Problems for 2.6 and 2.7

Transcription

1 UC Berkeley Department of Electrical Engineering and Computer Science EE 6: Probability and Random Processes Practice Problems for Midterm: SOLUTION # Fall 7 Issued: Thurs, September 7, 7 Solutions: Posted on Tues, October, 7 Reading: Bertsekas & Tsitsiklis, Chapters and Problems for.6 and.7 Problem 3. Let X and Y be independent random variables that take values in the set {,,3}. Let V X + Y, and W X Y. (a Assume that P(X k and P(Y k are positive for any k {,,3}. Can V and W be independent? Explain. (No calculations needed. For the remaining parts of this problem, assume that X and Y are uniformly distributed on {,,3}. (b Find and plot p V (v, and compute E[V ] and var(v. (c Find and show in a diagram p V,W (v,w. (d Find E[V W > ]. (e Find the conditional variance of W given the event V 8. Solution:. V and W cannot be independent. Knowledge of one random variable gives information about the other. For instance, if V we know that W.. We begin by drawing the joint PMF of X and Y. y 3 (/9 (/9 (/9 v (/9 (/9 (/9 (/9 (/9 (/9 v4 v6 v v8 3 x

2 X and Y are uniformly distributed so each of the nine grid points has probability /9. The lines on the graph represent areas of the sample space in which V is constant. This constant value of V is indicated on each line. The PMF of V is calculated by adding the probability associated with each grid point on the appropriate line. p (v V /3 /9 / v By symmetry, E[V ] 8. The variance is: var(v ( ( ( ( Alternatively, note that V is twice the sum of two independent random variables, V (X + Y, and hence var(v var ( (X+Y var(x+y 4 ( var(x+var(y 4 var(x 8var(X (note the use of independence in the third equality; in the fourth one we use the fact that X, Y are identically distributed, therefore they have the same variance. Now, by the distribution of X, we can easily calculate that var(x 3 ( + 3 ( + 3 (3 3, so that in total var(v 6 3, as before. 3. We start by adding lines corresponding to constant values of W to our first graph in part b:

3 y 3 w- w- w v w v v4 v6 w v8 3 x Again, each grid point has probability /9. Using the above graph, we get p V,W (v,w. w - - (/9 (/9 (/9 (/9 (/9 (/9 (/9 (/9 (/ v 4. The event W > is shaded below: w - - (/9 W> (/9 (/9 (/9 (/9 (/9 (/9 (/9 (/ v By symmetry, E[V W > ] The event V 8 is shaded below: 3

4 w - - V v When V 8, W can take on values in the set {,,} with equal probability. By symmetry, E[W V 8]. The variance is: var(w V 8 ( 3 + ( 3 + ( Problem 3. Joe Lucky plays the lottery on any given week with probability p, independently of whether he played on any other week. Each time he plays, he has a probability q of winning, again independently of everything else. During a fixed time period of n weeks, let X be the number of weeks that he played the lottery and Y the number of weeks that he won. (a What is the probability that he played the lottery any particular week, given that he did not win anything that week? (b Find the conditional PMF p Y X (y x. (c Find the joint PMF p X,Y (x,y. (d Find the marginal PMF p Y (y. Hint: One possibility is to start with the answer in (c, but the algebra can be messy. But if you think intuitively about the procedure that generates Y, you may be able to guess the answer. (e Find the conditional PMF p X Y (x y. In all parts of this problem, make sure to indicate the range of values to which your PMF formula applies. Solution:. Let L i be the event that Joe played the lottery on week i, and let W i be the event that he won on week i. We are asked to find P(L i W c i P(Wi c L ip(l i P(Wi c L ip(l i + P(W c i Lc i P(Lc i ( qp ( qp + ( p p pq pq 4

