* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download APM 504 - PS7 Solutions 3.4) Suppose that X1 and X2 are
Survey
Document related concepts
Functional decomposition wikipedia , lookup
Georg Cantor's first set theory article wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Vincent's theorem wikipedia , lookup
History of the function concept wikipedia , lookup
Dirac delta function wikipedia , lookup
Nyquist–Shannon sampling theorem wikipedia , lookup
Continuous function wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Law of large numbers wikipedia , lookup
Transcript
APM 504 - PS7 Solutions 3.4) Suppose that X1 and X2 are independent normal random variables with mean 0 and variances σ12 and σ22 . Noting that the characteristic function of Xi is φi (t) = exp(−σi2 t2 /2), it follows that the characteristic function of X1 + X2 is 1 2 2 2 φ(t) = φ1 (t)φ2 (t) = e− 2 (σ1 +σ2 )t which shows that X1 + X2 is normally distributed with mean 0 and variance σ12 + σ22 . 3.8) Now suppose that X1 , X2 , · · · are independent Cauchy random variables. In this case the characteristic function of Xi is exp(−|t|) and so the characteristic function of Tn = (X1 + · · · + Xn )/n is n E etTn = e−|t/n| = e−|t| . This shows that Tn also has the Cauchy distribution. 3.11) Let X1 , X2 , · · · be independent random variables with characteristic functions φi (t) and suppose that Sn = X1 + · · · + Xn → S almost surely. Then Sn also converges to S in probability and in distribution, and so it follows from the continuity theorem that ∞ n Y Y tS tSn φk (t). E e ] = lim E e = lim φk (t) = n→∞ n→∞ k=1 k=1 3.12) Let Xn be a sequence of independent random variables such that 1 2 n ). If S = X + · · · + X , and notice that the characteristic function of X is φ (t) = cos(t/2 n n n 1 n P P∞ ∞ −2n < ∞, Theorem (1.8.3) tells us that 2 Var(X ) = then because EX = 0 and n n n=1 n=1 P Sn → S ≡ ∞ n=1 Xn almost surely. I claim that S is uniformly distributed on (−1, 1). Let D be the collection of dyadic rationals in (−1, 1) and let Id be the collection of half open intervals with endpoints in D. Since D is dense in (−1, 1), an argument analogous to that given in exercises (1.1.3) and (1.1.5) shows that the Borel σ-algebra on (−1, 1) is generated by Id . However, Id is also a π-system and so by Theorem (2.2) of the Appendix it is enough to show that P(S ∈ (a, b]) = 21 (b − a) for every interval (a, b] ∈ Id . P(Xn = 2−n ) = P(Xn = −2−n ) = To do so, observe that each dyadic rational x ∈ D has a unique decomposition x= n X xk 2−k i=k where xk ∈ {−1, 1} and n ≥ 1. Since it follows that P∞ k=n+1 |Xn | ≤ 2−n and since P(X = y) = 0 for every y, P S ∈ (x − 2−n , x + 2−n ] = P(X1 = x1 , X2 = x2 , · · · , Xn = xn ) n Y = P(Xk = xk ) = 2−n . k=1 1 The claim then follows upon noting that every interval (a, b] ∈ Id has a unique decomposition into disjoint intervals of this form and that the total length of the interval is equal to the sum of the lengths of the sub-intervals. Combining the fact that the characteristic function of S is sin(t)/t with the result of exercise (3.11) allows us to conclude that ∞ Y sin(t) = cos t/2n . t n=1 3.14) We proceed by induction. For the case n = 0, the claim amounts to the requirement that φ(t) is continuous atR all t, which we know to be true. Suppose that the claim is true for some n and assume that |x|n+1 µ(dx) < ∞. To show that the claim is also true for n + 1, we will use Theorem (9.1) in the Appendix which comes with four conditions that need to be verified. We first define the function f (t, x) = (ix)n eitx and observe that the induction hypothesis asserts that Z φ(n) (t) = f (t, x)µ(dx) and that Z Z |f (t, x)|µ(dx) = |x|n µ(dx) < ∞. This