and her place would be filled up by advertisement.<br />Miss Gavan would be glad. She had always had an edge on<br />her, especially whenever there were people listening.<br />‘Miss Hill, don’t you see these ladies are waiting?’<br />‘Look lively, Miss Hill, please.’<br />She wouId not cry many tears at leaving the Stores.<br />But in her new home, in a distant unknown country,<br />it would not be like that. Then she would be married—she,<br />Eveline. People would treat her with respect then. She<br />would not be treated as her mother had been. Even now,<br />though she was over nineteen, she sometimes felt herself<br />in danger of her father’s violence. She knew it was that<br />that had given her the palpitations. When they were growing<br />up he had never gone for her, like he used to go for Harry<br />and Ernest, because she was a girl; but latterly he had<br />begun to threaten her and say what he would do to her<br />only for her dead mother’s sake. And now she had nobody<br />to protect her. Ernest was dead and Harry, who was in the<br />church decorating business, was nearly always down<br />somewhere in the country. Besides, the invariable squabble<br />for money on Saturday nights had begun to weary her<br />unspeakably. She always gave her entire wages—seven<br />shillings—and Harry always sent up what he could but<br />the trouble was to get any money from her father. He said<br />she used to squander the money, that she had no head,<br />that he wasn’t going to give her his hard earned money to<br />throw about the streets, and much more, for he was usually<br />fairly bad on Saturday night. In the end he would give her the money and ask her had she any intention of buying<br />Sunday’s dinner. Then she had to rush out as quickly as<br />she could and do her marketing, holding her black leather<br />purse tightly in her hand as she elbowed her way through<br />the crowds and returning home late under her load of<br />provisions. She had hard work to keep the house together<br />and to see that the two young children who had been left<br />to her charge went to school regularly and got their meals<br />regularly. It was hard work—a hard life—but now that she<br />was about to leave it she did not find it a wholly undesirable<br />life.She was about to explore another life with Frank.<br />Frank was very kind, manly, open-hearted. She was to go<br />away with him by the night boat to be his wife and to live<br />with him in Buenos Aires where he had a home waiting for<br />her. How well she remembered the first time she had seen<br />him; he was lodging in a house on the main road where<br />she used to visit. It seemed a few weeks ago. He was<br />standing at the gate, his peaked cap pushed back on his<br />head and his hair tumbled forward over a face of bronze.<br />Then they had come to know each other. He used to meet<br />her outside the Stores every evening and see her home. He<br />took her to see The Bohemian Girl and she felt elated as<br />she sat in an unaccustomed part of the theatre with him.<br />He was awfully fond of music and sang a little. People<br />knew that they were courting and, when he sang about<br />the lass that loves a sailor, she always felt pleasantly<br />confused. He used to call her Poppens out of fun. First of<br />all it had been an excitement for her to have a fellow and<br />then she had begun to like him. He had tales of distant<br />countries. He had started as a deck boy at a pound a month<br />on a ship of the Allan Line going out to Canada. He told<br />her the names of the ships he had been on and the names<br />of the different services. He had sailed through the Straits<br />of Magellan and he told her stories of the terrible<br />Patagonians. He had fallen on his feet in Buenos Aires, he<br />said, and had come over to the old country just for a holiday.<br />Of course, her father had found out the affair and had<br />forbidden her to have anything to say to him.<br />‘I know these sailor chaps,’ he said.<br />One day he had quarrelled with Frank and after that<br />she had to meet her lover secretly.<br />The evening deepened in the avenue. The white of two<br />letters in her lap grew indistinct. One was to Harry; the<br />other was to her father. Ernest had been her favourite but<br />she liked Harry too. Her father was becoming old lately,<br />she noticed; he would miss her. Sometimes he could be<br />very nice. Not long before, when she had been laid up for a<br />day, he had read her out a ghost story and made toast for<br />her at the fire. Another day, when their mother was alive,<br />they had all gone for a picnic to the Hill of Howth. She<br />remembered her father putting on her mother’s bonnet to<br />make the children laugh.<br />Her time was running out but she continued to sit by<br />the window, leaning her head against the window curtain,<br />inhaling the odour of dusty cretonne. Down far in the<br />avenue she could hear a street organ playing. She knew<br />the air. Strange that it should come that very night to<br />remind her of the promise to her mother, her promise to<br />keep the home together as long as she could. She<br />remembered the last night of her mother’s illness; she was<br />again in the close dark room at the other side of the hall<br />and outside she heard a melancholy air of Italy. The organ<br />player had been ordered to go away and given six-pence.<br />She remembered her father strutting back into the sickroom<br />saying: ‘Damned Italians! coming over here!’<br />As she mused—the pitiful vision of her mother’s life<br />laid its spell on the very quick of her being that life of<br />commonplace sacrifices closing in final craziness. She<br />trembled as she heard again her mother’s voice saying<br />constantly with foolish insistence: ‘Derevaun Seraun!<br />She stood up in a sudden impulse of terror. Escape! She<br />must escape! Frank would save her. He would give her life,<br />perhaps love, too. But she wanted to live. Why should she be<br />unhappy? She had a right to happiness. Frank would take<br />her in his arms, fold her in his arms. He would save her.<br />She stood among the swaying crowd in the station at<br />the North Wall. He held her hand and she knew that he<br />was speaking to her, saying something about the passage<br />over and over again. The station was full of soldiers with<br />brown baggages. Through the wide doors of the sheds she<br />caught a glimpse of the black mass of the boat, lying in<br />beside the quay wall, with illumined portholes. She<br />answered nothing. She felt her cheek pale and cold and,<br />out of a maze of distress, she prayed to God to direct her,<br />to show her what was her duty. The boat blew a long<br />mournful whistle into the mist. If she went, tomorrow she<br />would be on the sea with Frank, steaming toward Buenos<br />Aires. Their passage had been booked. Could she still draw<br />back after all he had done for her? Her distress awoke a<br />nausea in her body and she kept moving her lips in silent<br />fervent prayer.<br />A bell clanged upon her heart. She felt him seize her<br />hand:<br />‘Come!’<br />All the seas of the world tumbled about her heart. He<br />was drawing her into them: he would drown her. She<br />gripped with both hands at the iron railing.<br />‘Come!’<br />No! No! No! It was impossible. Her hands clutched the<br />iron in frenzy. Amid the seas she sent a cry of anguish.<br />‘Eveline! Evvy!’<br />He rushed beyond the barrier and called to her to<br />follow. He was shouted at to go on but he still called to her. She set her white face to him, passive, like a helpless<br />animal. Her eyes gave him no sign of love or farewell or<br />recognition.</div>

