Section 4.1 Random Variables

Transcription

1 Secton 4.1 Random Varables Oftentmes, we are not nterested n the specfc outcome of an experment. Instead, we are nterested n a functon of the outcome. EXAMPLE: Consder rollng a far de twce. S = {(, j) :, j {1,..., 6}} Suppose we are nterested n computng the sum,.e. we have placed a bet at a craps table. Let X be the sum. Then X {2, 3,..., 12} s random as t depends upon the outcome of the experment. It s a random varable. We can compute probabltes assocated wth X. Can wrte succnctly P (X = 2) = P {(1, 1)} = 1/36 P (X = 3) = P {(1, 2), (2, 1)} = 2/36 P (X = 4) = P {(1, 3), (2, 2), (1, 3)} = 3/36 Sum, P (X = ) 1/36 2/36 3/36 4/36 5/ /36 DEFINITION: Let S be a sample space. Then a functon X : S R s a random varable. EXAMPLE: Consder a bn wth 5 whte and 4 red chps. Let S be the sample space assocated wth selectng three chps, wth replacement, from the bn. Let X be the number of whte chps chosen. The sample space s We have We now fnd probabltes: S = {(a, b, c) : a, b, c {W, R}}. X(R, R, R) = 0 X(W, R, R) = X(R, W, R) = X(R, R, W ) = 1 X(W, W, R) = X(W, R, W ) = X(R, W, W ) = 2 X(W, W, W ) = 3 P (X = 0) = P (R, R, R) = ( ) P (X = 1) = P ((W, R, R) (R, W, R) (R, R, W )) = P (W, R, R) + P (R, W, R) + P (R, R, W ) = ( ) 2 5 P (X = 2) = ( ) 3 5 P (X = 3) = 9 1 ( ) 2 4 9

2 Note that ( ) ( ) ( ) ( ) 3 5 = 1 9 What f no replacement? Then, P (X = 0) = ( ) 4 3 ( ), P (X = 1) = 9 3 ( )( ) 5 4 ( )( ) ( ) 9 3 ( ) 5 P (X = 2) = 2 1 ( ) 9, P (X = 3) = 3 3 ( ) 9 3 Agan, we can show that ths sums to one. REMARK: Note that f X and Y are random varables on the sample space S, then so are X + Y, X Y, ax + by (for a, b R), XY, X/Y (so long as Y 0). Also, f f : R R, then f(x) s also a random varable. So, X 2, X 3, sn(x), etc. are all random varables. Also have mx of two deas: X 2 + Y 2, X 2 + Y 3Z 4, etc. EXAMPLE: Consder choosng a cube wth sde-length chosen randomly from the nterval (0, 1). Let X be the sde-length of the cube chosen. Let Y = X 2 be the sde surface area of the cube chosen, and let Z = X 3 be the volume of the cube chosen. By assumpton, P (X > 1/2) = 1/2. What s P (Y > 1/4) and P (Z > 1/8)? We have Also, P (Y > 1/4) = P (X 2 > 1/4) = P (X > 1/2) = 1/2 P (Z > 1/8) = P (X 3 > 1/8) = P (X > 1/2) = 1/2 P (Z < 1/2) = P (X 3 < 1/2) = P (X < ) = Typcally we want the probabltes assocated wth a random varable, whch can be captured wth the dstrbuton functon of X. DEFINITION: If X s a random varable, then the functon F X, or F, defned on (, ) by F X (t) = F (t) = P (X t) s called the dstrbuton functon of X. Sometmes called cumulatve dstrbuton functon. Note that f a b, then {X a} {X b}. Therefore, for a b we have F (a) = P {X a} P {X b} = F (b) and we see that the dstrbuton functon of a random varable s non-decreasng. We wll see other propertes of F n Secton

3 EXAMPLES: 1. Consder rollng a de. Let S = {1, 2, 3, 4, 5, 6}. Let X : S R be gven by X(s) = s 2. Then, P (X = ) = 1/6 for each {1, 4, 9, 16, 25, 36}. Note, however, that F s defned for all t R. So functon s constant between numbers and F (1) = P (X 1) = P (X = 1) = 1/6 F (4) = P (X 4) = P ( X = 1 or X = 4) = P (X = 1) + P (X = 4) = 2/6, etc. 2. Reconsder cube example. We have P (X t) = t for all t (0, 1), so 0, t 0 F X (t) = t, 0 < t 1 1, 1 < t For the sde surface area, Y = X 2 and so P (Y t) = P (X 2 t). So, Smlarly for volume. 0, t 0 F Y (t) = t, 0 < t 1 1, 1 < t Secton 4.2 Dscrete Random Varables Consder a random varable X : S R. Let R(X) be the range of X. If R(X) s fnte or countably nfnte, then X s sad to be dscrete. For a dscrete random varable X, we defne the probablty mass functon p(a) or p X (a) of X by p(a) = P {X = a} The probablty mass functon p(a) s postve for at most a countable number of values of a. That s, f R(X) = {x 1, x 2,...} s the range of X, then p(x ) 0, for = 1, 2,... p(x) = 0 for all other values of x Because X must take one of those values we also see that p(x ) = 1 Often helpful to present probablty mass functons graphcally. 3

