Fifth Problem Assignment

Transcription

1 EECS 40 PROBLEM (24 points) Discrete random variable X is described by the PMF { K x p X (x) = 2, if x = 0,, 2 0, for all other values of x Due on Feb 9, 2007 Let D, D 2,..., D N represent N successive independent experimental values of random variable X. () Determine the numberical value of K. For p X (x) to be a valid probability measure, it must sum to. Thus, K + K 2 + K 2 2 = (2) Determine the probability that D > D 2. = K = 2 Pr(D > D 2 ) = Pr ( {(, 0), (2, 0), (2, )} ) = Pr ( (, 0) ) + Pr ( (2, 0) ) + Pr ( (2, ) ) = = 47 (3) Determine the probability that D + D D N.0 Pr(D + D d N <= ) = Pr(All D i s are zero) + Pr(One D i is one and rest are zeros) = ( ) N ( 2 + N N 3 2) (4) Define R = max(d, D 2 ) and S = min(d, D 2 ). Determine the following PMF s for all values of their arguments: 44 r = max(d, D 2 ) and s = min(d, D 2 ). We have Due on Feb 9, 2007

2 s Hence (a) p S (s) D D 2 Pr(D, D 2 ) r s t p S (s) = 9 44, s = 0 8, s = 6, s = 2 (b) P R S (r 0) p R S (r 0) = 9, r = 0 8 9, r = 6 9, r = 2 (c) P R,S (r, s) p R,S (r, s) = , (r, s) = (0, 0), (r, s) = (, 0), (r, s) = (2, 0), (r, s) = (, ), (r, s) = (2, ), (r, s) = (2, 2) (d) P T (t) with T = ( + D )/( + S) 2 Due on Feb 9, 2007

3 s p T (t) = 97 44, t = 2, t = , t = 2, t = 3 () Determine the expected value and variance of random variable S defined above. Using the PMF of S given by part (i) above, we have E[S] = (6) Given D + D 2 2.0, determine the conditional expected value and conditional variance of random variable S defined above. If D + D 2 2, then S can only take values 0 and. Let A be the event that D + D 2 2. We have and Thus Pr(A) = 37 p S A (s) = Pr(min(D, D 2 ) = s, A) Pr(A) E[S A] = 6 and Var(S) = , s = 0 = 6, s = PROBLEM 2 (2 points) A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability 0.3, and his second appointment will lead independently to a sale with probability 0.6. Any sale made is equally likely to be either for the deluxe model, which costs $000, or the standard model, which costs $00. Determine the PMF of X, the total dollar value of the sales this salesman makes. Let (S, S 2 ) denote the amounts of sale of first and the second appointment. Thus we have, 3 Due on Feb 9, 2007

4 s Sale T otal Probability (0, 0) = 0.28 (00, 0) = 0.06 (0, 00) = 0.2 (00, 00) = 0.04 (000, 0) = 0.06 (0, 000) = 0.2 (00, 000) = 0.04 (000, 00) = 0.04 (000, 000) = 0.04 Thus the PMF is given by 0.28, x = , x = x = 000 p X (x) = 0.09 x = x = 2000 PROBLEM 3 (2 points) Let X, Y, and Z be independent geometric random variables with the same PMF: p X (k) = p Y (k) = p Z (k) = p( p) k, where p is a scalar with 0 < p <. Find P(X = k X + Y + Z = n). Hint: Try thinking in terms of coin tosses. We have Note that Pr(X = k X + Y + Z = n) = Pr(X = k, X + Y + Z = n) Pr(X + Y + Z = n) Pr(X = k, X + Y + Z = n) = Pr(X = k, Y + Z = n k) = Pr(X = k) Pr(Y + Z = n k) where the last equality follows from the fact that X, Y and Z are independent. Consider the following experiment. We repeatedly toss a coin with probability of head equal to p. Let X be the number of tosses up to and including the first head, Y be the number of tosses following the first head up to and including the second head, and Z be the number of tosses following the second head up to and including the third head. The random variables X, Y and Z are independent having the distribution as given above. 4 Due on Feb 9, 2007