5 . Conditioned on X, Y is binomial ( x q p Y X (y x y y ( q (x y, y x;, otherwise. 3. Realizing that X has a binomial PMF, we have p X,Y (x,y p Y X (y xp X (x ( ( x n q y y ( q (x y x 4. Using the result from (c, we could compute 5. p x ( p (n x, y x n;, otherwise. p Y (y n p X,Y (x,y, xy but the algebra is messy. An easier method is to realize that Y is just the sum of n independent Bernoulli random variables, each having a probability pq of being. Therefore Y has a binomial PMF: p Y (y ( n (pq y y ( pq (n y, y n;, otherwise. p X Y (x y p X,Y (x,y p Y (y Aq y ( q (x n Ap y x x ( p (n n, y x n; A(pq y y ( pq (n y, otherwise. Problem 3.3 Suppose you and your friend play a game where each of you throws a 6-sided die, each of your throws being independent. Each time you play the game, if the largest of the two values you obtained from each die is greater than 4, then you win dollar; otherwise, you lose dollar. Suppose that you play the game n 3 times, each game being independent of the others. (a What is the amount of money you expect to win on the first and last game combined? 5

6 (b How much do you expect to win in your last game given that you lost in the first game? (c How much do you expect to have won in your last game given that you won the first game and you won a total of m dollars at the end? (d What is the probability that you won both the first and last game given that you won a total of m dollars at the end? Solution: (a We can define the collection of random variables {X i } i I for I {,,...,n} with each X i representing the amount that is won in game i. For this question, we want to calculate E [X + X n ] E [X ] + E [X n ] E [X ], where we have used linearity of expectation and the fact that {X i } i I are identically distributed. To compute E [X i ], we note that X i takes value whenever max {roll #, roll #} > 4 and it takes on otherwise. If we name the two dice rolls as A and B and consider the joint PMF p A,B (a,b p A (a p B (b 36, we can evaluate the expectation directly: E [X i ] 9 max{a,b}>4 max{a,b}>4 p A,B (a,b max{a,b} 4 max{a,b} 4 36 ( p A,B (a,b So the solution to this question is $ 9. (b By independence, we can ignore the information about losing the first game, and simply calculate the expected value of the last game as above to get $ 9. (c Easier Solution: The given information provides E [ n i X i] m, and by linearity of expectation and the fact that all X i are identical, we then have (n E [X i ] m E [X i ] m Longer Solution: n. We want to compute E [X n C], where our conditioning event C is defined as C {X } D and D is defined as D { won m dollars total }. The expectation is given by: 6

7 E [X n C] x n {,} x n p xn C(x n P ({X n } {X } D P ({X } D P ({X n } {X } D P ({X } D + ( P ({X n } {X } D P ({X } D To carry out the computation, we need to translate our event statements into familiar forms. First, we consider the event {X n } {X } D. This event can be translated to we win the first game AND the second game AND m- dollars in the other games. Since we have split the sub-events across games, we now have independet events, and we notice that the last event can be expressed as a binomial; to win m dollars in n games, we must have won (m + (n (m / games total and lost (n (m / games total. Therefore, we have: P ({X n } {X } D P ({X n } P ({X } P (D 36 ( ( n m ( n 6 36 n m (n n m To calculate the probability of the event {X } D is similar; we translate into won the first game AND won m- dollars in n- other games, and we carry out the computation with independence and a binomial. Note that for both of the probability expressions, expression is only valid when n m is even. When n m is odd, the probability is zero, because it is impossible to win m dollars in n games in that case (since we must have split the other n m games evenly between wins and losses. Our final expression, neglecting the trivial case of n m odd and after performing some algebra and cancellations, is: E [X n C] ( n n m ( n n m n + m (n m n (d Now we want to compute P({X } {X n } D. As before, we can expand the conditional by the definition of conditional probability, and carry out the computation as above: 7