4 EXAMPLES: 1. Consder rollng a far de. Let X be a value rolled. Then R(X) = {1, 2, 3, 4, 5, 6}, and for each x R(X), p(x ) = P {X = x } = 1/6. Plot as vertcal bars. 2. Suppose we are gamblng on a game n whch we wn each round wth a probablty of We wll play untl we wn one round. Let X be the number of rounds we play. What s the probablty mass functon for X? Soluton: We see that p(1) = P {X = 1} = P {wn frst} = 0.25, p(2) = P {frst lose, then wn} = In general, for any nteger k 1, we have Note that (by geometrc seres) p(k) = P {k 1 losses, then 1 wn} = 0.75 k p(k) = k= k = k = 0.25 k=1 k= = 1 Ths s an example of a geometrc random varable. 3. Let X be a dscrete random varable wth probablty mass functon cλ, f {0, 1, 2, 3,...} p() =! 0, else Ths s an example of a Posson random varable wth parameter λ. For ths to be a probablty mass functon, what must c be? We must have p() = 1. Thus, =0 Thus, c = e λ. Thus, for example () P {X = 0} = e λ c =0 λ! = c eλ = 1 () P {X 2} = 1 P {X 1} = 1 P {X = 0} P {X = 1} = 1 e λ λe λ Note connecton wth dstrbuton functons. Dscrete random varable f and only f dstrbuton functon has only jumps. So t s a step functon. Note that F (a) = x a p(x ) 4

5 4. Consder a random varable X wth probablty mass functon p(1) = 1 4, p(2) = 1 2, p(3) = 1 8, p(4) = 1 8 Its dstrbuton functon s F (a) = 0, a < 1 1 4, 1 a < 2 3 4, 2 a < 3 7 8, 3 a < 4 1, 4 a Secton 4.3 Expected Value Consder a random varable X wth range {x 1, x 2,...}. Imagne each x as a possble wnnngs (n dollars) from the occurrence of a game of chance (X represents the outcome of one such game). Suppose that the probablty mass functon for X s gven by P {X = x } = p(x ) Suppose you play ths game many tmes. Intutvely, how much do you expect to have won? What, then, s your average wnnngs per turn? Sad another way, how much do you expect to wn n a sngle occurrence of the game? Most reasonable people would probably argue as follows: If n s bg, after n plays I wll wn x dollars approxmately p(x ) n tmes. (Ths wll be shown to be true, n some sense, later n the class. It s called the law of large numbers.) Therefore, my total wnnngs are approxmately x p(x )n Dvdng by n gves us the average wnnngs per game to be x p(x ) DEFINITION: Let X be a dscrete random varable wth probablty mass functon p(x). Then the expectaton or the expected value of X, denoted by E[X], s E[X] = {x:p(x)>0} xp(x) 5

6 EXAMPLES: 1. Let X be the outcome of a roll of a far de. Fnd E[X]. Soluton: We know that p(1) = p(2) =... = p(6) = 1/6, thus E[X] = = 7 2 Ths result can be generalzed n the followng natural way: 2. Consder a random varable takng values n {1,..., n} wth P {X = } = 1/n for each {1,..., n}. We say that X s dstrbuted unformly over {1,..., n}. What s the expectaton? Soluton: We have E[X] = P {X = } = 1 n = 1 n n(n + 1) 2 = n Suppose that X takes values on {0, 1, 2,...} wth probablty mass functon λ λx P {X = x} = e x! Then X has a Posson dstrbuton wth parameter λ. The expected value s E[X] = x=0 λ λx xe x! = λe λ x=1 λ x 1 (x 1)! = λe λ x=0 λ x x! = λe λ e λ = λ Note that f X 0,.e. the range only takes nonnegatve values, then E[X] Consder a Bernoull random varable X that takes a value of 0 or 1 and the p.m.f. s p(0) = 1 p and p(1) = p for some 0 < p < 1. In ths case E[X] = 0 (1 p) + 1 p = p 5. We say that I = I A = 1 A (all these notatons are used) s an ndcator functon for the event A f { 1, f A occurs I A = 0, f A c occurs (example: rollng a de and lettng A = {1, 2, 5}, then I A gves the value 1 to the outcomes are 1, 2, or 5 and zero otherwse). Let s fnd E[I A ]. We know that the range of I A s smply 1 and 0. We also know that p(1) = P (A) and p(0) = P (A c ). Therefore, E[I A ] = 1 p(1) + 0 p(0) = P (A) 6. A school class of 120 students s drven somewhere n 3 buses. There are 36, 40, and 44 students on the bus 1, 2, and 3, respectvely. Upon arrval, one of the 120 students s chosen randomly and we let X be the number of students on the chosen students bus. What s E[X]? 6

