Good luck! Problem 1. (b) The random variables X 1 and X 2 are independent and N(0, 1)-distributed. Show that the random variables X 1

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1 Avd. Matematisk statistik TETAME I SF2940 SAOLIKHETSTEORI/EXAM I SF2940 PROBABILITY THE- ORY, WEDESDAY OCTOBER 26, 2016, Examinator : Boualem Djehiche, tel , boualem@kth.se Tillåtna hjälpmedel/permitted means of assistance: Appendix 2 in A. Gut: An Intermediate Course in Probability, Formulas for probability theory SF2940, L. Råde & B. Westergren: Mathematics Handbook for Science and Engineering and pocket calculator. All used notation must be explained and defined. Reasoning and the calculations must be so detailed that they are easy to follow. Each problem yields max 10 p. You may apply results stated in a part of an exam question to another part of the exam question even if you have not solved the first part. 25 points will guarantee a passing result. If you have received 5 bonus points from the home assignments, you may skip Problem 1(a. If you have received 10 bonus points, you may skip the whole Problem 1. Solutions to the exam questions will be available at starting from Friday 28 October Good luck! Problem 1 (a The random variables X 1, X 2,..., are non-negative integer-valued and independent and identically distributed (i.i.d.. The random variable Po(a, a > 0, is independent of X 1, X 2,...,. Set S = X 1 + X X. If we know that S Po(b, where 0 < b < a, show that for each k = 1, 2, 3,..., X k Be( b i.e. Bernoulli-distributed with parameter b/a. (5 p a (b The random variables X 1 and X 2 are independent and (0, 1-distributed. Show that the random variables X 1 X 2 and X1 2 + X2 2 are independent and determine their distributions. (5 p (Hint: Use polar coordinates. Problem 2 X 1, X 2,... is a sequence of independent and identically distributed random variables with mean zero and variance σ 2. The random variable is independent of the sequence (X n n 1 and Po (, > 0. Show that

2 cont. exam sf2940 Fall (a P 1, when. (2 p (b d (0, 1, when. (2 p (c X 1 + X X d (0, σ 2,. (6 p Let X and Y be random variables such that Problem 3 Y X = x (x, a 2 with X (µ, b 2. ( X (a Determine the distribution of the vector. (5 p Y (b Determine the distribution of X Y = y. (5 p Problem 4 Let X and Y be two random variables with finite second moment (square-integrable defined on the probability space (Ω, F, P. Let G be a sub-σ-algebra of F. (a Prove (or find a counter-example that if X and Y are equally distributed (have the same probability distribution then E[X G] = E[Y G] almost surely. (3 p (b Prove that if E[X Y ] = Y and E[Y X] = X, then X = Y almost surely. (3 p (c Let X 1, X 2,..., X n be identically distributed and positive random variables. Show that for any k n [ ] X1 + X X k E = k X 1 + X X n n. (4 p Problem 5 On the space of random variables on (Ω, F, P we set [ ] X Y d(x, Y := E. 1 + X Y

3 cont. exam sf2940 Fall (a Show that d is a metric, that is d(x, X = 0, d(x, Y = d(y, X and d(x, Y d(x, Z + d(z, Y. Let X 1, X 2,..., be a sequence of random variables on the probability space (Ω, F, P. (2 p (b Show that if d(x n, X 0 as n, then X n converges to X in probability as n. (4 p (c Let X 1, X 2,..., be random variables such that P (X n = 1 = 1 1 n, P (X 1/5 n = n = 1, n 2. n1/5 Determine the limit X such that d(x n, X 0 and show that X n does not converge to X in L 1 i.e. E[ X n X ] 0 as n. (4 p (Hint: the function f(t := t, t 0, may be useful. 1+t