8 P ({X } {X n } D P ({X } {X n } D P (D Sample midterm problems (n n m (n n m n(n Problem 3.4 Since there is no direct flight from San Diego (S to New York (N, every time Alice wants to go to the New York, she has to stop in either Chicago (C or Denver (D. Due to bad weather conditions, both the flights from S to C and the flights from C to N have independently a delay of hour with probability p. Similarly, at Denver airport, both incoming and outgoing flights are independently subject to a hour delay with probability q. On any given occasion, Alice chooses randomly between the Chicago or Denver routes with equal probability. (a What is the average total delay (across both legs of the overall trip that she experiences in going from S to N? (b Suppose Alice arrives at N with a delay of two hours. What is the probability that she flew through C? (c Suppose that Alice wants to maximize the probability that she arrives in New York with a total delay < hours. Under what conditions on p and q is going via Chicago a better choice than going via Denver? (d Suppose now that Alice always flies through C. On average, how many trips does she make before experiencing a hour delay? (e Suppose now that the flight between S and D is known to be delayed, but Alice still randomly flies either via C or D with equal probability. With what delay should she expect to arrive at N? Solution: The problem can be modeled as a network having four nodes (S,D,C,S, where SD, SC, CD and CN are linked. We can define four independent random variable indicating the delay on each of the link: X SC is wp p and is wp p. X SD is wp q and is wp q. X CN is wp p and is wp p. X DN is wp q and is wp q. Also, let us define D as the event that Alice flies through Denver, and C as the event that Alice flies through Chicago. 8

9 (a There are two possible ways to go to N from S, so using Total Probability law we have E(delay E(delay CP(C + E(delay DP(D E(X SC + X CN P(C + E(X SD + X DN P(D (E(X SC + E(X CN + E(X SD + E(X DN p + q (b Using Total Probability law and Bayes rule, and since P(D P(C we have P(C delay P(delay CP(C P(delay DP(D + P(delay CP(C P(X SC + X CN P(D P(X SD + X DN P(D + P(X SC + X CN P(C P(X SC + X CN P(X SD + X DN + P(X SC + X CN p q( q + p (c Flying via Denver P(delay < D P(X SD + X DN < ( q and flying via Chicago P(delay < C P(X SC + X CN < P(X SC + X CN p This solution has been updated: > should have been <. Alice should fly via Chicago when ( q < p. This is the case when q? p or, equivalently, p < q( q. (d A delay of two hours happens with probability p on each trip. We are asked for the mean of a geometric random variable with parameter p. Thus, the average number of trips is p. (e From the independence of the four random variables E(delay X SD E(delay X SD,DP(D X SD +E(delay X SD,CP(C X SD ( + E(X DN + E(X SC + E(X CN ( + q + p + q + p 9

10 Problem 3.5 We transmit a bit of information which is with probability p and with p. Because of noise on the channel, each transmitted bit is received correctly with probability ǫ. (a Suppose we observe a at the output. Find the conditional probability p that the transmitted bit is a. (b Suppose that we transmit the same information bit n times over the channel. Calculate the probability that the information bit is a given that you have observed n s at the output. What happens when n grows? (c For this part of the problem, we suppose that we transmit the symbol a total of n times over the channel. At the output of the channel, we observe the symbol three times in the n received bits, and that we observe a at the n-th transmission. Given these observations, what is the probability that the k-th received bit is a? (d Going back to the situation in part (a: some unknown bit is transmitted over the channel, and the received bit is a. Suppose in addition that the same information bit is transmitted a second time, and you again receive another. We want to find a recursive formula to update p to get p, the conditional probability that the transmitted bit is a given that we have observed two s at the output of the channel. Show that the update can be written as Solution: p ( ǫp ( ǫp + ǫ( p (a Let A be the event that is transmitted, A c be the event that is transmitted, B n be the event the n-th bit we received is. Then using Total Probability law and Bayes rule: p P(A B P(B AP(A P(B AP(A + P(B A c P(A c ( ǫp ( ǫp + ǫ( p (b As in the previous part, and assuming that the uses of the channel are independent, we have P(A B,...,B n As n, we have P(B,...,B n AP(A P(B,...,B n AP(A + P(B,...,B n A c P(A c ( ǫ n p ( ǫ n p + ǫ n ( p p p + ( ǫ ǫ n ( p ǫ < p p ǫ ǫ >

