kp(x = k) = λe λ λ k 1 (k 1)! = λe λ r k e λλk k! = e λ g(r) = e λ e rλ = e λ(r 1) g (1) = E[X] = λ g(r) = kr k 1 e λλk k! = E[X]

Transcription

1 Problem 1: (8 poits) Let X be a Poisso radom variable of parameter λ. 1. ( poits) Compute E[X]. E[X] = = kp(x = k) = k=1 λe λ λ k 1 (k 1)! = λe λ ke λλk λ k k! k =0 2. ( poits) Compute g(r) = E [ r X], for ay real umber r > 0. E[r X ] = r k e λλk = e λ g(r) = e λ e rλ = e λ(r 1) (rλ) k. (2 poits) Take for a fact that g (1) = E[X]. Verify your aswer to questio 1. g (r) = λe λ(r 1) g (1) = E[X] = λ (you ll get 2 extra poits if you actually prove that g (1) = E[X]) g(r) = r k e λλk g (r) = d dr g(r) = g (1) = kr k 1 e λλk ke λλk = E[X]

2 Problem 2: coutig squirrels (9 poits) We wat to estimate the umber N of squirrels o campus. To do that, we capture k squirrels, we put a mark o oe of their leg, ad release them. The, we wait a little bit, capture a squirrel, see if it has a mark, ad release it. We wait a little bit, capture aother squirrel, see if it has a mark, ad release it. etc... We do that l times, ad we cout the umber of marks that we saw. We assume that whe we capture a squirrel, we pick it uiformly at radom amog the N squirrels, idepedetly of the other time we have captured a squirrel (ote that amog the l oes that we checked, we could have picked twice, three time, etc.. the same oe). 1. ( poits) What is the probability that the first squirrel we capture is marked? Fid the probability p(m) that we fid m marks (it should deped o N,K,l). There are k marked squirrels amog N oes. Probability that the first oe I pick is marked = k/n. Probability to fid m marks amog l picks = ( ) l m (k/n) m (1 k/n) l m 2. (4 poits) We proceeded the experimet: k,l,m are data. To estimate N, we choose the value that maximizes p(m) as a fuctio of N. Fid this value, for ay give k,l,m. Hit: compute the derivative of p(m) with respect to N. We give the derivative of f(x) = (1 x) l m x m with respect to x: f (x) = (m lx)(1 x) l m 1 x m 1. Write y = k/n. p(m) = ( l m) y m (1 y) l m. ( ) d l dy p(m) = (m ly)(1 y) l m 1 y m 1 m p(m) maximal for y = l/m, that is for N = kl/m.. (2 poits) If we marked 20 squirrels, captured 20 squirrels, ad that we fid oly 5 marked squirrels, what is the estimated umber of squirrels o campus? k = 20, l = 20, m = 5. Estimatio by questio 2: N = kl/m = 80

3 Problem : (2 pages!) gamblig or ot gamblig (16 poits) There are 10 balls i a ur: 6 Gree balls, Red balls, ad 1 Black ball. You pick balls i the ur, ad your gai depeds o the umber of Red balls you get: with oe Red ball you wi $10, with two red balls $25, ad if you get the three red balls you wi the jackpot, $240! But careful, if you get the Black ball, you do t wi aythig (2 poit) What is the probability of wiig the jackpot? 1 ) = ( poits) What is the probability of gettig the black ball? B = gettig the black ball : pick the black ball, ad choose the two remaiig balls: ( ) 9 2 ( 9 P(B) = ) = 10. (4 poits) Show that the probability that you do t wi aythig is 7 otatios: for example B is the evet that you get the black ball, etc...) N = gettig NO red ball. Dot wi aythig: N B 15 (give evets some P(N B) = P(N)+P(B) P(N B) NO red balls: all the balls are picked i the remaiig 7: ( ) 7 No red ball ad the black ball: pick the black, choose the remaiig two outside the red oes (6 choices): ( ) 6 2 ( 7 P(N) = ) ) = 7 24, P(B) = ( 6, P(N B) = ) = the = 7 15

4 Problem (page 4. (5 poits) We ote G the gai oe gets at this game. Fid the probability mass fuctio of the radom variable G. (Verify that is sums up to 1...) Questio 1 ad give, respectiveley P(G = 0) = 7/15 ad P(G = 240) = 1/120. P(G = 10) = P(1 red ball ad o black ball) = ( )( 6 1 ( 10 ) = 8 (choose the red ball you pick ( 1), the two other balls ( 6 ) P(G = 25) = P(2 red balls ad o black ball) = ( )( 6 2 ( 1) 10 ) = 40 (choose the two red balls you pick (, the other ball ( 6 1) ) Easy to verify it sums up to 1 (put everythig o 120) 5. ( poits) What is the expected value of the Gai? E[G] = 0.P(G = 0)+10.P(G = 10)+25.P(G = 25)+240.P(G = 240) = = 9.5 4

5 Problem 4: (7 poits) There are 5 Red balls, Gree balls ad 2 Yellow balls i a ur (agai!). Some boucig balls may have bee itroduced... The Red balls have a probability.2 to be boucig, the Gree balls a probability.4, ad the Yellow balls a probability.9. You pick 1 ball at radom. 1. ( poits) What is the probability that you get a boucig ball? B = boucig, R = Red, G = Gree, Y = Yellow P(B) = P(R)P(B R)+P(G)P(B G)+P(Y)P(B Y) = = = (4 poits) Give that you got a boucig ball, what is the coditioal probability that it was a Red ball? A Gree ball? A Yellow ball? P(R B) = P(R)P(B R) P(B) P(G B) = P(G)P(B G) P(B) P(Y B) = P(Y)P(B Y) P(B) =.10.4 = 1 4 =.12.4 = 10 =.18.4 = 9 20

6 Problem 5: collectig toys (10 poits) Each week, you buy your favorite pack of cereals. It is your favorite because i each pack, there is a toy, ad you try to collect all the differet toys. We assume that the toy you get each week is chose uiformly amog the possible, idepedetly of what happes the other weeks. We deote T i the umber of weeks it takes to reach a collectio of i differet toys, ad X i = T i T i 1 the umber of weeks it takes to get from a collectio of i 1 differet toys to a collectio of i differet toys. 1. ( poits) Show that X i is a Geometric radom variable. What is its parameter? You have i 1 differet toys, there are (i 1) that you do t have. You get a ew toy ext week with probability p := (i 1). The probability that you wait exactly k weeks before you get a ew toy is therefore: (1 p) k 1 p (you fail the k 1 first week, ad you success the last oe, the weeks beig idepedet). This is exactly the probability mass fuctio of a geometric radom variable, with parameter p = (i 1). 2. ( poits) T /2 is the time to complete a half collectio. Write T /2 as a sum of X i s, ad compute E[T /2 ] (you do t have to simplify the sum). T /2 = X 1 +X X /2 = /2 X i, so that /2 E[T /2 ] = E[X i ] = /2 i+1. We used that the expectatio of a geometric radom variable of parameter p is 1/p (here X i is a geometric r.v. with parameter ( i+1)/). (4 poits) Show that E[T /2 ] < (Hit: give a boud for each term i the sum). Deduce that, o average, gettig the last missig toy (it takes a time X ) is loger tha completig a half collectio. Sice the idex i the sum is i /2, oe gets that i+1 > /2, so that Oe gets /2 /2 E[T /2 ] = i+1 < 2 i+1 < /2 = 2. The time to get the last missig toy is X. It s a geometric radom variable with parameter 1/. The average time to get the last missig toy is E[X ] = > E[T /2 ], the average time to get a half collectio.