5 s Then {Y + Z = n k} is the event that the second head occurs on the (n k)th toss. For this event to occur, we need to get two heads and n k 2 tails, and the second head must occur on toss n k, but the first head could occur at any of the previous n k tosses. Therefore Pr(Y + Z = n k) = (n k )p 2 ( p) n k 2 Similarly, {X + Y + Z} is the event that the third head occurs on the nth toss. For this to occur, we need to get 3 heads and n 3 tails. The third head must occus on the nth toss, but the first two heads could occur at any of the previous n tosses. The total number of ways two heads can occur in n tosses is ( ) n 2. Therefore ( ) n Pr(X + Y + Z = n) = p 3 ( p) n 3, n 3. 2 Combining the preceeding equations we have (n ( k )p 3 ( p) n 3 if n 3, k =, 2,..., n 2, Pr(X = k X + Y + Z = n) = n )p 3 ( p) n 3 2 { 2(n k ) = (n )(n 2) if n 3, k =, 2,..., n 2, PROBLEM 4 (30 points) Nick shops for probability books for K hours, where K is a random variable that is equally likely to be, 2, 3, or 4. The number of books N that he buys is random and depends on how long he shops according to the conditional PMF p N K (n k) =, for n =,..., k. k and We are given that { /4 if k =, 2, 3, 4, p K (k) = (a) Find the joint PMF of K and N. { /k if n =,..., k, p N K (n k) = Due on Feb 9, 2007

6 s Applying the chain rule, we have p N,K (n, k) = p K (k)p N K (n K), Substituting p K (k) and p N K (n k) we obtain { /(4k) if k =, 2, 3, 4 and n =,..., k, p N,K (n, k) = (b) Find the marginal PMF of N. The marginal PMF p N (n) is given by p N (n) = k p N,K (n, k) = 4 k=n 4k that is, = 2, if n =, p N (n) = = 3, if n = = 7, if n = 3 6 = 3, if n = 4 (c) Find the conditional PMF of K given that N = 2. We have p K N (k 2) = p N,K(2, k) p N (2) 6 3 if k = 2, 4 = 3 if k = 3, 3 3 if k = 4, (d) Find the conditional mean and variance of K, given that he bought at least two but no more than three books. Let A be the event 2 N 3. We know that p K A (k) = Pr(K = k, A), Pr(A) Pr(A) = p N (2) + p N (3) = 2, 8 if k = 2, Pr(K = k, A) = if k = 3, if k = 4, 6 Due on Feb 9, 2007

7 s and finally 3 0 if k = 2, 2 p K A (k) = if k = 3, 3 0 if k = 4, The conditional PMF of K given A is symmetric around k = 3, so E[K A] = 3 The conditional variance of K given A is given by var (K A) = E [ (K E[K A]) 2 A ] = 3 0 (2 3) (4 3)2 = 3 (e) The cost of each book is a random variable with mean $30. What is the expected value of his total expenditure? Hint: Condition on the events {N = },..., {N = 4}, and use the total expectation theorem. We are given that E[C i ] = 30, where C i is the cost of book i. Let T be the total cost, so that T = C C N. (Notice that N here is a random variable.) Using the total expectation theorem we have E[T] = E [ E[T N] ] PROBLEM (2 points) = E [E [ N C i N ]] i= [ ] = E N 30 = 30E[N] ( = ) = 2. = (a) Let X be a random variable that takes on nonnegative integer values. Show that E[X] = P(X k). k= Hint: Start with the above formula and interchange the order of the summation. 7 Due on Feb 9, 2007

8 s Consider the right hand side Pr(X k) = k= = k= i=k i= k= = E[X] p X (X = i) i p X (X = k) = ip X (X = i) (b) Use the formula in the previous part to compute the expectation of a random variable Y with the geometric PMF p Y (l) = p( p) l, l =, 2,.... (Here p is a constant between 0 and.) For a geometric distribution Pr(X k) = i= p( p) l = p( p) k ( p) i = ( p) k l=k From the previous part we know that E[X] = Pr(X k) = k= i=0 ( p) k = p PROBLEM 6 (0 points) Consider the following two cellphone billing plans. (a) You are charged a monthly fee of $20, which includes thirty free minutes. Beyond this you pay $0.0 for each extra minute. k= Then Let c(x) denote the amount of money that one has to pay for using X minutes. c(x) = { 20 x (x 30) 0. x > 30 Thus, E[X] = c(x)p X (x) = 20 + x= = 20 + x=3 0.(x 30)p( p) x 0.yp( p) y+29 = ( p) 30 y= = ( p)30 p y= yp( p) y 8 Due on Feb 9, 2007

9 s (b) You are charged a monthly fee of $, which has no free minutes. You pay $ for each minute you use. In this case we have c(x) = + x. Thus, E[c(X)] = E[ + X] = + E[X] = + p Suppose that the number of minutes you use every month is a geometric random variable X with mean /p. What is the expected monthly cost of each of the two plans? 9 Due on Feb 9, 2007