Transforms. Convergence of probability generating functions. Convergence of characteristic functions functions
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1 Transforms For non-negative integer value ranom variables, let the probability generating function g X : [0, 1] [0, 1] be efine by g X (t) = E(t X ). The moment generating function ψ X (t) = E(e tx ) is efine as long this expectation is efine an there exists h > 0 such that E(e tx ) < for all t [ h, h]. The characteristic function is given by ϕ X (t) = E(e itx ) = E(cos[tX ]) + ie(sin[tx ]) an is efine for all ranom variables. We will use these transforms for proving convergence results. In particular, we will investigate which conitions are neee for equivalence between convergence in istribution an convergence of the transforms. Convergence of probability generating functions All convergence results are for n. Theorem (4.1) Let X, X 1, X 2, be non-negative integer value ranom variables, an suppose that g Xn (t) g X (t) for 0 t 1, then X n X Exercise: prove that if X is almost surely finite, then X n X implies g Xn (t) g X (t) for 0 t 1. Convergence of moment generating functions Theorem (4.2) Let X 1, X 2, be ranom variables, such that the moment generating function ψ Xn (t) exists for t h for some h > 0 an all n. Suppose further that X is a ranom variable, for which ψ X (t) exists for t h 1 < h for some h 1 > 0 an that ψ Xn (t) ψ X (t) for t h 1, then X n X Convergence of characteristic functions functions Theorem (4.3) Let X, X 1, X 2, be ranom variables an suppose that ϕ Xn (t) ϕ X (t) for t R, then X n X. We can relax the assumptions. If ϕ Xn (t) ϕ(t) for some ϕ(t) which is continuous in t = 0, then ϕ(t) is the characteristic function of some ranom variable X an X n X It is also true that if X n X then ϕxn (t) ϕ X (t) for t R. Since transforms are transforming istributions, the convergence results are only about convergence in istribution, not about almost sure convergence, convergence in probability or r-th mean.
2 Proof of X n X ϕxn (t) ϕ X (t) Consequence Let h(x) = 1(a < x b), where a an b are points of continuity of F X. Then E(h(X n )) = P(a < X n b) = F Xn (b) F Xn (a) F X (b) F X (a) = E(h(X )) Approximate e itx by m w kh k (x) where w k C an h k (x) = 1(a k < x b k ) for points of continuity a k an b k. It follow (after some (in this course) unjustifie playing with the orer of limits) that lim E(e itxn ) E(e itx ) will become arbitrary small. Let X, X 1, X 2, be ranom variables an suppose that ϕ Xn (t) e itc for some c R an all t R, then X n c. If follows from Theorem 4.3 that X n c, which, since c is P constant, implies X n c We can now easily prove the Weak Law of Large Numbers: Let X 1, X 2, be i.i.. ranom variables with E(X 1 ) = µ < then n 1 S n = n 1 n X k P µ Using (in orer of occurrence) Theorem 4.4.9, inepenence, ientical istributions an Theorem gives n ϕ Sn/n(t) = ϕ Sn (t/n) = ϕ Xk (t/n) = [ϕ X1 (t/n)] n = [1 + iµt/n + o(t/n)] n [e iµt/n ] n = e iµt Consequence Example Let X, X 1, X 2, be ranom variables an suppose that X n X an Xn Y. Now ϕ Xn (t) ϕ X (t) an ϕ Xn (t) ϕ Y (t) for t R ϕ X (t) ϕ Y (t) ϕ X (t) ϕ Xn (t) + ϕ Xn (t) ϕ Y (t) 0 by Remark 4.2 This gives ϕ X (t) = ϕ Y (t) an by Theorem that X an Y have the same istribution. Assume P(X n = n) = 1/n = 1 P(X n = 0), then ϕ Xn (t) = (1 1/n)e it0 + (1/n)e itn = 1 + (e itn 1)/n 1 this implies X n P 0 Note that ψ Xn = 1 1/n + (1/n)e tn for t > 0 We cannot use Theorem 4.2, since there is no open ball aroun 0 on which ψ X (t) is efine.