Sell my dream One morning at nine o’clock, while we were having breakfast on the terrace of the Havana Riviera Hotel under a bright sun, a h... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><h2>Equally Likely Outcomes</h2>The most commonly made assumption for finite sample spaces is that all outcomes are equally likely. This assumption can by no means be taken for granted, however; it must be carefully justified. There are many experiments for which such an assumption is warranted, but there are also many experimental situations in which it would be quite erroneous to make this assumption. For example, it would be quite unrealistic to suppose that it is as likely for no telephone calls to come into a telephone exchange between 1 a.m. and 2 a.m. as between 5 p.m. and 6p.m.<br />If all the k outcomes are equally likely, it follows that each p; = l/k. For the condition p_i + · · · + p_k=1 becomes kp; = 1 for all i. From this it follows<br />that for any event A consisting of r outcomes, we have<br /><br />p(A)=r/k<br />This method of evaluating P(A) is often stated as follows:<br /><br />P(A) = (number of ways in which e can occur favorable to A)/(total number of ways in which e can occur)<br /><br />It is important to realize that the above expression for P(A) is only a consequence<br />of the assumption that all outcomes are equally likely and is only applicable when<br />this assumption is fulfilled. It most certainly does not serve as a general definition<br />of probability.<br /><b><br /></b><b>EXAMPLE 1</b><br />A die is tossed and all outcomes are assumed to be equally likely. The event A occurs if and only if a number larger than 4 shows. That is, A = {5, 6}. Hence P(A) = 1/6 +1/6 = 2/6<br /><b><br /></b><b>EXAMPLE 2</b><br /> An honest coin is tossed two times. Let A be the event: {one head appears}. In evaluating P(A) one analysis of the problem might be as follows. The sample space is S = {0, 1, 2 }, where each outcome represents the number of heads that occur. Hence P(A) = 1/3! This analysis is obviously incorrect,<br />since for the sample space considered above, all outcomes are not equally likely. In order to apply the above methods we should consider, instead, the sample space S' = {HH, HT, TH, IT}. In this sample space all outcomes are equally likely, and hence we obtain for the correct solution of our problem, P(A) = 2/4 = 1/2 We could use the sample space S correctly as follows: The outcomes 0 and 2 are equally likely, while the outcome I is twice as likely as either of the others. Hence P(A) = 1/2, which checks with the above answer.<br />This example illustrates two points. First, we must be quite sure that all outcomes may be assumed to be equally likely before using the above procedure. Second, we can often, by an appropriate choice of the sample space, reduce the problem to one in which all outcomes are equally likely. Whenever possible this should be done, since it usually makes the computation simpler. This point will<br />be referred to again in subsequent examples.<br />Quite often the manner in which the experiment is executed determines whether or not the possible outcomes are equally likely. For instance, suppose that we choose one bolt from a box containing three bolts of different sizes. If we choose the bolt by simply reaching into the box and picking the one which we happen to touch first, it is obvious that the largest bolt will have a greater probability of being chosen than the other two. However, by carefully labeling each bolt with a number, writing the number on a tag, and choosing a tag, we can try to ensure that each bolt has in fact the same probability of being chosen. Thus we may have to go to considerable trouble in order to ensure that the mathematical assumption of equally likely outcomes is in fact appropriate. In examples already considered and in many subsequent ones, we are concerned with choosing at random one or more objects from a given collection of objects. Let us define this notion more precisely. Suppose that we have N objects, say a₁, a₂, • • • , a_N.<br />(a) To choose one object at random from the N objects means that each object has the same probability of being chosen. That is,<br />                          Prob (choosing a;) = 1/N, i = I, 2, ... , N.<br />(b) To choose two objects at random from N objects means that each pair of objects (disregarding order) has the same probability of being chosen as any other pair. For example, if we must choose two objects at random from (a₁, a₂, a₃, a₄), then obtaining a₁ and a₂ is just as probable as obtaining a₂ and a₃, etc. This formulation immediately raises the question of how many different pairs there are. For suppose that there are K such pairs. Then the probability of each pair would be l/K. We shall learn shortly how to compute K.<br />(c) To choose n objects at random (n <span style="font-family: "Cambria Math", serif;">≤</span> N) from the N objects means that each n-tuple, say a_i1, a_i2, • • • , a_in, is as likely to be chosen as any other n-tuple.<br /><br /><b>Note:</b> We have already suggested above that considerable care must be taken during the experimental procedure to ensure that the mathematical assumption of "choosing at random" is fulfilled.</div>