7 6. A school class of 120 students s drven somewhere n 3 buses. There are 36, 40, and 44 students on the bus 1, 2, and 3, respectvely. Upon arrval, one of the 120 students s chosen randomly and we let X be the number of students on the chosen students bus. What s E[X]? Soluton: We know that all 120 students are equally lkely to be chosen. We also know that the range of X s {36, 40, 44}. Fnally, the probablty mass functon s gven by P {X = 36} = , P {X = 40} =, P {X = 44} = Thus, 36 E[X] = = Obvously, the average number of students per bus s 120/3 = 40, showng the expected number of students on a randomly chosen students bus s larger than the average. Ths s a general property. Bascally, buses wth many students are weghted heavly. Ths s easly seen f there were 118 students on one bus and 1 n each of the others. The expected number of students on a randomly chosen students bus s then easly seen to be around 118. Secton 4.4 Expectaton of a Functon of a Random Varable Suppose that we have a dscrete random varable, X, and a functon g, and suppose that we want to understand the random varable g(x). More to the pont, we want to be able to compute the expected value of X. Note that g(x) s tself a random varable, so t has a probablty mass functon. Once we know t, we can compute the expected value. EXAMPLE: Let Y = X 2, where X s a random varable takng values 1, 0, 1 wth probabltes The range of Y s {0, 1}. Also, P {X = 1} = 0.2, P {X = 0} = 0.5, P {X = 1} = 0.3 P {Y = 0} = P {X = 0} = 0.5 P {Y = 1} = P {X = 1} + P {X = 1} = 0.5 Therefore, the expected value can be computed as normal Note that E[X 2 ] = E[Y ] = 0 P {Y = 0} + 1 P {Y = 1} = = = E[X 2 ] (E[X]) 2 = ( ) 2 = 0.01 There s a nce way to make ths computaton. THEOREM: If X s a dscrete random varable wth range x for 1, wth respectve probabltes p(x ), then for any real-valued functon g we have E[g(X)] = g(x )p(x ) 7

8 Proof: The proof proceeds by groupng together terms of g(x )p(x ) that have the same value g(x ). Ths s just how we computed the last example! So, let y j, for j 1, represent all possble values of g(x ). Note that {y j } s just the range of g(x). We have g(x )p(x ) = g(x )p(x ) j :g(x )=y j = j = j y j p(x ) :g(x )=y j y j p(x ) = y j P {g(x) = y j } :g(x )=y j j = E[g(X)] Propertes of Expectatons: 1. For any α 1, α 2 R we have E[α 1 g 1 (X) + α 2 g 2 (X)] = α 1 E[g 1 (X)] + α 2 E[g 2 (X)] In fact, E[α 1 g 1 (X) + α 2 g 2 (X)] = (α 1 g 1 (x ) + α 2 g 2 (x ))p(x ) = α 1 g 1 (x )p(x ) + α 2 g 2 (x )p(x ) = α 1 E[g 1 (X)] + α 2 E[g 2 (X)] In partcular, for any a, b R we have E[aX + b] = ae[x] + b 2. If X 0, then E[X] If f g, then E[f(X)] E[g(X)]. EXAMPLE: Suppose X has R(X) = { 2, 4, 6} and p( 2) = 1/7, p(4) = 2/7, and p(6) = 4/7. What s E[X(X 2)]? Soluton: We have E[X(X 2)] = ( 2) ( 4) p( 2) p(4) p(6) = = 56 7 = or E[X(X 2)] = E[X 2 2X] = E[X 2 ] 2E[X] = ( ) =

9 Secton 4.5 Varance Varance gves a measure on the spread of a random varable around the mean, whereas expected value gave the mean. Let µ = E[X] denote the mean of a random varable X. We defne the varance and standard devaton of X to be Var(X) = E[(X µ) 2 ] σ X = Var(X) Sometmes more convenent alternatve arses from lnearty of expectatons: Thus, Var(X) = E[(X µ) 2 ] = E[X 2 2µX + µ 2 ] = E[X 2 ] 2µE[X] + µ 2 = E[X 2 ] µ 2 E[X 2 ] = Var(X) + µ 2 µ 2 = E[X] 2 where E[X 2 ] s called second moment. In general, E[X n ] s nth moment of the random varable. EXAMPLE: Consder rollng a far de. What s the varance of X, the value rolled? Soluton: We know from before that E[X] = 7/2. Now we need E[X 2 ]. Agan, R(X) = {1, 2, 3, 4, 5, 6} and p() = 1/6 for all = 1, 2,..., 6. Thus, Thus, E[X 2 ] = 1 p(1) + 4 p(2) p(6) = 1 6 ( ) = 91 6 Var(X) = E[X 2 ] E[X] 2 = = THEOREM: For any constants a and b we have Var(aX + b) = a 2 Var(X) Proof: Let µ = E[X]. We have Var(aX + b) = E[(aX + b aµ b) 2 ] = E[a 2 (X µ) 2 ] = a 2 Var(X) EXAMPLE: Consder agan a Posson random varable wth parameter λ > 0. That s R(X) = {0, 1, 2,...} and for k {0, 1, 2,...}, λ λk p(k) = e k! 9

