ISyE 6761 Fall 2012 Homework #2 Solutions

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1 1 1. The joint p.m.f. of X and Y is (a) Find E[X Y ] for 1, 2, 3. (b) Find E[E[X Y ]]. (c) Are X and Y independent? ISE 6761 Fall 212 Homework #2 Solutions f(x, ) x 1 x 2 x 3 1 1/9 1/3 1/9 2 1/9 1/18 3 1/6 1/9 Solution: (a) B definition of conditional expectation, E[X Y ] x xpr(x x Y ) x xf(x, ). f Y () Now, f Y (1) x f(x, 1) 5/9. Similarl, f Y (2) 1/6 and f Y (3) 5/18. Plug these results into the first equation to get E[X Y 1] 2, E[X Y 2] 5/3, and E[X Y 3] 12/5. (b) E[E[X Y ]] E[X ]f Y () 37/18. Check: Note that f X (1) f(1, ) 2/9. Similarl, f X (2) 1/2 and f X (3) 5/18. Thus, E[E[X Y ]] E[X] x xf X (x) 37/18, which matches the answer above! (c) Adding in the p.m.f. s, the original joint table becomes

2 2 f(x, ) x 1 x 2 x 3 f Y () 1 1/9 1/3 1/9 5/9 2 1/9 1/18 1/6 3 1/6 1/9 5/18 f X (x) 2/9 1/2 5/18 1 Notice, e.g., that f(1, 3) f X (1)f Y (3). Thus, X and Y are not independent. 2. Show in the discrete case that if X and Y are independent, then E[X Y ] E[X] for all. Solution: Since X and Y are independent, we have E[X Y ] xpr(x x Y ) xf X (x) E[X]. x x 3. Suppose the joint p.d.f. of X and Y is Find E[X Y ]. Solution: First of all, Now, f Y () f(x, ) e x/ e, for x and >. f(x, ) dx e E[X Y ] R xf(x ) dx 1 f Y () e x/ dx e. xf(x, ) dx e xe x/ e dx xe x/, since the integral is the expected value of an Exp(1/) p.d.f. dx

3 3 4. Suppose X Exp(λ). Find E[X X > x] for x >. Solution: I ll use as a dumm variable. The conditional p.d.f. of X given that X > x is This implies that f X X>x () d Pr(X X > x) d for d d Pr(x < X ) Pr(X > x) d (F () F (x)) d Pr(X > x) f() Pr(X > x) λe λ e λx. x E[X X > x] x f X X>x () d (Could ou have gotten the same thing via inspection?) x λe λx e λ d x + 1 λ. 5. Suppose X Unif(, 1). Find E[X X < x] for < x < 1. Solution: Again, I ll use as a dumm variable. Similar to the above work, we get f() f X X<x () Pr(X < x) 1, for < < x. x This implies that E[X X < x] x f X X<x () d x (Could ou have gotten the same thing via inspection?) d x/2. x 6. Suppose the joint p.d.f. of X and Y is f(x, ) e, for < x < <.

4 4 Find E[X 2 Y ]. Solution: First of all, f Y () f(x, ) dx e dx e. Then Thus, f(x ) E[X 2 Y ] f(x, ) f Y () 1, for < x < <. x 2 f(x ) dx x 2 dx Suppose Y Gamma(α, λ). Also suppose that the conditional distribution of X given that Y is Pois(). Find the conditional distribution of Y given that X x. Solution: We have that the required conditional p.d.f. is f Y X ( x) f(x, ) f X (x) f X Y (x )f Y () f X (x) 1 f X (x) e x Ce λ α 1 x! Ke x e λ α 1 Ke (1+λ) α+x 1, where C and K don t depend on. This is simpl the Gamma(α + x, 1 + λ) p.d.f. 8. A miner is trapped in a mine with three doors. The first door leads to a tunnel that takes him to safet after two hours. The second leads back to the mine after three hours. The third leads back to the mine after five hours. (a) Assuming that he s equall likel at an time to choose an one of the three doors, what s the expected length of time X until he reaches safet?