Equally Likely Outcomes The most commonly made assumption for finite sample spaces is that all outcomes are equally likely. This assumption ... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><h2>Relative Frequency</h2>In order to motivate the approach adopted for the solution of the above problem,<br />consider the following procedure. Suppose that we repeat the experiment e<br />n times and let A and B be two events associated with e. We let nA and nn be the<br />number of times that the event A and the event B occurred among then repetitions,<br />respectively.<br /><b>Definition</b>.<br />f_A = n_A/n is called the relative frequency of the event A .in the n<br />repetitions of 8. The relative frequency f_A has the following important properties,<br />which are easily verified.<br />(1) 0 <span style="font-family: "cambria math" , serif;">≤</span> f_A<span style="font-family: "cambria math" , serif;">≤ 1</span><br />(2) f_A = 1 if and only if A occurs every time among the n repetitions.<br />(3) f_A = 0 if and only if A never occurs among the n repetitions.<br />( 4) If A and B are two mutually exclusive events and if  f_A∪B  is the relative frequency associated with the event A <span style="font-family: "cambria math" , serif;">∪ </span>B, then f_A∪B = f_A + f_B.<br />(5) f_A, based on n repetitions of the experiment and considered as a function<br />of n, "converges" in a certain probabilistic sense to P(A) as n <span style="font-family: "cambria math" , serif;">⟶∞</span><br /><div class="MsoNormal"><o:p></o:p></div><br /><b>Note: </b>Property (5) above is obviously stated somewhat vaguely at this point. Only later (Section 12.2) will we be able to make this idea more precise. For the moment let us simply state that Property (5) involves the fairly intuitive notion that the relative frequency based on an increasing number of observations tends to "stabilize" near some definite value. This is not the same as the usual concept of convergence encountered elsewhere in mathematics. In fact, as stated here, this is not a mathematical conclusion at all but simply an empirical fact. Most of us are intuitively aware of this phenomenon of stabilization although we may never have checked on it. To do so requires a considerable amount of time and patience, since it involves a large number of repetitions of an experiment. However, sometimes we may be innocent observers of this phenomenon as the following example illustrates.<br /><br /><b>EXAMPLE 1</b><br /> Suppose that we are standing on a sidewalk and fix our attention on two adjacent slabs of concrete. Assume that it begins to rain in such a manner that we are actually able to distinguish individual raindrops and keep track of whether these drops land on one slab or the other. We continue to observe individual drops and note their point of impact. Denoting the ith drop by X_i, where X_i = 1 if the drop lands on one slab and 0 if it lands on the other slab, we might observe a sequence such as 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1. Now it is clear that we are not able to predict where a particular drop will fall. (Our experiment consists of some sort of meteorological situation causing the release of raindrops.) If we<br />compute the relative frequency of the event A = {the drop lands on slab 1}, then the above sequence of outcomes gives rise to the following relative frequencies<br />(based on the observance of 1, 2, 3, . . . drops): 1 , 1,2/3 , 3/4 , 3/5 , 3/6 , 3/7 ,4/8 , 4/9 , 4/10 , 5/11 ,....<br />These numbers show a considerable degree of variation, particularly at the beginning. It is intuitively clear that if the above experiment were continued indefinitely, these relative frequencies would stabilize near the value t. For we have every reason to believe that after some time had elapsed the two slabs would be equally wet.<br />This stability property of relative frequency is a fairly intuitive notion as yet,<br />and we will be able to make it mathematically precise only later. The essence of<br />this property is that if an experiment is performed a large number of times, the<br />relative frequency of occurrence of some event A tends to vary less and less as the<br />number of repetitions is increased. This characteristic is also referred to as<br />statistical regularity.<br />We have also been some what vague in our definition of experiment. Just when<br />is a procedure or mechanism an experiment in our sense, capable of being studied<br />mathematically by means of a nondeterministic model? We have stated previously<br />that an experiment must be capable of being performed repeatedly under essentially<br />unchanged conditions. We can now add another requirement. When the<br />experiment is performed repeatedly it must exhibit the statistical regularity<br />referred to above. Later we shall discuss a theorem (called the Law of Large<br />Numbers) which shows that statistical regularity is in fact a consequence of the<br />first requirement: repeatability.</div>

Relative Frequency In order to motivate the approach adopted for the solution of the above problem, consider the following procedure. Suppos... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><h2 style="text-align: left;">Events</h2>Another basic notion is the concept of an event. An event A (with respect to a particular sample space S associated with an experiment <span style="font-family: "cambria math" , serif;">ε</span>) is simply a set of possible outcomes. In set terminology, an event is a subset of the sample space S. In view of our previous discussion, this means that S itself is an event and so is the empty set <span style="font-family: "cambria math" , serif;">ϕ</span>. Any individual outcome may also be viewed as an event. The following are some examples of events. Again, we refer to the above-listed experiments: A_i will refer to an event associated with the experiment E_i.<br />A₁ : An even number occurs; that is, A₁ = {2, 4, 6}.<br />A₂: {2}; that is, two heads occur.<br />A₃: {HHHH , HHHT , HHTH , HTHH, THHH}; that is, more heads than tails showed.<br />A₄: {0} ; that is, all items were non defective.<br />A₅: {3, 4, . . . , M} ; that is, more than two rivets were defective.<br />A₆: {t | t < 3}; that is, the bulb burns less than three hours.<br />A₁₄: {(x, y) | y = x + 20}; that is, the maximum is 20° greater than the minimum.<br /><br />When the sample space S is finite or countably infinite, every subset may be considered as an event. [It is an easy exercise to show, and we shall do it shortly, that if S has n members, there are exactly 2ⁿ subsets (events).] However, if S is non countably infinite, a theoretical difficulty arises. It turns out that not every conceivable subset may be considered as an event. Certain "non admissible" subsets must be excluded for reasons which are beyond the level of this presentation. Fortunately such non admissible: sets do not really arise in applications and hence will not concern us here. In all that follows it will be tacitly assumed that whenever we speak of an event it will be of the kind we are allowed to consider.<br /><br />We can now use the various methods of combining sets (that is, events) and obtain the new sets (that is, events) which we introduced earlier.<br />(a) If A and Bare events, A U Bis the event which occurs if and only if A or B (or both) occur.<br />(b) If A and B are events, A n B is the event which occurs if and only if A and B occur.<br />(c) If A is an event, A is the event which occurs if and only if A does not occur.