10 What s the varance? Recall, we know that E[X] = λ. We need E[X 2 ]. Therefore, E[X 2 ] = k 2 λ λk e k! = e λ k 2 λk k! = e λ λ k λk 1 (k 1)! = e λ λ (k + 1) λk k! k=0 k=1 k=1 k=0 ( ) λ λk = λ ke k! + λ k e λ = λ(e[x] + 1) = λ 2 + λ k! k=0 k=0 So, Var(X) = E[X 2 ] E[X] 2 = λ 2 + λ λ 2 = λ = E[X]. Secton 4.6 The Bernoull and Bnomal Random Varables Bernoull random varable: Suppose we have an experment and care about 2 events: success or falure (s or f). A random varable X s a Bernoull random varable f X(s) = 1 for s success and X(f) = 0 for f falure. Thus, for some 0 < p < 1, p(1) = P {X = 1} = p and p(0) = P {X = 0} = 1 p = q (p(x) = 0 otherwse) Propertes: E[X] = 1 p(1) + 0 p(0) = p E[X 2 ] = 1 2 p(1) p(0) = p Var(X) = E[X 2 ] E[X] 2 = p p 2 = p(1 p) σ X = p(1 p) EXAMPLE: There are 100 tagged and numbered (1-100) rabbts. You set a trap at nght that s guaranteed to catch one. You need one that s numbered Let X be 1 f you catch a desred rabbt and zero otherwse. Then X s Bernoull wth P {X = 1} = 14/100, P {X = 0} = 86/100 Bnomal random varable: Now consder n ndependent repeated trals of a Bernoull random varable. Let X be the number of successes n the n trals. Clearly, R(X) = {0, 1,..., n} What s P {X = k} for k n? Each specfc sequence of trals that has k successes and n k falures has a probablty of happenng of p k (1 p) n k. How many ways can we get k successes and n k falures? It s the same queston as how many ways can we choose k slots from n. Thus, the answer s ( n k ). Therefore, p(k) = P {X = k} = ( ) n p k (1 p) n k f k {0, 1, 2,..., n} k 0 otherwse 10

11 DEFINITION: Any random varable wth probablty mass functon gven above s a bnomal random varable wth parameters (n, p). Note that by bnomal theorem (TYPO IN THE BOOK IN THIS EQUATION ON PAGE 135) ( ) n p (1 p) n = (p + (1 p)) n = 1 =0 EXAMPLE: There s a product that s defectve wth a probablty of They are sold n packages of 10. The company offers money back f two or more are defectve. What s the probablty that a gven package wll be returned for cash? Soluton: Let X denote the number of defectve products n a gven package. We want to know P {X 2}. Clearly, X s a bnomal(10, 0.01) random varable. It s obvously easer to compute P {X 2} = 1 P {X 1}. Thus, we need ( ) ( ) P {X = 0} = = , P {X = 1} = = Therefore, the desred probablty s P {X 2} = 1 P {X = 0} P {X = 1} Thus, 0.4 percent of the packages are elgble for recall. Expectatons and Varance of Bnomal random varable. Let X be bnomal(n, p). We wll compute the moments of X, E[X k ]. We have ( ) n ( ) n E[X k ] = k p (1 p) n = k p (1 p) n =0 Note that the followng dentty holds ( ) ( ) n n! =!(n )! = n (n 1)! n 1 ( 1)!(n )! = n 1 Therefore, E[X k ]=np ( ) n 1 n 1 ( ) n 1 k 1 p 1 (1 p) n =np (j+1) k 1 p j (1 p) n 1 j =npe[(y +1) k 1 ] 1 j where Y s a bnomal(n 1, p) random varable. Lettng k = 1, we see that j=0 E[X] = np Settng k = 2 and usng the precedng formula yelds Therefore, E[X 2 ] = npe[y + 1] = np[(n 1)p + 1] = n 2 p 2 np 2 + np Var(X) = E[X 2 ] E[X] 2 = n 2 p 2 np 2 + np n 2 p 2 = np np 2 = np(1 p) Interestng: Note that ths s n tmes the varance of the Bernoull random varable. 11