5 5 (b) Let N denote the total number of doors (attempts) before he reaches safet. Let T i be the travel time corresponding to the ith choice i 1. Give an identit that relates X to N and the T i s. (c) What is E[N]? (d) What is E[T N ]? (e) What is E[ N i1 T i N n]? (f) Using the above, what is E[X]? Solution: (a) Let Y denote the door he initiall chooses. Then E[X] E[E[X Y ]] 3 E[X Y ]Pr(Y ) 1 3 E[X Y ] Meanwhile, E[X Y 1] 2, E[X Y 2] 3 + E[X], and E[X Y 2] 5 + E[X]. Plugging into the above, we have leading to E[X] 1. E[X] 1 + 2E[X], 3 (b) X N i1 T i. (c) N Geom(1/3). Thus, E[N] 3. (d) T N is the travel time corresponding to the final choice, i.e., the choice leading to freedom. Therefore, T N 2, and so E[T N ] 2. (e) Since N n, this means that the travel times T 1, T 2,..., T n 1 are each equall likel to be 3 or 5, while T n 2. Therefore, [ N ] E T i N n i1 [ n 1 ] E T i N n + E[T n N n] i1 (3 + 5) (n 1) n 2.

6 6 (f) B (b), (c), and (e), we have [ ( N )] E[X] E E T i N i1 E[4N 2] 4E[N] A coin having probabilit p of heads is flipped until the two most-recent flips are heads, Let N be the number of flips. Find E[N]. Solution: E[N] E[N H]p + E[N T ]q E[N HH]p 2 + E[N HT ]pq + E[N T ]q 2p 2 + (E[N] + 2)pq + (E[N] + 1)q where q 1 p. After a little algebra, we get E[N] 2p2 + 2pq + q 1 pq q 2 q p You have two opponents with whom ou alternate pla. Whenever ou pla A, ou win with probabilit p A ; whenever ou pla B, ou win with probabilit p B, where p B > p A. If our objective is to minimize the expected number of games ou need to pla to win two in a row, should ou start with A or B? Solution: Let N A [N B ] denote the number of games needed if ou start with A [B]. We ll condition on the outcome of the first game, where w denotes a win and l denotes a loss. E[N A ] E[N A w]p A + E[N A l](1 p A ). We can also condition on the outcome of the next game, E[N A w] E[N A ww]p B + E[N A wl](1 p B ) 2p B + (2 + E[N A ])(1 p B ) 2 + (1 p B )E[N A ]. Furthermore, E[N A l] 1 + E[N B ].

7 7 Plugging these results into the first equation gives E[N A ] ( ) 2 + (1 p B )E[N A ] p A + (1 + E[N B ])(1 p A ). B smmetr, E[N B ] ( ) 2 + (1 p A )E[N B ] p B + (1 + E[N A ])(1 p B ). Subtract, and plow thru some algebra. Since p B > p A, ou ll eventuall see that E[N A ] < E[N B ] (so plaing A first is better). 11. A manuscript is sent to a tping firm that has tpists A, B, and C. If it s tped b A, then the number of errors made is Pois(2.6). If B, then the number is Pois(3). If C, then the number is Pois(3.4). Let X denote the number of errors in the tped manuscript, and assume that each tpist is equall likel to do the work. What are E[X] and Var(X)? Solution: (a) E[X] ( )/3 3. (b) Before we begin, recall that if Y Pois(λ), then E[Y 2 ] λ + λ 2. This implies that E[X 2 ] [( ) + ( ) + ( )]/ , and so Var(X) E[X 2 ] (E[X]) Suppose U Unif(, 1). Suppose that n trials are to be performed and that, conditional on U u, these trials will be independent with a common success probabilit u. Compute the mean and variance of the number of successes that occur in these trials. Solution: Suppose X is the number of successes. Then (X U u) Bin(n, u). So E[X] E[E(X U)] E[nU] n/2.