<br />(d) If A₁. . . . , A_n is any finite collection of events, then <span style="font-family: "cambria math" , serif;">∪ⁿ_i=1</span> A₁ is the event<br />which occurs if and only if at least one of the events Ai occurs.<br />(e) If A₁, . .. , A_n is any finite collection of events, then <span style="font-family: "cambria math" , serif;">⋂ⁿ_i=1</span> A₁ is the event<br />which occurs if and only if all the events A; occur.<br />(f) If A₁. ... , An, ... is any (countably) infinite collection of events, then<br /><span style="font-family: "cambria math", serif;">∪</span><span style="font-family: "cambria math", serif; font-size: 13.3333px;">∞</span><sub style="font-family: "cambria math", serif;">i=1</sub> A; is the event which occurs if and only if at least one of the events A₁ occurs.<br />(g) If A₁. .. . A_n, . . . is any (countably) infinite collection of events, then <span style="font-family: "cambria math" , serif;">⋂^∞</span><sub style="font-family: "cambria math", serif;">i=1</sub> A₁ is the event which occurs if and only if all the events A₁ occur.<br /><div class="MsoNormal"><o:p></o:p></div>(h) Suppose that S represents the sample space associated with some experiment S and we perform S twice. Then S x S may be used to represent all outcomes of these two repetitions . .That is, (s₁. s₂) <span style="font-family: "cambria math" , serif;">ϵ</span>S*S means that s₁  resulted when S was performed the first time and s₂ when S was performed the second time.<br />(i) The example in (h) may obviously be generalized. Consider n repetitions of an experimert S whose sample space is S. Then S*S* · · · *S = {(s₁. s₂, • • • , s_n), Si <span style="font-family: "cambria math" , serif;">ϵ</span>S, i = 1, ... , n} represents the set of all possible outcomes when <span style="font-family: "cambria math" , serif;">ε</span> is performed n times. In a sense, S * S * · · · * S is a sample space itself, namely the sample space associated with n repetitions of S.<br /><br /><b>Definition. </b><br />Two events, A and B, are said to be mutually exclusive if they cannot occur together. We express this by writing A  <span style="font-family: "cambria math" , serif;">⋂</span> B = 0; that is, the intersection of A and B is the empty set.<br /><br /><b>EXAMPLE 1 </b><br />An electronic device is tested and its total time of service, say t, is recorded. We shall assume the sample space to be {t | t <span style="font-family: "Cambria Math", serif; font-size: 11pt; line-height: 115%;">≥ </span>0}. Let the three events A, B, and C be defined as follows:<br /><br />A = {t | t < 100};                  B = {t | 50 <span style="font-family: "Cambria Math", serif;">≤ </span>t <span style="font-family: "Cambria Math", serif;">≤ </span>200};        C = {t | t > 150};<br />Then<br />A <span style="font-family: "Cambria Math", serif; font-size: 11pt;">∪</span> B = {t | t <span style="font-family: "Cambria Math", serif;">≤</span> 200};                       A <span style="font-family: "cambria math", serif;">⋂</span> B = {t | 50 <span style="font-family: "Cambria Math", serif;">≤</span> t < 100};<br />B <span style="font-family: "Cambria Math", serif; font-size: 14.6667px;">∪</span> c = {t | t <span style="font-family: "Cambria Math", serif; font-size: 11pt;">≥</span> 50};                          B <span style="font-family: "cambria math", serif;">⋂</span> c = {t | 150 < t <span style="font-family: "Cambria Math", serif;">≤</span> 200};          A <span style="font-family: "cambria math", serif;">⋂</span> c = <span style="font-family: "Cambria Math", serif;">∅</span><br /><div class="MsoNormal"><o:p></o:p></div>A <span style="font-family: "Cambria Math", serif; font-size: 14.6667px;">∪</span> C = {t I t < 100 or t > 150};                   A = {t | t <span style="font-family: "Cambria Math","serif"; font-size: 11.0pt; line-height: 115%; mso-ansi-language: EN-US; mso-bidi-font-family: Arial; mso-bidi-language: AR-SA; mso-bidi-theme-font: minor-bidi; mso-fareast-font-family: Calibri; mso-fareast-language: EN-US; mso-fareast-theme-font: minor-latin;">≥</span>100};                 C = {t | t <span style="font-family: "Cambria Math", serif;">≤</span> 150}.<br /><br />One of the basic characteristics of the concept of "e'.'periment" as discussed in<br />the previous section is that we do not know which particular outcome will occur<br />when the experiment is performed. Saying it differently, if A is an event associated<br />with the experiment, then we cannot state with certainty that A will or will not<br />occur. Hence it becomes very important to try to associate a number with the<br />event A which will measure, in some sense, how likely it is that the event A occurs.<br />This task leads us to the theory of probability.</div>

Events Another basic notion is the concept of an event. An event A (with respect to a particular sample space S associated with an experimen... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><h2>The Sample Space</h2><b>Definition.</b><br />With each experiment <span style="font-family: "cambria math" , serif;">ε</span> of the type we are considering we define the sample space as the set of all possible outcomes of <span style="font-family: "cambria math" , serif;">ε</span>. We usually designate this set by S. (In our present context, S represents the universal set described previously.)<br />Let us consider each of the experiments above and describe a sample space for<br />each. The sample space Si will refer to the experiment E_i.<br />S₁.: { l ,2,3,4,5,6}.<br />S₂: {0, 1, 2, 3, 4}.<br />S₃: {all possible sequences of the form a₁, a₂, a₃, a₄, where each a_i = H or T depending on whether heads or tails appeared on the ith toss.}<br />S₄ {0, 1, 2, ... , N}, where N is the maximum number that could be produced in 24 hours.<br />S₅: {0, 1 , 2, ... , M}, where M is the number of rivets installed.<br />S₆: {t | t2 <span style="font-family: "cambria math" , serif;">≥ </span>0}.<br />S₇: {3, 4, 5, 6, 7, 8, 9, 10}.<br />S₈: {10, 11, 12, ... }.<br />S₉: {v_x, v_y, v_x | v_x, v_y, v_z real numbers}.<br />S₁₀: { h_i . ... ' h_n | h_i  <span style="font-family: "cambria math" , serif;">≥</span> 0, i = 1, 2, ... 'n}.<br />S₁₁: {T | T₂  <span style="font-family: "cambria math" , serif;">≥ </span>0}.<br />S₁₂: {black ball}.<br />S₁₃: This sample space is the most involved of those considered here. We may realistically suppose that the temperature at a specified locality can never get above or below certain values, say M and m. Beyond this restriction, we must allow the possibility of any graph to appear with certain qualifications. Presumably the graph will have no jumps (that is, it will represent a continuous function ). In addition, the graph will have certain characteristics of smoothness which can be summarized mathematically by saying that the graph represents a differentiable function. Thus we can finally state that the sample space is<br /><br />{f | fa differentiable function, satisfying m <span style="font-family: "cambria math" , serif;">≤</span> f(t) <span style="font-family: "cambria math" , serif;">≤</span> M, all t}.