Fourth Problem Assignment

Transcription

1 EECS 401 Due on Feb 2, 2007 PROBLEM 1 (25 points) Joe and Helen each know that the a priori probability that her mother will be home on any given night is 0.6. However, Helen can determine her mother s plan for the night at 6 P.M., and then, at 6:15 P.M., she has only one chance each evening to shout one of two code words across river to Joe. He will visit her with probability 1.0 if he thinks Helen s message means Ma will be away, and he will stay home with probability 1.0 if he thinks the message means Ma will be home. But Helen has a meek choice and the river is channeled for heavy barge traffic. Thus she is faced with the problem of coding for a noisy channel. She has decided to use a code containing only the code words A and B. The channel is described by P(a A) = 2 3, P(a B) = 1 4, P(b A) = 1 3, P(b B) = 3 4 where a is the event that Joe think message is A and b is the event that Joe thinks message is B. (a) In order to minimize the probability of error between transmitted and received messages, should Helen and Joe agree to use code I or code II? Code I A = Ma away B = Ma home Code II A = Ma home B = Ma away Using Code I Pr(error) = Pr(A, b) + Pr(B, a) = Pr(A) Pr(b A) + Pr(B) Pr(a B) = = Using Code II 1 Due on Feb 2, 2007

2 s Pr(error) = Pr(A, b) + Pr(B, a) = Pr(A) Pr(b A) + Pr(B) Pr(a B) Thus Joe and Helen should code I = = 0.3 (b) Helen and Joe put the following cash values (in dollars) on all possible outcomes of a day Ma home and Joe comes -30 Ma home and Joe doesn t come 0 Ma away and Joe comes +30 Ma away and Joe doesn t come -5 Joe and Helen make their plans with the objective of maximizing the expected value of each day of their continuing romance. Which of the above codes will maximize the expected cash value per day of this romance? Using code I Using code II E[value] = Pr(A, a)(30) + Pr(A, b)( 5) + Pr(B, a)( 30) + Pr(B, b)(0) = = E[value] = Pr(A, a)( 30) + Pr(A, b)(0) + Pr(B, a)( 5) + Pr(B, b)(30) = = Thus to maximize the expected cash value of their romance, Joe and Helen should use code I. (c) Clara isn t quite so attractive as Helen, but at least she lives on the same side of the river. What would be the lower limit of Clara s expected value per day which would make Joe decide to give up Helen? 2 Due on Feb 2, 2007

3 s Clara s expected value per day must be at least $2.833 for Joe to decide to give up Helen. (d) What would be the maximum rate which Joe would pay the phone company for a noiseless wire to Helen s house which he could use once per day at 6:15 P.M.? Then, Suppose Helen uses the telephone line. Let M be the event that Ma is home. E[value] = 0 Pr(M) + 30 Pr(M ) = 12 Thus Joe will be willing to give $( ) = $9.16. (e) How much is it worth to Joe and Helen to double her mother s probability of being away from home? Would this be a better or worse investment than spending the same amount of money for a telephone line (to be used once a day at 6:15 P.M.) with the following probabilities. P(a A) = P(b B) = 0.9, P(b A) = P(a B) = 0.1 Suppose that the probability of Ma begin away from home is doubled to 0.8. Then using code I, Using code II E[value] = Pr(A, a)(30) + Pr(A, b)( 5) + Pr(B, a)( 30) + Pr(B, b)(0) = = E[value] = Pr(A, a)( 30) + Pr(A, b)(0) + Pr(B, a)( 5) + Pr(B, b)(30) = = 15 4 Thus, if Ma s probability of being away is doubled, Joe and Helen will use code II with an expected value of $15. Thus, they will be willing to give $( ) = $ The telephone line is symmetric, so both codes will have the same expected value of (asumming A is the event Ma is home E[value] = Pr(A, a)( 30) + Pr(A, b)(0) + Pr(B, a)( 5) + Pr(B, b)(30) = = 8.8 This is less than the expected return when Ma is away with a probability of 0.8. Thus spending in doubling Ma s probability from being away from home is better. 3 Due on Feb 2, 2007

4 s PROBLEM 2 (18 points) Hypothesis testing May B. Lucky is a compulsive gambler who is convinced that on any given day she is either lucky, in which case she wins each red/black bet she makes in roulette with probability p L > 0.5, or she is unlucky, in which case she wins each red/black bet she makes in roulette with probability p U < 0.5. May visits the casino every day, and believes that she knows the a priori probability that any one given visit is a lucky one (i.e., corresponds to p L rather than p U ). To improve her chances, May adopts a system whereby she estimates on-line whether she is lucky or unlucky on a given day, by keeping a running count of the number of bets that she wins and loses. In particular, she continues to play until the conditional odds in favor of the event lucky on the current day} given the number of wins and losses so far, fall below a certain threshold. As soon as this happens, she stops playing. Provide a simple algorithm for updating May s conditional odds with each play. Note that if A and B are events with P(A) > 0 and P(B) > 0, the odds in favor of A given B are defined as O(A B) = P(A B) P(A c B). Let A be the event that May is lucky on the current day and let B n,m be the event that n wins and m losses have occurred so far. Assume independence of the results of different spins/plays. Let p be the a prior probability that she is lucky on the current day. Then we have, O(A B n,m ) = Pr(A B n,m) Pr(A c B n,m ) = Pr(A) Pr(B ( n+m ) n,m A) Pr(A c ) Pr(B n,m A c ) = n p n L (1 p L ) m ) p n U (1 p U ) m = ( pl p U ) n ( 1 pl 1 p U ) m p 1 p ( n+m m From this formula, a recursive algorithm can be obtained. Let q(n + m) be the odds after n + m games. Then q(n + m + 1) = q(n + m) p L p U, q(n + m) 1 p L 1 p U, if she wins the next play if she loses the next play The initial condition is O(A B 0,0 ) which is equal to the initial (unconditional) odds O(A), which May knows by assumption. PROBLEM 3 (12 points) Fischer and Spassky play a sudden-death chess match whereby the first player to win a game wins the match. Each game is won by Fischer with probability p, by Spassky with probability q, and is a draw with probability 1 p q. 4 Due on Feb 2, 2007

