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Third Assignment: Solutions 1. Since P (X(p) > n) = (1 − p)n , n = 0, 1, 2, . . ., we have, for x ≥ 0, P (X(p)/p > x) = P (X(p) > [px] + 1) = (1 − p)[px]+1 → e−x , as p → 0. The conclusion follows from the fact that e−x is continuous for x > 0. 2. (i) We only give sufficient conditions. First assume that the probability measure Q has density, with respect to the Lebesgue measure, f . Then Z 1 f (x) = e−iθ·x Φ(θ)dθ. (2π)d Rd Second, assume that f (x) is positive definite: N N X X j=1 k=1 cj ck f (xj − xk ) ≥ 0, for all N ∈ N, all c1 , . . . , cN ∈ C, and all x1 , . . . , xN ∈ Rd . Then Φ(θ) ≥ 0, by Bochner’s theorem. (ii) If R is a rotation and if X is a random variable with values in Rd and law Q, then T Φ(θ) = Eeiθ·X . Hence Φ(Rθ) = Eei(Rθ)·X = Eeiθ·(R X) = Eeiθ·X = Φ(θ). (iib) No: take, e.g., the probability measure Q on R2 with density f (x, y) = c(x2 + y 2 )e−(x 2 +y 2 ) . Then Q is invariant under rotations and has characteristic function 2 2 Φ(θ1 , θ2 ) = cπ(1 − (θ12 + θ22 )/4)e−(θ1 +θ2 )/4 , which takes both positive and negative values. 3. If there are two probability measures Q′ , Q′′ corresponding to the same characteristic function, then the alternating sequence Q′ , Q′′ , Q′ , Q′′ , . . . satisfies the assumptions of Lévy’s theorem and so it converges weakly to a probability measure. The only way for this sequence to converge is if Q′ = Q′′ . √ 4. (i) Since (X1 +· · ·+Xn )/ n is a linear combination of (X1 , . . . , Xn ) [which has normal √ law on Rn ], it follows immediately that Z := (X1 + · · · + Xn )/ n has normal law on R1 . That it is standard is clear from the fact that E(Z) = 0, E(Z 2 ) = (1/n) × n = 1. (ii) We have Z ∞ 1/π eiθx Φ(θ) = dx = e−|θ| , 1 + x2 −∞ as can be easily established by a contour integral. Therefore, E[eiθXj ] = e−|θ| , for all j. Let W := (X1 + · · · + Xn )/n. Then E[eiθW ] = E n Y eiθXj /n = n Y j=1 j=1 1 E[ei(θ/n)Xj ], by independence, and so W has characteristic function E[eiθW ] = e−|θ/n| )n = e−|θ| , which is identical to the characteristic function of X1 . Hence W and X1 have the same law. (iii) Indeed, the function α Φ(θ) := e−|θ| is positive semidefinite and hence a characteristic function of some probability measure Qα . The stated property follows exactly in the same manner as above [part (i) corresponds to α = 2, and part (ii) to α = 1]. If α > 2, the function above is not positive semidefinite. 5. (i) As in the hint, let U be uniform in (−1, 1). Then, for all n, n X ξj U + n. U = j 2 2 (d) j=1 P ξj The right-hand side converges weekly to ∞ j=1 2j . Hence the later has the law of U . (ii) The left-hand side is the characteristic function of U . The characteristic function P n ξj P ξj of ∞ j=1 2j is the limit of the characteristic function of j=1 2j , as n → ∞. The Qn Q Pn ξ j j characteristic function of j=1 2j is j=1 cos(x/2 ). Hence limn→∞ nj=1 cos(x/2j ) exists and is equal to (sin x)/x. (iii) Set x = π/2. Then (sin x)/x = 2/π. From the formula cos(2x) = 1 + 2 cos2 (x), it follows that q p √ 2 + 2 + ··· + 2 j+1 cos(π/2 ) = , 2 where the square root symbol appears j times. Hence 2 = π ∞ Y j=1 q p √ 2 + 2 + ··· + 2 2 (j times) . 