Y = Alice's spending X < Y X + Y = 150. X = Bob's spending

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1 Department of EECS- University of California at Berkeley EECS16 - Probability and Random Processes - Spring Final Examination: 5/15/ SOLUTIONS Problem 1: Bob and Alice go Christmas shopping and agree to spend independent amounts uniformly distributed between and 1. What it the probability that Bob spent less than 5 given that together they spent less than 15 and that Bob spent less than Alice? Y = Alice's spending 1 X < Y ;;A B X + Y = 15 X < 5 1 X = Bob's spending The figure shows the sample space and the various possible inequalities. The requested conditional probability isthe ratio of the area of A over the sum of the areas of A and B. The areas of A and B are area(a) = 1 (1 + 5) 5 = 375 and area(b) = = 65: Hence, the requested probability is equal to ß :86. Problem : Let fx n ;n g be independent standard Gaussian random variables. Define Y n+1 = ay n + bx n ;n where Y is N(;ff ) and independent of fx n ;n g and a ( 1; 1);b6= are known numbers. a. Find ff so that the probability distribution function of Y n does not depend on n. b. Calculate E[Y n jy ]. a. Taking the square of both sides of the equation that defines Y n, we find Y n+1 = a Y n + b X n +aby n X n : Taking expectations and using stationarity offy n g and the independence of Y n and X n, we find that ff = a ff + b ; so that ff = b 1 a :

2 b. We have so that Y n = a n Y + n 1 X m= E[Y n jy ]=a n Y : a n 1 m bx m ; Problem 3: A power line is made of N parallel lines. Each line is down with probability p (; 1), independently of the others. When alineisup,it can carry a current of 1 Ampere. When one line is down, the whole power line is shut down for repair. Find the value of N that maximizes the average power that the line carries. The power that the line carries is N Amperes with probability (1 p) N. Otherwise, the line does not carry any current. Consequently, the expected power is f(n) =N(1 p) N : It is now a matter of optimizing over N. Let N Λ be the best value of N. Then, f(n Λ 1)» f(n Λ ) and f(n Λ +1)» f(n Λ ). The first inequality gives (with q := 1 p to simplify the notation) The second inequality gives These inequalities show that N Λ ß 1=p. (N Λ 1)q N Λ 1» N Λ q N Λ ; or, N Λ» 1 p : (N Λ +1)q N Λ +1» N Λ q N Λ ; or, N Λ 1 p 1: Problem 4: Let X be a random variable uniformly distributed in [; ß]. Find the estimator Z = a + bx that minimizes E(sin(X) Z) ). By definition, we are looking for the LLSE of Y =sin(x) given X. We know that L[Y jx] =E(Y )+ cov(x; Y ) (X E(X)): vax(x) Now, E(Y )=by symmetry; E(X) =ß; var(x) =E(X ) (E(X)) =4ß (1=3 1=4) = ß =3; cov(x; Y ) = E(XY ) E(X)E(Y )=E(X sin(x)) = 1 ß = 1 Z ß xd cos(x) =[ 1 ß ß x cos(x)]ß + 1 ß Putting these values together, we find = 1+ 1 ß [sin(x)]ß = 1: L[Y jx] = 1 ß =3 (X ß) = 3 (ß X): ß Z ß Z ß x sin(x)dx cos(x)dx

3 Problem 5: For n 1, let X n be a Poisson random variable with mean n. Use the CLT to calculate lim n!1 e n nx k= n k k! : We can write Now, for n fl 1, Xn n p n e n n X k= ß N(; 1). Therefore, n k k! = P (X n» n) =P ( X n n p n» ): P ( X n n p n» )! 1 as N!1: Problem 6: Let X; Y; Z be independent and N(; 1). Calculate E[XjaX + Y; ax + Z]: This is all standard. Let V = ax + Y and W = ax + Z. Let also U =(V; W) T. We have E[XjU] =E(X)+± X;U ± 1 U;U(U E(U)) = ± X;U ± 1 U;U U: Now, Also, so that ± X;U = E(XU T )=E(X[aX + Y; ax + Z]) = (a; a): ax + Y ± U;U = E(UU T )=E( ax + Z ± 1 U;U = a +1 a [ax + Y; ax + Z]) = a a +1 1 a +1 a a +1 a a +1 Substituting these values in the expression for E[XjU], we find E[XjV; W] = a (V + W ): a +1 ): ); Problem 7: Let fx n ;n g be a discrete time Markov chain on the state space f1; ; 3; 4g with the following transition probability matrix: P = 6 4 :5 : a. Is this Markov chain irreducible? b. Is the Markov chain aperiodic?

4 c. Find the stationary distribution of fx n ;n g. d. Calculate E[T 4 jx = 3] where T 4 =minfn jx n =4g. a. Yes. The graph shows that the Markov chain can go from every state to every other state. b. The Markov chain is periodic with period 3. For instance, d(1) = g.c.dfn 1jP n (1; 1) > g = g.c.df3; 6; 9;:::g =3: c. We must solve the balance equations: ß(1) = ß(); ß() = ß(3) + ß(4); ß(3) = ß(1)=; ß(4) = ß(1)=: These equations show that ß(1) = ß() = ff, say, andß(3) = ß(4) = ff=. Since ß(1) + ß() + ß(3) + ß(4) = 1, one finds that ff + ff + ff=+ff= = 1, so that ff =1=3. Hence, ß(1) = ß() = 1=3 andß(3) = ß(4) = 1=6: d. We write down the first step equations for fi(i) :=E[T 4 jx = i]: fi(1) = 1+:5fi(3) + :5 fi() = 1+fi(1) fi(3) = 1+fi() fi(4) = From the third and second equations we find that fi(3)=1+fi() = + fi(1): Substituting this expression in the first equation we find that fi(1) = 1 + :5( + fi(1))=+:5fi(1): Solving for fi(1) we find Using fi(3) = + fi(1) we conclude that fi(1) = 4: fi(3) = E[T 4 jx =3]=6: Problem 8 (1pts/1) Let X; Y be independent and exponentially distributed with mean 1. Given X and Y, the random variables fz n ;n 1g are i.i.d. and uniformly distributed on [X; X + Y ]. Showthat(minfZ 1 ;:::;Z n g;maxfz 1 ;:::;Z n g) are sufficient statistics for estimating X and Y given fz 1 ;:::;Z n g. Let U = fz 1 ;:::;Z n g;v = minfz 1 ;:::;Z n g; and W = maxfz 1 ;:::;Z n g. Let also v = minfz 1 ;:::;z n g and w = maxfz 1 ;:::;z n g. We first observe that

5 for some function h. Consequently, f Uj(X;Y ) [ujx; y] = Π n 1fx» z k» x + yg k=1 y = 1 1fx» v and w» x + yg yn = h(x; y; v; w); f (X;Y )ju [x; yju] = f X;Y (x; y)f Uj(X;Y ) [ujx; y] f U (u) = f X;Y (x; y)h(x; y; v; w) r(v; w) where r(v; w) is the integral of f X;Y (x; y)h(x; y; v; w) over all possible values of (x; y). We conclude that f (X;Y )ju [x; yju] =g(x; y; v; w) which proves that (V; W) is sufficient to estimate (X; Y )given U.