Probability A exam solutions

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1 Probability A exam solutions David Rossell i Ribera th January 005 I may have committed a number of errors in writing these solutions, but they should be ne for the most part. Use them at your own risk! Probability January 00 Problem a) Usual denitions n b) X n = wp/n 0wp /n converges a.s. to zero by Borel-Cantelli and hence also in probability, but not in L since E(X n 0) = E(X n ) =. wp/n X n = 0wp /n with X n independent converges in L and thus in probability to zero, but not to zero by Borel-Cantelli. wp/n X =, Y = X. Obviously Y and X have the same 0wp /n distribution but P ( X Y > 0) =, so we don't have convergence in probability. Problem a) (i) P (A B) 0) (ii) A...A n disjoint,

2 Q ( i= A i ) = P (( i= A i ) B) =P ( i= (A i B)) = disjoint} = i= P (A i B) = i= Q(A i ) For Q to be a probability measure, we need Q(Ω) = P (Ω B) = P (B) =. b) Start with indicators (i) X = I A. Then E(X; B) = B I AdP = P (A B) = Q(A) = Ω XdQ (ii) X = n i= a k I Ak. Then Ω XdQ = ni= Ω a k I Ak dq = n i= a k Ω I A k dq = ak P (A k B)= ni= B a k I Ak dp = E(X; B). (iii) X 0. Let 0 X n X with X n simple and Ω XdQ =...= B XdP (iv) General X. Proceed as usual. Problem a) It means that (i) Y n is σ(x...x n ) measurable and (ii) E(Y n+ σ(x...x n )) = Y n. b) ) Y n is a function of X...X n and thus σ(x...x n ) measurable, and E(Y n+ σ(x...x n )) = independent} =E(Y n+ ) = 0 Y n, so it's not a martingale. ) Y n is a function of X...X n and thus σ(x...x n ) measurable, and E(Y n+ σ(x...x n )) = nk= X k + E(X n+ ) = n k= X k = Y n, so it's a martingale. ) Y n is NOT a function of X...X n and thus NOT σ(x...x n ) measurable, so it's not a martingale. 4) Y n is a function of X...X n and thus σ(x...x n ) measurable and E(Y n+ σ(x...x n )) = ni= X i E(X n+ ) = 0 Y n, so not a martingale. 5) Y n is a function of X...X n and thus σ(x...x n ) measurable and E(Y n+ σ(x...x n )) = ni= X i E( X n+ ), and E( X n+ ) = +, so it's not a martingale. Problem 4 By the Strong Law of Large Numbers, X i /n converges to / almost surely. For logg n = n logxi, since E(logX) = (can be found integrating the pdf as usual) by SLLN again we have that logg n converges almost surely to

3 . Hence by the Continuous Mapping Principle G n converges almost surely to e. Now since E(/X) =, by SLLN n n X i a.s. and by CMP /Xi a.s. 0. (note that SLLN also applies when the expectation is. Problem 5 ) The chain is irreducible since all states communicate and the period of all states is since p > 0 and periodicity is a class property. To nd E(τ 0 ) we'll nd the stationary distribution rst. π = πp gives π 0 = /4, π = /4, π = / and thus E(τ 0 ) = /π 0 = 4. ) a)this is a branching process, so E(Z n ) = (p) n. b) We know that π = i p so p /. Hence for p > / we got π <, and π is the smallest solution to π = G(π), where G is the pgf of a binomial(,p). Solve for s ing(s) = ( p) + p( p)s + p s = s to get π = ( p p ). c) By independence, it's π N. Problem 6 b) For X continuous we know F (X) is Unif(0, ) and hence E(X) = /. For X categorical consider Y = F x (X), and let Z be any continuous random variable such that it's cdf matches the cdf of X in each of its jumps, i.e. F z (z) = P (Z z) = P (X z) = F x (z) for all z discontinuity point. Note that F z (z) F x (z), with strict equality in the discontinuity points (if this is confusing, plotting both cdfs may help to see it clearer). That gives us that E(F X (z)) E(F Z (z)) = /, the latter equality due to Z being a continuous random variable. c) I don't know how to do it.

4 Probability January 004 Problem a) Usual denition b) (i) Let x (0, ). P (m n x) = P (m n > x) = (P (X > x)) n = ( x) n, i.e. m n D 0 (ii) P (nm n x) = (P (X > x/n)) n = ( x/n) n e x, which is the cdf of an exp(). c) Theorem. P ( X n X > ɛ) < X n a.s. X Since P ( m n 0 > ɛ) = P (m n > ɛ) = ( ɛ) n <, we have m n a.s. 0. In general nm n doesn't converge almost surely to X, although by Skorohod's theorem we know that Y n = D nm n, Y = D X such that Y n Y. Problem a) Delta method. b) Let's nd rst the asymptotic distribution of logg n = logxi. It can n be seen that E(logX i ) = and V (logx i) =, so by the Central Limit 4 Theorem n ( logg n + ) D N(0, /4), and since g(x) = e x g (x) = e x the delta method gives that n ( G n e /) D N(0, e/4) Now, for H n let's rst nd the asymptotic distribution of Hn Since E(/X) =, V (/X) = the CLT doesn't apply to H = n n. X k. Problem a) BCI and BCII n wp/n b) (i) X n = 0 wp /n } independent. 4

5 } n wp/n (ii) X n = doesn't converge in L 0 wp /n nor a.s., but it does in probability. n wp /n (iii) X n = } n(n ) wp/n, by BCI X n a.s., but E(X n ) = 0. } n wp /n (iv) X n = n 4 wp/n. By BCI X n a.s. 0 but E(X n ). Problem 4 ) (a) 0} is transient and }, } are recurrent. (b) We can nd the stationary distribution of the Markov Chain dened by, }, P = ( ) = /. ) and π = πp gives π = / and π = /. Hence E(N X 0 = (c) Let Y be the number of transitions until we leave 0. P (Y = y) = for y =,,... i.e. Y Geom(/). Hence E(Y ) = /. ) Not necessary ( ) y Problem 5 a) State MCT and DCT b) (i) I [n,n+] (x)dx = n+ n dx = n (ii) Since sin(nx) e nx x e x x lim sin(nx) dx = 0 x e nx (iii) x/ndx = n xdx = n whic is integrable, DCT gives sin(nx) e nx x dx (iv) For n odd, x/n dx = = n, and for n even n x/n dx = = n, so the limit doesn't exist. n 5

6 (v) The function inside the integral is positive and decreasing with n, with its limit being 0 so for n (+nx ) (+x ) < +x which is integrable since the degree of the denominator is n (+x ) 4 and for the numerator it's. Hence DCT applies to give lim (+nx ) dx = 0. (+x ) n (+nx ) (+x ) n dx Problem 6 a) Y n is a function of X...X n σ(x...x n ) measurable. Some algebra shows that E(S 4 n+ σ(x...x n )) = S 4 n + 6S n +, E(S n+ σ(x...x n )) = S n +. Thus E(Y n+ σ(x...x n )) =E(S 4 n+ 6(n + )S n+ + (n + ) + (n + ) σ(x...x n ))=... = S 4 n 6nS n + n + n = Y n. Hence Y n }is a martingale. b) The optional stopping theorem gives that Y o, Y ν is a martingale so in particular E(Y ν ) = E(Y o ) = 0. Since E(Y ν ) = E(S 4 ν 6νS ν + ν + ν) =... = 5a 4 + a + E(ν ), we get that E(ν ) = 5a4 a and V (ν) = a (a ). 6