12 PROPOSITION: If X s a bnomal random varable wth parameters (n, p), then as k goes from 0 to n, P {X = k} frst ncreases monotoncally, and then decreases monotoncally, reachng ts largest value when k s the largest nteger less than or equal to (n + 1)p. Proof: Consder the rato P {X = k}/p {X = k 1} and ask when s ths greater than or less than 1. We have P {X = k} P {X = k 1} = Thus, P {X = k} P {X = k 1} f and only f n! (n k)!k! pk (1 p) n k = n! (n k + 1)!(k 1)! pk 1 (1 p) n k+1 (n k + 1)p k(1 p) (n k + 1)p k(1 p) = np pk + p k kp = (n + 1)p k Note that we have the followng corollary whch shows how to compute the probabltes of a bnomal(n, p) teratvely: P {X = k + 1} = p 1 p n k P {X = k} k + 1 Secton 4.7 The Posson Random Varable Recall that a random varable X s sad to be a Posson random varable wth parameter λ > 0 f for k {0, 1, 2,...}, λ λk P {X = k} = e k! One of the reasons that Posson random varables are so useful s that they are an approxmaton to the bnomal random varable n the followng sense. Consder a bnomal(n, p) random varable Y wth bg n and small p, but np = λ moderate. Thus, there are many ndependent experments, each wth a small probablty of occurrng, but expected number of occurrences s moderate. Then the probablty of k successes s approxmately P {Y = k} = ( ) n p k (1 p) n k = k n! (n k)!k! ( ) k ( λ 1 λ ) n k n n = n(n 1)... (n (k 1)) n k λk k! ( 1 λ ( 1 λ n ) n n ) k = [take lmt as n ] nk n λk k k! e λ 1 = e λ λk k! EXAMPLE: A communty of 100,000 people s struck wth a fatal dsease n whch the chance of death s What s the probablty that at most 8 people wll de? 12

13 EXAMPLE: A communty of 100,000 people s struck wth a fatal dsease n whch the chance of death s What s the probablty that at most 8 people wll de? Soluton: Ths s bnomal wth parameters n = 10 5 and p = So, f X s the random varable gvng number of deaths, then P {X 8} = 8 ( ) 10 5 (10 4 ) k ( ) 105 k = k k=0 8 k=0 10 5! k!(10 5 k)! (10 4 ) k ( ) 105 k Computatonally ths formula s too complcated. Another opton s to use the Posson approxmaton wth λ = np = 10: P {X 8} 8 k= k e k! = The Actual answer computed wth Maple s But the Posson approach was (relatvely) nstant, whereas the Bnomal computaton took 43.5 seconds. λ λk Recall, we ve already seen that for k {0, 1, 2,...}, p(k) = e k! functon. We also know that E[X] = λ, Var(X) = λ s a probablty mass EXAMPLES: 1. Number of msprnts on a page. 2. The number of people n a group who survve to age Number of certan natural dsasters (earthquake/volcano) happenng n a gven year. Another beneft: you only need the average occurrence rate of event, not n or p. EXAMPLES: 1. Dsney World averages 5 njures per day. What s the probablty that there wll be more than 6 njures tomorrow? 13

14 1. Dsney World averages 5 njures per day. What s the probablty that there wll be more than 6 njures tomorrow? Soluton: It seems lke there s not enough nformaton to be able to answer ths queston. However, we know that E[X] = np = 5. We also know that n s bg, p s small and np = λ = 5. Thus, X s approxmately Posson wth parameter 5. Therefore, P {X 7} = 1 P {X < 7} 1 6 k=0 k=0 5 5k e k! = Note that t would be mpossble to solve ths usng the bnomal vew because 6 ( ) n P {X 7} = 1 P {X < 7} = 1 p k (1 p) k k but we don t know p or n. 2. Suppose that there are, on average, 1.2 typos on a page of a gven book. What s the probablty that there are more than two errors on a gven page? Soluton: We model the number of errors on a gven page, X, as a Posson random varable wth parameter 1.2. Then the desred probablty s P {X 3} = 1 P {X 2} = 1 e k=0 1.2 k k! = Not even the assumptons lad our above are strctly necessary. We have the followng Posson Paradgm: Consder n events, wth p beng the probablty that the th event occurs. If all the p are small and the trals are ndependent or weakly dependent, then the number of these events that occur approxmately has a Posson dstrbuton wth mean p. Posson Processes Suppose that startng at tme 0 we start countng events (earthquakes, arrvals at a post-offce, number of meteors). For each t, we obtan a number N(t) gvng the number of events up to tme t. N(t) s a random varable. So we have an nfnte (uncountable) number of random varables. Let N(t) be a dscrete random varable wth range {0, 1, 2, 3,...}. We make the followng assumptons: 1. The probablty of exactly one event occurrng n a gven tme nterval of length h s equal to λh + o(h), where o(h) stands for any functon g that satsfes g(h)/h 0 as h 0. So, g(h) = h 2 s o(h), but g(h) = h s not. 2. The probablty that 2 or more events occur n an nterval of length h s o(h). 3. For any ntegers n, j 1,..., j n and any set of n nonoverlappng ntervals, f we defne E to be the event that exactly j of the events under consderaton occur n the th of these ntervals, then the events E 1, E 2,..., E n are ndependent. 14