6. If X(n), n = 1, 2, . . . is a tight family of random variables in Rd then it is obvious that, for each j, Xj (n), n = 1, 2, . . . is a tight family of random variables in R. For the converse, fix ε > 0 and pick Mj > 0 such that P (|Xj (n)| > Mj ) ≤ ε/d, for Pd all n. Let M := max(M1 , . . . , Md ). Then P (max(|X1 (n)|, . . . , |X(n)|) > M ) ≤ j=1 P (|Xj (n)| > M ) ≤ ε, for all n. Hence, for all n, the probability that X(n) belongs to the compact set K := [−1, 1]d is at least 1 − ε, for all n, and hence the sequence if tight. 7. Let f : Rd → R be continuous and bounded. Since g is continuous, it follows that (d) f ◦g : S → R is continuous and bounded. Therefore, Qn ◦(f ◦g)−1 −−→ Q◦(f ◦g)−1 . But Qn ◦(f ◦g)−1 = (Qn ◦g −1 )◦f −1 , and Q◦(f ◦g)−1 = (Q◦g −1 )◦f −1 . 2 8. (i) If ϕ ∈ C[0, 1] then ϕ(g(ϕ)) = 0. Let τ := g(ϕ). If τ < 1 and there is ε > 0 such that ϕ > 0 on (τ − ε, τ ) and ϕ < 0 on (τ, τ + ε) then g is continuous at ϕ. To see this, let ϕn be a sequence of continuous functions such that ϕn → ϕ, uniformly. Then ϕn will eventually be positive on (τ − ε, τ ) and negative on (τ, τ + ε), implying that all the zeros of ϕn on (τ − ε, τ + ε) converge to τ . So, necessarily, any discontinuity point ϕ of g must have the property that it has constant sign (or zero) on a small neighborhood of its last zero τ . Intuitively, a Brownian motion cannot do this, almost surely. The event that a Brownian motion B touches 0 at some point t0 but is positive (respectively, negative) on a neighborhood of t0 implies that the infimum (respectively, supremum) of the Brownian motion B on the same interval equals 0. By time scaling and space translation, it suffices to prove that P (sup0≤t≤1 B(t) = x) = 0 for all x. This follows because sup0≤t≤1 B(t) is a random variable with continuous distribution function. (ii) Since Sn = 0 if and only if n is even, it suffices to compute P (T2m = 2k). We have P (T2m = 2k) = P (S2k = 0, S2k+1 6= 0, . . . , S2m 6= 0) = P (S2k = 0)P (S2k+1 6= 0, . . . , S2m 6= 0 | S2k = 0) = P (S2k = 0)P (S1 6= 0, . . . , S2m−2k 6= 0) = P (S2k = 0)P (S2m−2k = 0). 2k −2k 2m − 2k −(2m−2k) 2 2 = m−k k 2k 2m − 2k −2m = 2 . k m−k (Alternatively, you can count the number of paths of length 2m, starting from 0 and which hit 0 for the last time at time 2k, and find that this is equal to the product of the binomial coefficients above.) √ (iii) Let Zn (t) := B(nt)/ n if t ∈ (1/n)Z, and let Zn (t) to be obtained by linear interpolation for other values of t. We know that the random sequence Zn converges weakly to B. Since g is continuous at B, almost surely, we have that g(Zn ) converges weakly to T = g(B) and so P (g(Zn ) ≤ x) → P (g(B) ≤ x), for all x such that P (g(B) = x) = 0. But g(Zn ) = sup{t ≤ 1 : S[nt] = 1} = 1 Tn . n Therefore P (Tn ≤ nx) → P (T ≤ x), (*) for all x such that P (T = x) = 0. From the formula for P (T2m = 2k) and Stirling’s approximation we have 2/π =: f (x), 2mP (T2m = 2km ) → p x(1 − x) if km /m → x, In particular, 2mP (T2m = 2[mx]) → f (x), 3 as m → ∞. as m → ∞. By the dominated convergence theorem, Z x Z 2mP (T2m = 2[mt])dx → 0 x f (t)dt, 0 as m → ∞. But the left-hand side equals P (T2m ≤ 2mx) (here, we use the fact that T2m takes only even values). Hence Z x P (T2m ≤ 2mx) → f (t)dt. (**) 0 Comparing (*) and (**), we have Z Z x √ dt 2 x 2 p f (t)dt = P (T ≤ x) = = arcsin x. π 0 π t(1 − t) 0 (Moreover, since P (T = x) = 0, for all x, convergence (*) holds for all x.) 4