11 These results meet the intuitions. When ǫ <, it means the observations are positively correlated with the bit transmitted, thus knowing the observations B n n,,... will increase the conditional probability of A given the observations, until it hits. When ǫ, it means the observations are independent to the bit transmitted, thus given a bunch of observations B n n,,... won t change the conditional probability of A given the observations, which is the same as the prior probability p. When ǫ >, it means the observations are negatively correlated with the bit transmitted, thus knowing the observations B n n,,... will decrease the conditional probability of A given the observations, until it hits. (c For j < n P(B j n B i 3,B n,a i P( n i B i 3 B j,b n,ap(b B n,a P( n i B i B n,a ( n ( ǫǫ n 3 ( ǫ ( ǫ ǫ n 3 ( n n while if j n P(B n n i B i 3,B n,a. (d Assuming the uses of the channel are independent, we have p P(A B,B P(B A,B P(A B P(B B P(B,B AP(A P(B AP(A P(A B P(B B ( ǫ ( ǫ p P(B B And since We have P(B B P(B,B P(B P(B,B AP(A + P(B,B A C P(A C P(B AP(A + P(B A c P(A c ( ǫ p + ǫ ( p ( ǫp + ǫ( p ( ǫp ( ǫ ( ǫp + ǫ( p + ǫ ǫ( p ( ǫp + ǫ( p ( ǫp + ( p ǫ p ( ǫp ( ǫp + ( p ǫ

12 One might have noticed that, if we let p p, the functions that we use to upgrade p to p (part (a, and p to p are the same. Intuitively, for the first observation, p is the prior probability of A before the observation, and p is the posterior probability of A after the observation. Similarly, for the second observation, p is the prior probability, and p is the posterior probability. Thus the ways to upgrade two prior probabilities to the two posterior probabilities should be the same. Problem 3.6 You play the lottery by choosing a set of 6 numbers from {,,...,49} without replacement. Let X be a random variable representing the number of matches between your set and the winning set. (The order of numbers in your set and the winning set does not matter. You win the grand prize if all 6 numbers match (i.e., if X 6. (a What is the probability of winning the grand prize? Compute the PMF p X of X. (b Suppose that before playing the lottery, you (illegally wiretap the phone of the lottery, and learn that of the winning numbers are between and ; another are between and 4, and the remaining are between 4 and 49. If you use this information wisely in choosing your six numbers, how does your probability of winning the grand prize improve? (c Now suppose instead that you determine by illegal wiretapping that the maximum number in the winning sequence is some fixed number R (note that R must be 6 or larger. If you use this information wisely in choosing your 6 numbers, how does your probability of winning the grand prize improve? (d Use a counting argument to establish the identity Solution: ( n k n rk ( r. k (a Among the 49 numbers used at the lottery, only 6 correspond to the winning sequence. If the number of matches between our set and the winning set is k, then we must have selected (without replacement and without ordering exactly k elements from the winning set of size 6 and 6 k elements from the remaining set of 49 6 available numbers. Then, we have that ( k( P(X k 6 k ( 49 6 so the probability of winning the grand prize is ( 6 P(X 6 ( 6 49 (

13 (b In order to use wisely the information given to us, we select (without replacement and without ordering two numbers from the set {,...,}, two numbers from the set {,...,4}, and two numbers from the set {4,...,49}. Among the ( ( ( 9 ways of selecting the six numbers, there is only one that corresponds to the right sequence. So, the probability of winning is ( 9 ( ( Which corresponds to an improvement of (roughly a factor with respect to case when no information is available. (c Since we know that the maximum number in the winning sequence is some number r, we select r in our sequence of numbers. Next, we use the information that r is the maximum number, and we select the remaining 5 numbers in the set {,...,r }. The probability of winning the grand prize becomes ( r 5 (d We can imagine to number in increasing order the n elements from which we are sampling k elements. Let us denote with r the maximum number that we select. For a fixed r we can use the result in part c for determining in how many ways we can select the remaining r elements: ( r k. Summing over all the possible values of r, we have the result. 3