<br /><br />S₁₄: { (x, y) | m <span style="font-family: "cambria math" , serif;">≤</span> x <span style="font-family: "cambria math" , serif;">≤</span> y <span style="font-family: "cambria math" , serif;">≤</span> M}. That is, S₁₄ consists of all points in and<br />on a triangle in the two-dimensional x,y-plane.<br /><br />(In this book we will not concern ourselves with sample spaces of the complexity encountered in S₁₃. However, such sample spaces do arise, but require more advanced mathematics for their study than we are presupposing.)<br /><br />In order to describe a sample space associated with an experiment, we must have a very clear idea of what we are measuring or observing. Hence we should speak of "a" sample space associated with an experiment rather than "the" sample space. In this connection note the difference between S₂ and S₃<br /><b>Note </b>also that the outcome of an experiment need not be a number. For example, in E₃ each outcome is a sequence of H's and T's. In E₉ and E₁₀ each outcome consists of a vector, while in E₁₃ each outcome consists of a function.<br /><br />It will again be important to discuss the number of outcomes in a sample space. Three possibilities arise: the sample space may be finite, countably infinite, or non countably infinite. Referring to the above examples, we note that S₁, S₂, S₄<span style="font-size: 13.3333px;">, </span>S_{5 ,}S₇and S₁₂ are finite, S₈ is countably infinite, and S₆, S₉, S₁₀, S₁₁, S₁₃, and S₁₄ are non countably infinite.<br />At this point it might be worth while to comment on the difference between a mathematically "idealized" sample space and an experimentally realizable one. For this purpose, let us consider experiment E₆ and its associated sample space S₆• It is clear that when we are actually recording the total time t during which a bulb is functioning, we are "victims" of the accuracy of our measuring instrument. Suppose that we have an instrument which is capable of recording time to two<br />decimal places, for example, 16.43 hours. With this restriction imposed, our  sample space becomes countably infinite: {0.0, 0.0 1, 0.02, . .. } . Furthermore, it is quite realistic to suppose that no bulb could possibly last more than H hours, where H might be a very large number. Thus it appears that if we are completely realistic about the description of this sample space, we are actually dealing with a<br />finite sample space: {0.0, 0.0 1, 0.02, .. . , H}. The total number of outcomes would be (H/0.0 1) + 1, which would be a very large number if His even moderately large, for example, H = 100. It turns out to be far simpler and convenient, mathematically, to assume that all values of t <span style="font-family: "cambria math", serif;">≥</span> 0 are possible outcomes and hence to deal with the sample space S₆ as originally defined.<br />In view of the above comments, a number of the sample spaces described are idealized. In all subsequent situations, the sample space considered will be that one which is mathematically most convenient. In most problems, little question arises as to the proper choice of sample space.<br /><div class="MsoNormal"><o:p></o:p></div></div>

The Sample Space Definition. With each experiment  ε  of the type we are considering we define the sample space as the set of all possible o... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><h2>Functions of a Random Variable</h2>In defining a random variable X we pointed out, quite emphatically, that X is a function defined from the sample space S to the real numbers. In defining a two dimensional random variable (X, Y) we were concerned with a pair of functions X = X (s), Y = Y(s), each of which is defined on the sample space of some experiment and each of which assigns a real number to every s E S, thus yielding the two-dimensional vector [X(s), Y(s)].<br />Let us now consider Z = H₁( X, Y), a function of the two random variables X and Y. It should be clear that Z = Z(s) is again a random variable. Consider the following sequence of steps:<br /><br />(a) Perform the experiment e and obtain the outcome s.<br />(b) Evaluate the numbers X(s) and Y(s).<br />(c) Evaluate the number Z = H₁[X(s), Y(s)].<br /><br />The value of Z clearly depends on s, the original outcome of the experiment. That is, Z = Z(s) is a function assigning to every outcome s E S a real number, Z(s). Hence Z is a random variable. Some of the important random variables we shall be interested in are X + Y, XY, X/ Y, min (X, Y), max (X, Y), etc.<br />The problem we solved in the previous chapter for the one-dimensional random variable arises again: given the joint probability distribution of (X, Y), what is the probability distribution of Z = H₁ ( X, Y)? (It should be clear from the numerous previous discussions on this point, that a probability is distribution is induced on , R_zthe sample space of Z.)<br />If (X, Y) is a discrete random variable, this problem is quite easily solved.The following (one-dimensional) random variables might be of interest:<br /><br />U = min (X, Y) = least number of items produced by the two lines;<br />V = max (X, Y) = greatest number of items produced by the two lines;<br />W = X + Y = total number of items produced by the two lines.<br /><br />To obtain the probability distribution of U, say, we proceed as follows. The possible values of U are: 0, I, 2, and 3. To evaluate P( U = 0) we argue that U = 0 if and only if one of the following occurs: X = 0, Y = 0 or X = 0, Y = I or X = 0, Y = 2 or X = 0, Y = 3 or X = I, Y = 0 or X = 2, Y = 0 or X = 3,<br />Y = 0 or X = 4, Y = 0 or X = 5, Y = 0. Hence P( U = 0) = 0.28. The rest of the probabilities associated with U may be obtained in a similar way. Hence the probability distribution of U may be summarized as follows: u: 0, I, 2, 3; P( U = u): 0.28, 0.3p, 0.25, 0.17. The probability distribution of the random variables V and W as defined above may be obtained in a similar way.<br /><br />If (X, Y) is a continuous two-dimensional random variable and if Z = H₁ ( X, Y) is a continuous function of (X, Y), then Z will be a continuous (one-dimensional) random variable and the problem of finding its pdf is somewhat more involved. In order to solve this problem we shall need a theorem which we state and discuss below. Before doing this, let us briefly outline the basic idea. In finding the pdf of Z = H₁ ( X, YJ it is often simplest to introduce a second random variable, say W = H₂( X, Y), and first obtain the joint pdf of Z and W, say k(z, w). From a knowledge of k(z, w) we can then obtain the desired pdf of Z, say g(z), by simply integrating k(z, w) with respect to w. That is,<br /><br /><br />\[g\left( z \right)=\int_{-\infty }^{+\infty }{k\left( z,w \right)}dw\] <br /><br />The remaining problems are<br />(1) how to find the joint pdf of Z and W, and<br />(2)how to choose the appropriate random variable W = H₂(X, Y). To resolve the latter problem, let us simply state that we usually make the simplest possible choice for W. In the present context, W plays only an intermediate role, and we are not really interested in it for its own sake. In order to find the joint pdf of Z and W we need Theorem 1 .