Revision as of 13:50, 4 November 2014

Part 4

Consider a sequence of independent random variables $ X_1,X_2,... $, where $ X_n $ has pdf

$ \begin{align}f_n(x)=&(1-\frac{1}{n})\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\frac{n-1}{n}\sigma)^2]\\ &+\frac{1}{n}\sigma exp(-\sigma x)u(x)\end{align} $.

Does this sequence converge in the mean-square sense? Hint: Use the Cauchy criterion for mean-square convergence, which states that a sequence of random variables $ X_1,X_2,... $ converges in mean-square if and only if $ E[|X_n-X_{n+m}|] \to 0 $ as $ n \to \infty $, for every $ m>0 $.

Solution 1

$ E(X_n)=\frac{n-1}{n}E(Y)+\frac{1}{n}E(Z) $

Where

$ Y \sim N(\frac{n-1}{n}\sigma, \sigma^2) $

$ Z \sim EXP(\sigma) $

From the property of Normal distribution and exponential distribution,

$ E(Y)=\frac{n-1}{n}\sigma $

$ E(Z)=\frac{1}{\sigma} $.

Therefore,

$ \lim_{x\to \infty}E(X_n)=\lim_{x\to \infty}(\frac{n-1}{n})^2\sigma+\frac{1}{n}\frac{1}{\sigma}=\sigma $.

Also,

$ \lim_{x\to \infty}E(X_{n+m})=\lim_{x\to \infty}(\frac{n+m-1}{n+m})^2\sigma+\frac{1}{n+m}\frac{1}{\sigma}=\sigma $.

Thus,

$ \lim_{x\to \infty}E(X_n-X_{n+m})=\lim_{x\to \infty}E(X_n)-\lim_{x\to \infty}E(X_{n+m})=0 $,

$ \lim_{x\to \infty}E(X_{n+m}-X_n)=\lim_{x\to \infty}E(X_{n+m})-\lim_{x\to \infty}E(X_n)=0 $.

So we have

$ \lim_{x\to \infty}E(|X_{n+m}-X_n|)=0 $

for every m.

From the Cauchy criterion for mean-square convergence, this sequence converges int he mean-square sense

Solution 2

$ \begin{align} E(X)&=\int_{-\infty}^{+\infty}xp(x)dx\\ &=\int_{0}^{\infty}x\lambda e^{-\lambda x}dx\\ &=-(xe^{-lambda x}|_0^{\infty}-\int_0^{\infty}e^{-\lambda x}dx)\\ &=\frac{1}{x} \end{align} $

$ \begin{align} E(X^2)&=\int_{-\infty}^{+\infty}x^2p(x)dx\\ &=\int_{0}^{\infty}x^2 \lambda e^{-\lambda x}dx\\ &=-(x^2e^{-lambda x}|_0^{\infty}-\int_0^{\infty}2xe^{-\lambda x}dx)\\ &=\frac{2}{x^2} \end{align} $

Therefore,

$ Var(X)=E(X^2)-E(X)^2=\frac{1}{\lambda^2} $

Critique on Solution 2:

Solution 2 is correct. In addition, calculating $ E(X) $ first is better since the result can be used in calculating $ E(X^2) $.

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