15 We now show that under the above assumptons the number of events happenng n any length of tme t has a Posson dstrbuton wth parameter λt. That s, we wll show that λt (λt)k P {N(t) = k} = e k! We begn by breakng the nterval [0, t] up n to n subntervals of length t/n (n s large). We have P {N(t) = k} = P {k of the subntervals contan exactly 1 event and other n k contan no events} +P {N(t) = k and at least one of the subntervals contans 2 or more events} Let A and B be the two events on the rght hand sde. We start by showng that P (B) 0 as n (so events happen one at a tme). We wll use Boole s nequalty P ( A ) P (A ) : P (B) P {at least one subnterval has 2 or more events} { n } = P {th subnterval contans 2 or more} = P {th subnterval contans 2 or more} [ ] o(t/n) o(t/n) = n o(t/n) = t t/n Thus, as n 0, P (B) 0. Now we just need to handle the set A. We see from assumptons 1 and 2 that P { 0 events occur n any nterval of length h} = 1 [λh + o(h) + o(h)] = 1 λh o(h) Now usng assumpton 3 (ndependence of ntervals), we see that P (A) = ( ) [ ] k [ ( )] n k n λt λt k n + o(t/n) 1 n + o(t/n) Usng the same argument as for the orgnal Posson approxmaton, as n, we fnd that Therefore, we see that λt (λt)k P (A) e k! λt (λt)k P {N(t) = k} = e, k = 0, 1, 2,... k! So, N(t) s a Posson random varable wth parameter λt. We say that N(t) s a Posson process wth rate (or ntensty) λ. Moreover, E[N(t)] = λt and so E[N(1)] = λ. So λ can be reconstructed through observaton (wat one unt of tme, record, then another, record, etc. usng statonarty). 15

16 EXAMPLE: Suppose that you are watchng people arrvng at a doctors offce and they are arrvng at an average rate of 3 per hour. What s the probablty that no one wll arrve n the next hour? What s the probablty that from 12pm 5pm there wll be 2 full hours (.e. 2pm 4pm) that are not necessarly contguous n whch no one wll arrve? Soluton: Let N(t) be the number of people to arrve n the next t hours. Then N(t) s Posson wth rate 3. Thus, 3 1 (3 1)0 P {N(1) = 0} = e = e 3 = ! For the second problem, by statonarty and ndependence of ncrements, the probablty of any hour long block havng n vsts s the same for each block. Therefore, f X s the number of hour long blocks n whch nobody shows up, X s bnomal wth parameters 5 and p = e 3. Therefore, ( ) ( ) 5 5 P {X = 2} = p 2 (1 p) 5 2 = e 3 2 (1 e 3 ) 3 = Computng the Posson random varable. Note that f X s Posson wth parameter λ > 0, then e λ λ +1 P {X = + 1} ( + 1)! = = λ P {X = } e λ λ + 1! Therefore, P {X = + 1} = λ P {X = } + 1 Secton 4.8 Other Dscrete Random Varables The geometrc random varable: Suppose that ndependent trals, each havng a probablty of p, 0 < p < 1, of success are performed untl a success occurs. Let X equal the number of trals requred. Then the range of X s {1, 2,...} and the probablty mass functon s gven by Note that by Geometrc seres we have P (X = n) = n=1 n=1 P (X = n) = (1 p) n 1 p, n = 1, 2,... (1 p) n 1 p = p (1 p) n 1 = p (1 p) n = p n=1 n=0 1 1 (1 p) = 1 A random varable wth ths probablty mass functon s called a geometrc random varable. EXAMPLE: What s the mnmum number of draws requred to get an Ace wth probablty 0.95? 16