<br /><br /><i><b>Theorem 1. </b></i><br />Suppose that (X, Y) is a two-dimensional continuous random variable with joint pdf f Let Z = H₁ ( X, Y) and W = H₂( X, Y), and assume that the functions H₁ and H₂ satisfy the following conditions:<br />(a) The equations z = H₁(x, y) and w = H₂(x, y) may be uniquely solved for x and y in terms of z and w, say x = G₁(z, w) and y = G₂(z, w).<br />(b) The partial derivatives ax/az, ax/aw, ay/az, and ayjaw exist and are continuous.Then the joint pdf of (Z, W ), say k(z, w), is given by the following expression: k(z, w) = f[G₁(z, w), G₂(t, w)] | J(z, w)|, where J(z, w) is the following 2 * 2<br />determinant:<br /><br />\[J\left( z,w \right)=\left| \begin{array}{*{35}{l}} \frac{dx}{dz} & \frac{dx}{dw} \\ dy & dy \\ dz & dw \\ \end{array} \right|\]<br /><br />This determinant is called the Jacobian of the transformation (x, y) = (z, w) and is sometimes denoted by d(x, y) / d(z, w). We note that k(z, w) will be nonzero for those values of (z, w) corresponding to values of (x, y) for which f(x, y) is nonzero.<br /><b><br /></b><b>Notes: </b>(a) Although we shall not prove this theorem, we will at least indicate what<br />needs to be shown and where the difficulties lie. Consider the joint cdf of the twodimensional<br />random variable (Z, W), say<br /><br /><br />\[k\left( z,w \right)=P(Z\le z,W\le w)=\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{k\left( s,f \right)}}dsdt\] <br /><br />where k is the sought pdf. Since the transformation (x, y) ----> (z, w) is assumed to be one<br />to one [see assumption (a) above], we may find the event, equivalent to {Z <span style="font-family: "cambria math" , serif;">≤</span> z, W <span style="font-family: "cambria math" , serif;">≤</span> w} ,<br />in terms of X and Y. Suppose that this event is denoted by C.it is, {(X, Y) <span style="font-family: "cambria math" , serif;">∈ </span>C} if and only if {Z <span style="font-family: "cambria math" , serif;">≤</span> z, W <span style="font-family: "cambria math" , serif;">≤</span> w} . Hence<br /><span style="font-family: "calibri" , "sans-serif"; font-size: 11.0pt; line-height: 115%;">\[\int_{-\infty }^{\infty }{\int_{-\infty }^{z}{k(s,t)}dsdt=\iint\limits_{c}{f\left( x,y \right)dxdy}}\]</span><br /><span style="font-family: "calibri" , "sans-serif"; font-size: 11.0pt; line-height: 115%;"><br /></span>Since f is assumed to be known, the integral on the right-hand side can be evaluated.<br />Differentiating it with respect to z and w will yield the required pdf. In most texts on<br />advanced calculus it is shown that these techniques lead to the result as stated in the<br />above theorem.<br />(b) Note the striking similarity between the above result and the result obtained in<br />the one-dimensional case treated in the previous chapter.  The<br />monotonicity requirement for the function y = H(x) is replaced by the assumption that<br />the correspondence between (x, y) and (z, w) is one to one. The differentiability condition<br />is replaced by certain assumptions about the partial derivatives involved. The<br />final solution obtained is also very similar to the one obtained in the one-dimensional<br />case: the variables x and y are simply replaced by their equivalent expressions in terms<br />of z and w, and the absolute value of dx/dy is replaced by the absolute value of the<br />Jacobian.<br /><br /><b>EXAMPLE 1</b>.<br />Suppose that we are aiming at a circular target of radius one which has been placed so that its center is at the origin of a rectangular coordinate system . Suppose that the coordinates ( X, Y) of the point of impact are uniformly distributed over the circle That is,<br />f(x,y) = l/<span style="font-family: "cambria math" , serif;">π</span>                     if (x, y) lies inside (or on) the circle,<br />        =0,                          elsewhere<br /><br />Suppose that we are interested in the random variable R representing the distance from the origin. That is, R = ( X² + Y² )^1/2We shall find the pdf of R, say g, as follows: Let  <span style="font-family: "cambria math" , serif;">ϕ</span><i>= tan-1 (Y/X). Hence X= </i>H₁<i>(R,</i><span style="font-family: "cambria math" , serif;">ϕ</span><i><i>) and </i></i>Y = H₂(R,<span style="font-family: "cambria math" , serif;">ϕ</span>  <i>) where x = </i>H₁<i> (r, </i><span style="font-family: "cambria math" , serif;">ϕ</span><i>) = r cos</i><span style="font-family: "cambria math" , serif;">ϕ </span><i> and  y = </i>H₂<i>(r, </i><span style="font-family: "cambria math" , serif;">ϕ</span><i>) = r sin </i><span style="font-family: "cambria math" , serif;">ϕ</span><i> (We </i>are simply introducing polar coordinates.)<br />The Jacobian is<br />\[J=\left| \begin{matrix} \frac{dx}{dr} & \frac{dx}{d\phi } \\ \frac{dy}{dr} & \frac{dy}{d\phi } \\ \end{matrix} \right|=\left| \begin{matrix} \cos \phi & -r\sin \phi \\ \sin \phi & r\cos \phi \\ \end{matrix} \right|=r{{\cos }^{2}}\phi +r{{\sin }^{2}}\phi =r\] <br /><br />Under the above transformation the unit circle in the xy-plane is mapped into<br />the rectangle in the cpr-plane in Fig. 6.11. Hence the joint pdf of {<i>, R) is given by</i><br />g(<span style="font-family: "cambria math", serif;">ϕ </span>, r)= r/x ,                               0 <span style="font-family: "cambria math", serif;">≤r </span><span style="font-family: "cambria math", serif;">≤ 1 ,        </span>0 <span style="font-family: "cambria math", serif;">≤ </span><span style="font-family: "cambria math", serif;">ϕ < 2r,</span><br /><br />Thus the required pdf of R, say h, is given by<br /><span style="font-family: "Calibri","sans-serif"; font-size: 11.0pt; line-height: 115%; mso-ansi-language: EN-US; mso-ascii-theme-font: minor-latin; mso-bidi-font-family: Arial; mso-bidi-language: AR-SA; mso-bidi-theme-font: minor-bidi; mso-fareast-font-family: Calibri; mso-fareast-language: EN-US; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">\[h(r)=\int_{0}^{2r}{g\left( \phi ,r \right)}d\phi =2r\]</span><br /><span style="font-family: "Calibri","sans-serif"; font-size: 11.0pt; line-height: 115%; mso-ansi-language: EN-US; mso-ascii-theme-font: minor-latin; mso-bidi-font-family: Arial; mso-bidi-language: AR-SA; mso-bidi-theme-font: minor-bidi; mso-fareast-font-family: Calibri; mso-fareast-language: EN-US; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin;">                                                                       </span>0 <span style="font-family: "cambria math", serif;">≤r </span><span style="font-family: "cambria math", serif;">≤ 1</span><br /><b>Note: </b>This example points out the importance of obtaining a precise representation<br />of the region of possible values for the new random variables introduced.</div>