4 Avd. Matematisk statistik Suggested solutions to the exam Wednesday October 26, Problem 1 (a Let ϕ X1 (t denote the characteristic function of X 1 and g (s = exp ((s 1 be the probability generating function of Po(a, a > 0. Using the composition formula with characteristic function (5.18 in the Lecture otes, we have ϕ S (t = g (ϕ X1 (t = exp (a(ϕ X1 (t 1. ow, if S Po(b, we should have ϕ S (t = exp (b(e it 1. Hence, which implies that (note that 0 < b/a < 1 exp (a(ϕ X1 (t 1 = exp ( b(e it 1, ϕ X1 (t = b a eit + (1 b a, that is X 1, X 2,... are all Bernoulli Be( b a -distributed. (b Introduce the transformation (X 1, X 2 := ϕ(r, Θ, R > 0, Θ (0, 2π, defined by { X1 = R cos Θ, X 2 = R sin Θ. Then, R = X1 2 + X2 2 and cotanθ = X 2 X 1. Furthermore, the corresponding Jacobian is x x r θ J ϕ := = cos θ r sin θ sin θ r sin θ = r. We have y r y θ f (R,Θ (r, θ = f (X1,X 2 (r cos θ, r sin θ J ϕ = 1 r 2 2π re 2 = fθ (θf R (r, r > 0, θ (0, 2π where, f Θ (θ = 1, θ (0, 2π which is the density of the uniform distribution over (0, 2π 2π and f R (r = re r2 2, r > 0 which is the density of the Rayleigh R(2 distribution. Θ U(0, 2π and R R(2 and are independent. Problem 2 The random variable Po( has E[] = Var( = and its characteristic function is ϕ (t = exp ((e it 1.

5 cont. exam sf2940 Fall (a This question can be solved either using Chebychev s inequality or the Cramér-Slutzky s theorem. Using Chebychev s inequality we have, for every fixed ɛ > 0, ( P 1 > ɛ = P ( > ɛ 1 ɛ 2 Var( = 2 ɛ 2 = 1 2 ɛ 2 0, as. Using the Cramér-Slutzky s theorem, then, when, We have P 1 if and only if ( t ϕ (t = ϕ d 1 if and only if ϕ = exp ((e i t 1. (t e it. But, by Lemma (in the course lecture notes we have, for every real x, where In particular, when x 0, we write e ix = n (ix k + R n (x, (1 k! k=0 ( x n+1 R n (x min (n + 1!, 2 x n. (2 n! e ix = n (ix k + o(x n, (3 k! k=0 where o(x n denotes the rest term R n (x. Hence, when, we obtain (taking n=2 e i t it = 1 + t2 t2 + o(. ( lim ϕ (t = lim exp (it t2 2 + o(t2 = e it. (b We have to show that lim ϕ (t = e t 2 2. Indeed, ( t ϕ (t = e it ϕ = exp ((e i t 1 it. We may use (3 to obtain (e i t t 2 1 = it 2 + o(t2,. oting that using (2, o( t2 0 as, we finally obtain lim ϕ (t = e t 2 2.

6 cont. exam sf2940 Fall (c Set S = X 1 + X X and let ϕ X (t denote the characteristic function of the i.i.d. variables X 1, X 2,.... We have Since S S = S /. P 1 when, by the Cramér-Slutzky s theorem it suffices to show that d (0, σ 2 as i.e. we have to show that lim ϕ S (t = e t 2 σ 2 2. Since Po(, we have ϕ S (t = ϕ S ( t ( = g ϕ X ( t ( ( = exp ϕ X ( t 1. ow, since X has finite variance σ 2, we use the fact that, when, ϕ X ( t = 1 + it E[X] t2 E[X2 ] + o( t2 and noting that E[X] = 0 and E[X 2 ] = Var(X + (E[X] 2 = σ 2, we obtain ( lim ϕ X ( t 1 = t2 2 σ2. or lim ϕ S (t = e t 2 σ 2 2. Problem 3 ( X (a We compute the characteristic function of Y. We have ϕ (X,Y (s, t = E [ e isx+ity ] = E [ E [ e isx+ity X ]] = E [ e isx E [ e ity X ]]. Since Y X (X, a 2 we have E [ e ity X ] = e itx t2 2 a2. [ ϕ (X,Y (s, t = E e isx e itx t2 a2] 2 = e t2 2 a2 E [ e i(s+tx]. Using the assumption that X (µ, b 2 we further obtain ( ϕ (X,Y (s, t = exp isµ + itµ 1 2 (s2 b 2 + 2stb 2 + (a 2 + b 2 t 2. ( X Y (( µ µ ( b 2 b, 2 b 2 a 2 + b 2.