proves condition (i). For (ii), we note that the partial derivative ∂t f (t, x) = (ix)n+1 eitx exists and is continuous in both arguments. Since |∂t f (t, x)| = |x|n+1 is an integrable function, the dominated convergence theorem implies that Z Z n+1 isx lim (ix) e µ(dx) = (ix)n+1 eitx µ(dx) s→t at every t and so (iii) is verified. Finally, for (iv), let δ > 0 and observe that Z Z δ Z Z δ Z n+1 ∂t f (t + s, x)dsµ(dx) = |x| dsµ(dx) = 2δ |x|n+1 µ(dx) < ∞. −δ −δ Since Theorem (9.1) allows us to interchange differentiation and integration, we have Z φ(n+1) (t) = (ix)n+1 eitx µ(dx), which completes the induction. 3.15) Let X be a standard normal random variable and observe that 2 /2 φ(t) = e−t = ∞ X (−1)n n=0 2 t2n 2n n! is the characteristic function of X. Since X has exponentially-decaying tails, it is clear that X has moments of all orders. In particular, by using (3.14) with t = 0, we can calculate EX 2n = i2n φ(2n) (0) = (2n)! . 2n n! Of course, symmetry considerations show that all of the odd moments vanish: EX 2n+1 = 0. 3.17) Let X1 , X2 , · · · be i.i.d. with characteristic function φ and let Sn = X1 + · · · + Xn . (i) Suppose that φ0 (0) = ia. By the definition of the derivative, this means that φ(t) = 1 + ia · t + o(t) as t → 0. Since Theorem (2.4.2) implies that EeitSn /n = φ(t/n)n = 1 + iat/n + o (t/n) n → eiat for all t ∈ R, we can use the continuity theorem along with exercise (2.2.9) to conclude that Sn /n → a in probability. (ii) Now suppose that Sn /n → a in probability and let δn (t) = n φ(t/n) − 1 . In this case, the continuity theorem gives EeitSn /n = φ(t/n)n = 1 + δn (t)/n n → eiat , which then (using (2.4.2) again) implies that limn→∞ δn (t) = iat, i.e. for all t 6= 0, lim n→∞ φ(t/n) − φ(0) = ai. t/n Since φ(·) is continuous, an application of the following lemma to the function ψ(t) ≡ φ(t) − 1 t shows that φ is differentiable at t = 0 with φ0 (0) = ai. Lemma: Suppose that ψ : R/{0} → R is a continuous function with the property that for all t 6= 0, lim ψ(t/n) = c (?). n→∞ Then lim ψ(h) = c. h→0 Proof: Suppose that the conclusion does not hold. Without loss of generality, we may assume that lim sup ψ(h) > c, h↓0 3 in which case there is a sequence hn ↓ 0 and numbers e > d > c such that ψ(hn ) > e for all n. Furthermore, since ψ is continuous on R/{0}, there exist numbers ln < rn with ln → 0 such that for every n ≥ 1, ψ(t) > d for all t ∈ In ≡ (ln , rn ). In light of (?), there can be no t > 0 such that the set O(t) ≡ {t/m; m ≥ 1} intersects infinitely many of the open intervals In . So, for n ≥ 1, let Un be the collection of t ∈ [1, 2] such that O(t)∩In 6= ∅ and O(t)∩In+k = ∅, and let U0 be the collection of t ∈ [1, 2] such that O(t)∩Im = ∅ for all m ≥ 1. Since [1, 2] is a complete metric space and since ∞ [ [1, 2] = Un , n=0 the Baire category theorem tells us that the Un cannot all be nowhere dense. In other words, there exists n ≥ 0 and some interval (a, b) ⊂ [1, 2] such that (a, b) is contained in the closure of Un . To arrive at a contradiction, let M be the smallest integer greater than a/(b − a) and notice that for all m ≥ M , a/m ≤ b/(m + 1) so that the successive intervals (a/m, b/m) and (a/(m + 1), b/(m + 1)) overlap and consequently (0, M ) = ∞ [ (a/m, b/m). m=M Then, either In+k ∩(0, M ) = ∅ for all k ≥ 1, which would contradict our assumption that ln → 0, or there exists an t ∈ (a, b) and integers m, k ≥ 1 such that t/m ∈ In+k . If the latter is true, then since Un is dense in (a, b), there is also a sequence (tj ; j ≥ 1) ⊂ Un such that tj → t. Since In+k is an open interval, this implies that tj ∈ In+k for all j sufficiently large, which contradicts the definition of Un . This completes the proof of the lemma. 4