Functions of a Random Variable In defining a random variable X we pointed out, quite emphatically, that X is a function defined from the sam... Read More »

<div dir="ltr" style="text-align: left;" trbidi="on"><div dir="ltr" style="text-align: left;" trbidi="on"><h2 style="text-align: left;">Independent Random Variables</h2>Just as we defined the concept of independence between two events A and B, we shall now define independent random variables. Intuitively, we intend to say that X and Y are independent random variables if the outcome of X, say, in no way influences the outcome of Y. This is an extremely important notion and there are many situations in which such an assumption is justified.<br /><br /><b>EXAMPLE 1</b><br /> Consider two sources of radioactive material at some distance from each other which are emitting a-particles. Suppose that these two sources are observed for a period of two hours and the number of particles emitted is recorded. Assume that the following random variables are of interest: X₁ and X₂,<br />the number of particles emitted from the first source during the first and second hour, respectively; and Y 1 and Y 2, the number of particles emitted from the second source during the first and second hour, respectively. It seems intuitively obvious that (X₁ and Y₁), or ( X₁ and Y₂), or ( X₂ and Y₁), or (X₂ and Y₂) are all pairs of independent random variables, For the X's depend only on the characteristics of source I while the Y's depend on the characteristics of source 2, and there is presumably no reason to assume that the two sources influence each other's behavior in any way. When we consider the possible independence of X₁ and X₂, however, the matter is not so clear cut. Is the number of particles emitted during the second hour influenced by the number that was emitted during the first hour? To answer this question we would have to obtain additional information about<br />the mechanism of emission. We could certainly not assume. a priori, that X1<br />and X₂ are independent.<br />Let us now make the above intuitive notion of independence more precise.<br /><br /><b>Definition</b>. (a) Let (X, Y) be a two-dimensional discrete random variable. We say that X and Y are independent random variables if and only if p(x_i, y_j) =p(x_i)q(y_j) for all i and j. That is,P(X = x_i, Y = y_j)= P(X = x_i)P(Y =y_j), for all i and j.<br /><br />(b) Let (X, Y) be a two-dimensional continuous random variable. We say that X and Y are independent random variables if and only if f(x, y) = g(x)h(y) for all (x, y), where f is the joint pdf, and g and h are the marginal pdf's of X and Y, respectively.<br /><b><br /></b><b>Note: </b>If we compare the above definition with that given for independent events, the similarity is apparent: we are essentially requiring that the joint probability (or joint pdf) can be factored. The following theorem indicates that the above definition is equivalent to another approach we might have taken.<br /><br /><b><i>Theorem 1.</i></b><br /> (a) Let (X, Y) be a two-dimensional discrete random variable. Then X and Y are independent if and only if p(x_i | y_j) = p(x_i) for all i and j (or equivalently, if and only if q(y_{j |}x_i) = q(y_i) for all i and j).<br /><br />(b) Let (X, Y) be a two-dimensional continuous random variable. Then X and Y are independent if and only if g(x | y) = g(x), or equivalently, if and only if h(y | x) = h(y), for all (x, y).<br /><br /><b>EXAMPLE 2</b><br /> Suppose that a machine is used for a particular task in the morning and for a different task in the afternoon. Let X and Y represent the number of times the machine breaks down in the morning and in the afternoon, respectively.<br />Table 1 gives the joint probability distribution of (X. Y).<br />An easy computation reveals that for all the entries in Table 1 we have<br /><br />                      <i>     p(x_i, y_j) =p(x_i)q(y_j) </i><br /><br />Thus X and Y are independent random variables.<br /><br /><br /><table style="width: 100%;"><tbody><tr><th>X \ Y</th><th>0</th><th>1</th><th>2</th><th>q(y_j)</th></tr><tr><th>0</th><td>0.1</td><td>0.2</td><td>0.2</td><td>0.5</td></tr><tr><th>1</th><td>0.04</td><td>0.08</td><td>0.08</td><td>0.2</td></tr><tr><th>2</th><td>0.06</td><td>0.12</td><td>0.12</td><td>0.3</td></tr><tr><th>p(x_i)</th><td>0.2</td><td>0.4</td><td>0.4</td><td>1.0</td></tr></tbody></table><b><br /></b><b>EXAMPLE 3</b><br /> Let X and Y be the life lengths of two electronic devices. Suppose that their joint pdf is given by<br /><br />                   f(x, y) = e^-(x+y)x<span style="font-family: "cambria math" , serif; font-size: 11pt;">≥0,       y</span><span style="font-family: "cambria math" , serif; font-size: 11pt;">≥0</span><br />Since we can factor f(x, y) = e^-x e^-x, the independence of X and Y is established.<br /><b><br /></b><b>EXAMPLE 4. </b><br />Suppose that f(x, y) = 8xy, 0 <span style="font-family: "cambria math" , "serif"; font-size: 11.0pt; line-height: 115%;">≤</span> x <span style="font-family: "cambria math" , "serif"; font-size: 11.0pt; line-height: 115%;">≤</span> y <span style="font-family: "cambria math" , serif; font-size: 11pt;">≤</span> l.  Although/ is (already) written in factored form, X and Y  are not independent, since the domain of definition<br />                                {(x. y) | 0 <span style="font-family: "cambria math" , "serif"; font-size: 11.0pt; line-height: 115%;">≤</span> x <span style="font-family: "cambria math" , "serif"; font-size: 11.0pt; line-height: 115%;">≤</span> y  l}<br />is such that for given x, y may assume only values greater than that given x and less than I. Hence<br />X and Y are not independent.<br /><br /><b>Note: </b>From the definition of the marginal probability distribution (in either the discrete or the continuous case) it is clear that the joint probability distribution determines, uniquely, the marginal probability distribution. That is, from a knowledge of the joint pdf f. we can obtain the marginal pdf's g and h. However, the converse is not true! That is, in general, a knowledge of the marginal pdf's g and h do not determine the joint pdf .f Only when X and Y are independent is this true, for in this case we have f(x, y) = g(x)h(y).<br />The following result indicates that our definition of independent random variables is consistent with our previous definition of independent events.