17 EXAMPLE: What s the mnmum number of draws requred to get an Ace wth probablty 0.95? Soluton: We want the smallest n such that P (X n) Therefore, we want We have P (X > n) = 1 P (X > n) 0.95 or P (X > n) 0.05 k=n ( ) k 1 12 = ( ) n =0 ( ) 12 = 13 ( ) n So we need (12/13) n 0.05 or n ln(12/13) ln(0.05) or n ln(0.05)/ ln(12/13) = Thus, n s 38. Expected Value and Varance: We can compute the expected value of a geometrc random varable. One proof usng summatons s n the book. Here we show another way. Set q = 1 p. Then, E(X) = q 1 p = p q 1 The radus of convergence of to see that Therefore q = 1/(1 q) s q < 1. Thus, you can dfferentate termwse =0 E(X) = p 1 (1 q) = 2 q 1 = p p = 1 2 p q 1 = E(X) = 1 p We can also use ths method to compute the varance. Takng a second dervatve yelds 2 (1 q) = ( 1)q 2 = ( + 1)q 1 = 2 q 1 + q 1 = 1 3 p E[X2 ] + 1 p E[X] =2 = 1 p E[X2 ] + 1 p 2 So, 2 p 3 = 1 p E[X2 ] + 1 p 2 = E[X 2 ] = 2 p 2 1 p = Var(X) = 2 p 2 1 p 1 p 2 = 1 p p 2 Memoryless Property: Let X be Geometrc. Then for all n, m 1 we have P (X > n + m X > m) = = P (X > n + m) P (X > m) = n + m falures to start m falures to start (1 p)n+m (1 p) m = (1 p)n = P (X > n) So probablty that next n wll be falures gven frst m were s same as probablty that frst n are falures. Ths s ntutve by ndependence. However, Geometrc s the only dscrete random varable wth ths property. 17

18 Negatve Bnomal: Generalzed geometrc. Consder a sequence of Bernoull trals wth probablty of success = p. Let X be the number of trals untl the rth success occurs. Then X s Negatve Bnomal wth parameters r, p. Note that NegBn(1, p)=geom(p). The range s R(X) = {r, r + 1, r + 2,...} and for n R(X) (usng ndependence) P (X = n) = P (r 1 successes n n 1 trals and then success on nth) = P (r 1 successes n n 1 trals)p (success) = P (bnom(n 1, p) = r 1)p ( n 1 = r 1 ( n 1 = r 1 ) p r 1 (1 p) n 1 (r 1) p ) p r (1 p) n r We have E[X] = r r(1 p), Var(X) = p p 2 EXAMPLE: The Red Sox and Brewers are playng for the World Seres. The probablty that the Red Sox wn each game s What s the probablty that the Brewers wn n 6? 18

19 EXAMPLE: The Red Sox and Brewers are playng for the World Seres. The probablty that the Red Sox wn each game s What s the probablty that the Brewers wn n 6? Soluton: Let X be the number of games needed untl Brewers wn 4th. Ths s NegBn(r = 4, p = 0.45). Therefore, ( ) 5 P (X = 6) = = Note also, that ( ) 3 P (X = 4) = = ( ) 4 P (X = 5) = = ( ) 6 P (X = 7) = = Therefore, the probablty that the Brewers wn the seres s P (Brewers Wn) = P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) One can check the followng = p P (Brewers Wn) Hypergeometrc random varable: Then, Suppose we have an urn wth N balls of whch m are whte and N m are black. Now we choose a sample sze of n from the N balls. Let X denote the number of whte balls selected. Note that we must have P {X = } = ( )( ) m N m n ( ) N, = 0, 1,..., n n mn{m, n} else the probablty s zero (whch stll holds for our formula). Also, as there are a total of N m black balls, we must also have n (N m) However, our formula s stll vald for f we ddn t have that, the second term s zero. 19

20 Any random varable wth ths probablty mass functon for some m, N, n s sad to be a hypergeometrc random varable. Lettng p = m/n, we have E[X] = nm N Var(X) = (= m N n = pn ) nm(n m) N 2 ( 1 n 1 ) ( ( = np(1 p) 1 n 1 )) N 1 N 1 EXAMPLE (Maxmum lkelhood estmate): Suppose we want to estmate the number of rabbts n a regon. How can we do so? 1. Catch m anmals and mark them. Then release. 2. At a later date, catch n anmals. Now let X denote the number of marked anmals n ths second capture. 3. Assume the populaton remaned fxed between two captures and each anmal s equally lkely to be caught. 4. Clearly, X s hypergeometrc wth parameters N, m, n. Thus, ( )( ) m N m P {X = } = n ( ) N P (N) n 5. Now we assume that that X s observed to be. What s our best guess at N? 6. The dea s to fnd the value of N that maxmzes P (N). Note that P (N) P (N 1) = ( m )( ) N m n ( ) N n ( m )( ) N 1 m n ( ) N 1 Therefore, the rato s greater than one f and only f (after reducng) n 1 = (N m)(n n) N(N m n + ) N nm Thus, P (N) frst ncreasng and then decreasng, and reaches ts maxmum value at the largest nteger less than or equal to nm/. Ths value s the maxmum lkelhood estmate of N. Does ths make sense? Thnk of ths way. The proporton of marked anmals s m/n. Assumng that /n s of the same proporton and then solvng yelds n = m N = N mn 20