7 cont. exam sf2940 Fall (b We have X Y = y (µ 1 2, σ1 2 2, where µ 1 2 = µ + b2 a 2 + b (y µ, 2 σ2 1 2 = b 2 (1 b4 a 2 + b = 2 b2 b2. a 2 + b 2 ote that σ > 0. Problem 4 (a Recall the definition of the conditional expectation of an integrable random variable X w.r.t. a sub σ-algebra G. Z := E[X G] is the unique G-measurable random variable for which E[Zξ] = E[Xξ], for all bounded G measurable random variables ξ. But, if X and Y have the same distribution then it is not always true that In particular, it is not always true that E[Xξ] = E[Y ξ]. E[E[X G]ξ] = E[Xξ] = E[Y ξ] = E[E[Y G]ξ], for all bounded G-measurable random variables ξ. Here is a counter-example. We know that if X U[0, 1] then Y := 1 X U[0, 1]. Let ζ be a bounded random variable such that E[ζE[X G]] 0, but E[ζ] = 0 and G = σ(ζ i.e. the σ-algebra generated by ζ. If E[Xζ] = E[Y ζ] = E[ζ] E[Xζ]. Then 0 = E[ζ] = 2E[Xζ] = 2E[ζE[X G]] 0, leading to a contradiction. (In a previous version, the suggested solution was wrong. (b To show that X = Y almost surely it suffices to show that E[(X Y 2 ] = 0 because both are square-integrable. We have E[(X Y 2 ] = E[X 2 2XY + Y 2 ] = E[X 2 ] + E[Y 2 ] E[XY ] E[XY ]. But, using the assumption that X = E[Y X] we obtain E[XY ] = E[E[XY X]] = E[XE[Y X]] = E[X 2 ], and using that Y = E[X Y ] we also obtain E[XY ] = E[E[XY Y ]] = E[Y E[X Y ]] = E[Y 2 ]. E[(X Y 2 ] = E[X 2 ] + E[Y 2 ] E[XY ] E[XY ] = E[X 2 ] + E[Y 2 ] E[X 2 ] E[Y 2 ] = 0. (c Set, for k n, S k = X 1 + X X k. Since X 1, X 2,..., X n are identically distributed it is easy to see that E[X 1 S n ] = E[X 2 S n ] =... = E[X n S n ] almost surely.

8 cont. exam sf2940 Fall But, since S n = E[S n S n ] = n j=1 E[X j S n ] = ne[x 1 S n ], it follows that This in turns yields that E[X j S n ] = S n n almost surely, j = 1,..., n. E[S k S n ] = k E[X j S n ] = k n S n j=1 almost surely. E [ Sk S n ] = E [ [ ]] Sk E S n = E S n [ ] E [Sk S n ] S n = k [ ] n E Sn = k S n n. Problem 5 The following properties of the function t f(t = the problem. (i 0 f(t < 1 and is strictly increasing since f (t = 1 (1+t 2 > 0. t, t 0 turn out useful for solving 1+t (ii f is invertible because it is continuous (in fact differentiable and strictly increasing. (iii f(t + s f(t + f(s. indeed, noting that 1 + t + s 1 + t, 1 + s, we have f(t + s = t + s 1 + t + s = t 1 + t + s + s 1 + t + s t 1 + t + s 1 + s = f(t + f(s. (iii From (i and (ii if follows that if a, b and c are non-negative and satisfy a b + c then f(a f(b + f(c. [ ] (a We have d(x, X = E X X = 0. Since X Y = Y X we get d(x, Y = d(y, X. 1+ X X Lastly, by the triangle inequality, it holds that X Z X Y + Y Z. We may apply (iii to obtain that Take expectation to get X Z 1 + X Z X Y Y Z X Y 1 + Y Z. d(x, Y d(x, Z + d(z, Y. (b If d(x n, X 0 as n we want to show that for every fixed ɛ > 0, P ( X n X > ɛ 0 as n. Indeed, using the fact that f is a strictly increasing function and invertible, we have P ( X n X > ɛ = P (f( X n X > f(ɛ 1 f(ɛ E[f( X n X ] = 1 f(ɛ d(x n, X 0, as n, noting that since ɛ > 0, f(ɛ > 0. (c We first show that d(x n, 1 0 as n. Indeed, we have [ ] Xn 1 d(x n, 1 = E = n X n n 1 n = n 1 1 0, 1/5 n n1/5

9 cont. exam sf2940 Fall as n. Hence, X = 1. On the other hand Thus, X n 1 in L 1. E [ X n 1 ] = (n 1 1 +, n. n1/5