5 s (a) What is the probability that Fischer wins the match? Let E be the event that Fischer wins the match. We can express E as E = n 0 E n where E n is the event that each of the first n games is a draw and the (n + 1)th game is won by Fischer. Since E n s are disjoint, we have Pr(E) = Pr(E n ) = (1 p q) n p = p p + q. n 0 n 0 (b) What is the PMF, the mean, and the variance of the duration of the match? The PMF of D is given by p D (d) = (1 p q) d 1 (p + q), d = 1, 2, Since the duration D of the match is a geometric random variable with parameter p + q, we obtain E[D] = 1 p + q, and var(d) = 1 p q (p + q) 2. 5 Due on Feb 2, 2007

6 s PROBLEM 4 (24 points) When you push the SEND button on your cell phone, the phone attempts to set up a call by transmitting a SETUP message to a nearby base station. The phone waits for a response and if none arrives within 0.5 seconds it tries again. If it doesn t get a response after N tries the phone stops transmitting and generates a busy symbol. Assume that all transmissions are independent and that with probability p the SETUP message will get through. Also assume that if the SETUP message gets through the response from the base station is always correctly received by the cell phone within 0.5 seconds. (a) What is the PMF of X, the number of times the SETUP message is transmitted in a call attempt? X can take values 1, 2,..., N. The PMF is given by p(1 p) x 1, x = 1,2,...,(N-1) p X (x) = (1 p) N 1 x = N 0 otherwise (b) What is the probability that the call will generate a busy signal? We get a busy signal if all N attempts are unsuccessful. Thus Pr(BUSY) = (1 p) N (c) Assuming that there is no limit on the number of tries, i.e., your phone will keep transmitting the SETUP message indefinitely until it gets through, what is the PMF of X, the number of transmissions in a call attempt? If the number of trials are unlimited, the PMF is given by p X (x) = p(1 p) x 1, x = 1,2,3,... 0 otherwise (d) Following the previous part, what is the expected number of transmissions of the SETUP message in a call attempt? Notice that this is the geometric distribution with parameter p. Thus, E[X] = 1 p 6 Due on Feb 2, 2007

7 s PROBLEM 5 (12 points) Let X be a random variable with PMF p X (a) Suppose g is a one-to-one function; i.e., g(x) g(y) if x y. Show that E(g(X)) = x g(x)p X (x). (1) Assume X takes n values x 1, x 2,..., x n. Define a r.v. Y = g(x), which also takes n distict values y 1 = g(x 1 ), y 2 = g(x 2 ),..., y n = g(x n ). By definition E[Y] = y i p Y (y i ) (2) But event Y = y i } = X = x i } since g is a one-to-one mapping, thus i=1 p Y (y i ) = Pr(Y = y i }) = Pr(X = x i }) = p X (x i ) Pluggin the above result into (2), we have E[Y] = y i p Y (y i ) = i=1 g(x i )p X (x i ). (b) Suppose now that g is general and can be many-to-one. Show that (1) still holds. In this case, the event Y = y l } = i=1 m l k=1 1, 2,..., m l are disjoint and g(x kl ) = y l, k = 1, 2,..., m l. Thus p Y (y l ) = Pr(Y = y l }) = Substituting this in (2) above, we have m l k=1 X = xkl }, where X = xkl }, k = Pr(X = x kl }) = m l k=1 p X (x kl ). E[Y] = y l p Y (y l ) = m l l=1 m l g(x kl )p X (x kl ), l=1 Since x kl are distinct for all possible l, k, the summation just enumerates all the x i, i = 1, 2,..., n and therefore we have E[Y] = g(x i )p X (s i ) i=1 7 Due on Feb 2, 2007

8 s PROBLEM 6 (9 points) The annual premium of a special kind of insurance starts at $1000 and is reduced by 10% after each year where no claim has been filed. The probability that a claim is filed in a given year is 0.05, independently of preceding years. What is the PMF of the total premium paid up to and including the year when the first claim is filed? A claim is first filed for the first time in year n with probability (0.05) (0.95) n 1, and the corresponding total premium is 1000 ( (0.9) n 1) = (0.9)n = (1 (0.9) n ). Thus the PMF of X, the total premium paid up to and including the year when the first claim was filed, is p X (x) = 0.05 (0.95) n 1, if x = (1 (0.95) n ), n = 1, 2,... 0, otherwise 8 Due on Feb 2, 2007