21 Secton 4.9 Expected Value of Sums of Random Varables In ths secton, we wll show, n dscrete case, that the expected value of a sum of random varables s the sum of the expected values. (ths s true n the contnuous case also, we ll show that later.) Consder a random varable X wth sample space S that s fnte or countably nfnte. We wll wrte the expected value n a dfferent way: 1. For s S, denote by X(s) the value that X takes at s. 2. For s S, denote p(s) = P ({s}). 3. Note that for two random varables X, Y we can let Z = X + Y by defnng Z(s) = X(s) + Y (s). For example, f an experment conssts of flppng a con fve tmes and X gave the number of heads n the frst three flps and Y gves the number of heads n the fnal four, then the outcome s = (h, t, h, t, h) has X(s) = 2 Y (s) = 2 Z(s) = X(s) + Y (s) = 4 THEOREM: We have E[X] = s S X(s)p(s) Proof: Let the range of X be denoted by {x } for 1. Also, for each such x let S = {s S X(s) = x } denote the event that X equals x (note that these partton S). We have E[X] = x P {X = x } = = = = x P (S ) x p(s) s S x p(s) s S X(s)p(s) s S = s S X(s)p(s) 21

22 COROLLARY: For random varables X 1, X 2,..., X n we have [ ] E X = E[X ] Proof: Let Z = X. Then, E[Z] = s S Z(s)p(s) = s S (X 1 (s) X n (s))p(s) = s S X 1 (s)p(s) s S X n (s)p(s) = E[X 1 ] E[X n ] EXAMPLES: 1. Consder n trals where the th has probablty p of success. What s the expected number of successes E[X]? Soluton: Let X = { 1, f tral s a success 0, f tral s a falure Then X = X and so E[X] = E[X ] = Note that ndependence was not requred here. As a specal case, p p, we get the mean of the bnomal. However, we also get the mean of the hypergeometrc. Note that n balls are chosen and the probablty that the th s whte s always m/n. That s, f Z s 1 when a whte ball s chosen on th choce and zero otherwse, and f X s a total number of whte balls chosen, then p X = Z and E[X] = E[Z ] = P {Z = 1} = np = n(m/n) Even though hypergeometrc trals are dependent, the result holds people workng n an offce have a holday gft exchange. Suppose that there s no organzaton of the exchange and each person buys a gft for one of the other 19 selected at random wthout knowledge of who the others select. What s the expected number of people who receve no gft? 22

23 2. 20 people workng n an offce have a holday gft exchange. Suppose that there s no organzaton of the exchange and each person buys a gft for one of the other 19 selected at random wthout knowledge of who the others select. What s the expected number of people who receve no gft? Soluton: For person, let Z be one f ths person does not receve a gft. Then, f we let N be 20 the number of people who receve no gft, N = Z. Therefore, 20 E[N] = P {Z = 1} Person does not receve a gft f each of the other 19 people ndependently choose someone else. Thus, ths happens wth probablty (18/19) 19 (bnomal(n = 19, p = 18/19) and askng for 19 successes). Thus, 20 ( ) E[N] = P {Z = 1} = Propertes of Dstrbuton Functons Recall that for a random varable X we defne The followng hold: F (t) = F X (t) = P {X t} 1. F s non-decreasng: a < b = F (a) F (b). 2. lm b F (b) = lm F (b) = 0. b 4. F s rght contnuous: Let b n > 0 be such that b n b, then lm n F (b n ) = F (b). The fnal three can be proven by the followng fact. A sequence of events {E n } s sad to be ncreasng f E 1 E 2... E n... and t s sad to be decreasng f For an ncreasng sequence we defne and for a decreasng sequence we defne E 1 E 2... E n... lm E n = n lm E n = n E E 23

24 See Secton 2.6 for a proof of the followng: THEOREM: If {E n } s ether ncreasng or decreasng, then ( ) lm P (E n) = P lm E n n n Thus, to prove property 2 we let b n denote an ncreasng sequence that goes to, then {X b n } s an ncreasng sequence whose unon s {X < }. So, lm P {X b n} = P {X < } = 1 n All probablty questons about X can be answered n terms of the c.d.f., F. For example, P {a < X b} = F (b) F (a) for all a < b ( ) Ths equaton can best be seen to hold f we wrte the event {X b} as the unon of the mutually exclusve events {X a} and {a < X b}. That s, so whch establshes equaton ( ). Also, say we want P {X < b}. Then, ( ) P {X < b} = P {X b 1/n} n=1 {X b} = {X a} {a < X b} P {X b} = P {X a} + P {a < X b} = lm n P {X b 1/n} = lm n F (b 1/n) = F (b ).e. the lmt from the left (whch s not necessarly the value F (b)). 24