3 Example Assume X n = Bin(n, λ/n) then n ( ) n g Xn (t) = (t λ k n )k (1 λ n )n k = (1 λ n + t λ n )n = (1 + k=0 λ(t 1) ) n e λ(t 1) = n (λt) k k=0 this implies Bin(n, λ/n) Poi(λ) k! e λ Central Limit Theorem Theorem Let X 1, X 2, be i.i.. ranom variables with µ = E(X 1 ) < an σ 2 = E((X 1 µ) 2 ) <. Let S n = n X k, then S n nµ σ 2 n N (0, 1) With Y k = X k µ. The statement of the theorem reas σ 2 1 n n Y k N (0, 1) Note that E(Y 1 ) = 0 an Var(Y 1 ) = E((Y 1 ) 2 ) = σ2 σ 2 = 1 Proof of Central Limit Theorem Let Z = Sn nµ σ 2 n = 1 n n Y k then ϕ Z (t) = n ϕ Yk (t/ n) = [ϕ Yk (t/ n)] n = [1 + 0 it 1 1 n 2 ( t ) 2 t + o( n 2 )]n e t2 /2 n which is the characteristic function of a stanar normal istribution (see page 74). Empirical istribution functions Let X 1, X 2, be i.i.. ranom variables with istribution function F. The (Ranom!!) empirical istribution function is efine by F n (x) = n 1 {i; X i x} = n 1 n i=1 1(X i x) note that nf n (x) = n i=1 1(X i x) = Bin(n, F (x)) E(F n (x)) = n 1 nf (x) = F (x) an Var(F n (x)) = n 2 nf (x)(1 F (x)) < 1/n 0 The weak law of large numbers gives F n (x) P F (x) the central limit theorem gives nf n(x) nf (x) nf (x)(1 F (x)) N (0, 1) This gives information about the spee of convergence of F n (x) to F (x)
4 Example: Stirling s formula n! Stirling s formula states that n n+1/2 e n 1 Γ(u) More general u u 1/2 e u 1, where Γ(u) = 0 x u 1 e x x. Remember Γ(u) = (u 1)! for u N Let Y be a ranom variable with Γ istribution with mean an variance both u, i.e. f (x) = x u 1 e x Γ(u) an characteristic function ϕ(t) = (1 it) u X = (Y u)/ u has istribution function f u (x) = u(x u + u) u 1 e (x u+u) /Γ(u) an characteristic function ϕ u (t) = e it u (1 it/ u) u by the inversion theorem for continuous ensities (f (x) = 1 e itx ϕ(t)t) we obtain f u (0) = 1 ϕ u(t)t Example: Stirling s formula continue We have f u (0) = u u 1/2 e u /Γ(u) an f u (0) = 1 ϕ u(t)t We also obtain (for u ) ϕ u (t) = e it u (1 it/ u) u = exp[ it u u log[(1 it/ u)] = exp[ it u u(it/ u + t 2 /(2u) + O(t 3 u 3/2 ))] u u 1/2 e u /Γ(u) = 1 lim u uu 1/2 e u /Γ(u) = lim = 1 = exp[ t 2 /2 + O(t 3 u 1/2 ))] e t2 /2 ϕ u(t)t gives u 1 lim ϕ u(t)t = 1 u ϕ u (t)t e t2 /2 t = 1 Remark Tail events Let X 1, X 2, be i.i.. ranom variables with µ = E(X 1 ) <, σ 2 = E((X 1 µ) 2 ) < an E( X 1 3 ) <. Furthermore, let S n = n X k an Φ(x) = x 1 e y 2 /2 y be the istribution function of a stanar normal istribution an let F n (x) be the istribution function of Sn nµ Then σ 2 n sup x F n (x) Φ(x) CE( X 1 3 )/ σ 6 n for some constant C We omit the proof. Let X 1, X 2, be ranom variables on (Ω, F, P) Let H n = σ(x n+1, X n+2, ) be the smallest σ-algebra in which all ranom variables X n+1, X n+2, are measurable H n H n+1, so H = n H n is well efine. H is calle the tail σ-algebra, an events containe in it are tail events Some examples of tail events are { n=1 1(X n > 0) = }, {lim sup X n = } an { n=1 X n converges}
5 Kolmogorov s zero-one law Remark Theorem If X 1, X 2, are inepenent, then all tail events H H satisfy either P(H) = 1 or P(H) = 0 The proof is left as an exercise In many situations, it is easy to apply Kolmogorov s zero-one law to show that some event has probability 0 or 1, but surprisingly har to etermine which of the two. Let X 1, X 2, be i.i.. ranom variables with µ = E(X 1 ) = 0 an σ 2 = E((X 1 µ) 2 ) = 1. Furthermore, let S n = n X k Observe that lim sup n S n / 2n log log[n] = z is a tail event for all z R an has probability 0 or 1. Without proof we state the following theorem Theorem (Law of the iterate logarithm) P(lim sup n S n / 2n log log[n] = 1) = 1
X n D X lim n F n (x) = F (x) for all x C F. lim n F n(u) = F (u) for all u C F. (2)
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