8 8 Similarl, since Y Bin(n, p) implies E[Y 2 ] np(1 p) + n 2 p 2, we have E[X 2 ] E[E(X 2 U)] E[nU(1 U) + n 2 U 2 ] E[nU + (n 2 n)u 2 ] ) ne[u] + (n 2 n) (Var(U) + (E[U]) 2 n ( (n2 n) ) n 6 + n2 3. This implies that Var(X) E[X 2 ] (E[X]) 2 n 6 + n The number of customers entering a store toda is Pois(1). the amount of mone spent b a customer is Unif(, 1). Find the mean and variance of the amount of mone that the store takes in on a given da. Solution: Let N Pois(1) denote the number of customers, and let X i Unif(, 1) denote the $ spent b person i. Assume that the X i s and N are independent. Then we have and [ N ] E X i i1 ( N ) Var X i i1 E[N]E[X i ] E[N]Var(X i ) + (E[X i ]) 2 Var(N) 1(1 2 /12) + (5) If E[Y X] 1, show that Var(XY ) Var(X). Solution: First of all, let s write out the following general result. For an reasonable functions h(x) and g(y ), we have [ ] E[h(X)g(Y )] E E[h(X)g(Y ) X]

9 9 Thus, we have E[h(X)g(Y ) X x]f X (x) dx h(x)e[g(y ) X x]f X (x) dx [ ] E h(x)e[g(y ) X]. E[XY ] E[XE[Y X]] E[X], since, for the current problem, E[Y X] 1. Similarl, since E[Z 2 ] (E[Z]) 2 for an random variable Z (even conditional ones), we have E[(XY ) 2 ] E[X 2 E[Y 2 X]] E[X 2 (E[Y X]) 2 ] E[X 2 ]. Putting this all together, we get the desired result. 15. A and B pla a series of games with A winning each game with probabilit p. The overall winner is the first plaer to have won two more games than the other. (a) Find the probabilit that A is the overall winner. (b) Find the expected number of games plaed. Solution: Let A be the event that A is the winner, and let X denote the number of games plaed. Further, let Y be the number of times A wins in the first two games. (a) Then Pr(A) This implies that (b) Here we have 2 Pr(A Y i)pr(y i) + Pr(A)2p(1 p) + p 2. i Pr(A) p 2 1 2p(1 p). E[X] 2 E[X Y i]pr(y i) i 2(1 p) 2 + (2 + E[X])2p(1 p) + 2p E[X]2p(1 p).

10 1 This implies that E[X] 2 1 2p(1 p). 16. Suppose X is Pois(λ). The parameter λ is itself Exp(1). Find the p.m.f. of X. Solution: X Pois(Λ), where Λ Exp(1). Then b total probabilit, Pr(X n) after a little calculus. Pr(X n Λ λ)f Λ (λ) dλ e λ λ n e λ dλ 1/2 n+1 n! 17. A coin is randoml selected from a group of 1 coins. The ith coin has probabilit i/1 of coming up heads. The coin is then flipped until a head appears. Let N be the number of flips necessar. What s the p.m.f. of N? Is N geometric? (No!) What would ou have to do in order to make N geometric? Solution: B total probabilit, Pr(N k) So N is not geometric. 1 n1 1 n1 Pr(N k coin n selected)pr(coin n selected) ( 1 n 1 ) k 1 n You could make N geometric if the coin were re-selected after each flip. probabilit of success has to be constant from flip to flip.) (The 18. The number of storms this ear is Poisson, but with a parameter that is itself Unif(,5). Find the probabilit that there are at least three storms this ear. Solution: Suppose X is the number of storms. Then P (X 3) 1 P (X 2) 1 5 P (X 2 Λ x) 1 5 dx

11 e 5 1 [e x + xe x + x2 e x ] dx The Continuous Mapping Theorem (CMT) sas that if Z n, n 1, 2,..., and Z are random variables such that Z D n Z and h( ) is a continuous function, then h(z n ) D h(z). If the Z i s are i.i.d. with mean µ and variance σ 2, use the CMT and WLLN to show that S 2 P σ 2, where S 2 is the sample variance of Z 1, Z 2,..., Z n. 2. In addition to the same conditions as in the previous problem, assume that h( ) is twice continuousl differentiable at µ. Show that ( n E[h( Z ) n )] h(µ) h (µ)σ 2, 2 where Z n is the sample mean of the Z i s.