<br /><br /><i><b>Theorem 1 </b></i><br />Let (X, Y) be a two-dimensional random variable. Let A and B be events whose occurrence (or nonoccurrence) depends only on X and Y, respectively. (That is, A is a subset of Rx, the range space of X, while B is a subset of Rv, the range space of Y.) Then, if X and Y are independent<br />random variables,<br />we have P(A n B) = P(A)P(B).<br />Proof (continuous case only):<br /><br />\[P\left( A\cap B \right)=\iint\limits_{\left( A\cap B \right)}{f\left( x,y \right)}dxdy=\iint\limits_{\left( A\cap B \right)}{g(x)h\left( y \right)dx}=\int_{A}{g\left( x \right)}dx\int_{B}{h\left( y \right)dy=p\left( A \right)p\left( B \right)}\]<br /><h2 style="text-align: left;"> Distribution of Product and Quotient of Independent Random Variables </h2>Among the most important functions of X and Y which we wish to consider are the sum S = X + Y, the product W= XY, and the quotient Z = X/ Y. We can use the method in this section to obtain the pdf of each of these random variables under very general conditions. We shall investigate the sum of random variables in much greater detail in Chapter 11. Hence we will defer discussion of the probability distribution of X + Y until then. We shall, however, consider the product and quotient in the following two theorems.<br /><br /><b><i>Theorem 2</i></b><br />Let (X, Y) be a continuous two-dimensional random variable and assume that X and Y are independent. Hence the pdf f may be written as f(x, y)= g(x)h(y). Let W= XY.<br /><br />Then the pdf of W, say p, is given by \[p(w)=\int\limits_{-\infty }^{+\infty }{g\left( u \right)h\left( \frac{w}{u} \right)}\left| \frac{1}{u} \right|du\] <br /><br />Proof: Let w = xy and u = x. Thus x = u and y = w/u. The Jacobian is<br /><br /><br />\[J=\left| \begin{matrix} 1 & 0 \\ \frac{-w}{{{u}^{2}}} & \frac{1}{0} \\ \end{matrix} \right|=\frac{1}{u}\] <br /><br />Hence the joint pdf of W = XY and U = X is<br />s(w,u) = g(u)h (w/h) |1/u| ·<br />The marginal pdf of W is obtained by integrating s(w, u) with respect to u, yielding the required result. The values of w for- which p(w) > 0 would depend on the values of (x, y) for which/(x, y) > 0.<br /><br /><b>Note: </b>In evaluating the above integral we may use the fact that<br />\[\int_{-\infty }^{+\infty }{g(u)h}\left( \frac{w}{u} \right)\left| \frac{1}{u} \right|\] <br /><br /><br /><b>EXAMPLE 5. </b><br />Suppose that we have a circuit in which both the current I and the resistance R vary in some random way. Specifically, assume that I and R are independent continuous random variables with the following pdf's.<br />I:             g(i) = 2i,                                                0 <span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ </span>i <span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ 1         </span> and 0 elsewhere;<br />R:            h(r) = r²/9                                              0<span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ r </span><span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ 3</span>         and 0 elsewhere<br />. Of interest is the random variable E = IR (the voltage in the circuit). Let p be the pdf of E. By Theorem 2 we have <br />\[p(e)=\int_{-\infty }^{+\infty }{g(i)h}\left( \frac{e}{i} \right)\left| \frac{1}{e} \right|di\] <br />Some care must be taken in evaluating this integral. First, we note that the variable<br />of integration cannot assume negative values. Second, we note that in order<br />for the integrand to be positive, both the pdf's appearing in the integrand must<br />be positive. Noting the values for which g and hare not equal to zero we find that<br />the following conditions must be satisfied:<br />0≤i≤1                 and                       0≤e/i≤3<br />These two inequalities are, in turn, equivalent to <br />e/3 ≤ i  ≤ I.                                    Hence the above integral becomes<br />\[p(e)=\int_{1/8}^{1}{2i\frac{{{e}^{2}}}{9{{i}^{2}}}}\frac{1}{i}di=-\frac{2}{9}{{e}^{2}}\frac{1}{i}|_{e/3}^{1}=\frac{2}{9}e\left( 3-e \right),\] div=""> <br /><br />                                                                 0<span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ e  </span><span style="font-family: "cambria math" , serif; font-size: 14.6667px;">≤ 3</span>  <br /><b><i>Theorem 3</i></b><br /><b><i><br /></i></b>Let (X, Y) be a continuous two-dimensional random variable and assume that X and Y are independent. [Hence the pdf of ( X, Y) may be written as f(x, y) = g(x)h(y).] Let Z = X/ Y. Then the pdf of Z, say q, is given by<br /><br /><br />\[q(z)=\int\limits_{-\infty }^{+\infty }{g(vz)h(v)}\left| v \right|dv\]<br /><b>Proof"</b><br />Let z = x/y and let v = y. Hence x = vz and y = v. The Jacobian is<br /><br /><div class="MTDisplayEquation">\[J=\left| \begin{matrix}<o:p></o:p></div><div class="MTDisplayEquation">   v & z  \\<o:p></o:p></div><div class="MTDisplayEquation">   0 & 1  \\<o:p></o:p></div><span style="font-family: "calibri" , "sans-serif"; font-size: 11.0pt; line-height: 115%;">\end{matrix} \right|=v\]</span><br /><span style="font-family: "calibri" , "sans-serif"; font-size: 11.0pt; line-height: 115%;"><br /></span>Hence the joint pdf of Z = X/ Y and V = Y equals<br />f(z, v) = g(vz)h(v) |v|.<br /><br />Integrating this joint pdf with respect to v yields the required marginal pdf of Z.<br /><br /><b>EXAMPLE 6</b><br />Let X and Y represent the life lengths of two light bulbs manufactured by different processes. Assume that X and Y are independent random variables with the pdf's f and g, respectively, where<br />f(x) = e-x,            x<span style="font-family: "cambria math" , serif; font-size: 11pt;">≥0</span>              and 0 elsewhere;<br />g(y) = 2e-2y,        y<span style="font-family: "cambria math" , serif; font-size: 11pt;">≥0            </span>    and 0 elsewhere.<br /><br />Of interest might be the random variable X/ Y, representing the ratio of the two life lengths. Let q be the pdf of Z.<br />By Theorem 6.5 we have<br />\[q(z)=\int_{-\infty }^{+\infty }{g(vz)h(v)\left| v \right|dv}\]</div><br /> Since X and Y can assume only non negative quantities, the above integration need only be carried out over the positive values of the variable of integration. In addition, the integrand will be positive only when both the pdf's appearing are positive. This implies that we must have v ≥ 0 and vz ≥ 0. Since z > 0, these inequalities imply that v ≥ 0.<br />Thus the above becomes<br />\[q(z)=\int_{0}^{\infty }{{{e}^{-vz}}2{{e}^{-2v}}dv}=2\int_{0}^{\infty }{v{{e}^{-v\left( 2+s \right)}}}dv\] <br />An easy integration by parts yields<br />q(z) = 2/(z + 2)² '              z≥ 0.</div>

Independent Random Variables Just as we defined the concept of independence between two events A and B, we shall now define independent rand... Read More »