Revision as of 07:48, 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

$ Var(X)=E(X^2)-E(X)^2 $

First,

$ E(X^2)=\int_0^{\infty}x^2\lambda{e}^{-\lambda{x}}dx $

Since

$ \begin{array}{l}\int{x}^2\lambda{e}^{-\lambda{x}}dx\\ =\int -x^2 de^{-\lambda x}\\ =-x^2e^{-{\lambda}x}+{\int}2xe^{-{\lambda}x}dx\\ =-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}+{\int}\frac{e^{-{\lambda}x}}{\lambda}2dx\\ =-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x} \end{array} $,

We have

$ E(X^2)=-x^2e^{-\lambda x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x}|_0^\infty $

By L'Hospital's rule, we have

$ \lim_{x\to \infty}x^2e^{-\lambda x} = \lim_{x\to \infty}\frac{x^2}{e^{-\lambda x}}=\lim_{x\to \infty}\frac{2x}{\lambda e^{\lambda x}}=\lim_{x\to \infty}\frac{2}{\lambda^2e^{\lambda x}}=0 $

and

$ \lim_{x\to \infty}xe^{\lambda x} = \lim_{x\to \infty} \frac{x}{e^{\lambda x}}=\lim_{x\to \infty} \frac{1}{\lambda e^{\lambda x}} = 0 $.

Therefore,

$ E(X) = \frac{2}{\lambda^2} $.

Then we take a look at $ E(X) $.

$ E(X)=\int_0^{\infty}x\lambda{e}^{-\lambda{x}}dx $

$ \begin{array}{l} \int x\lambda{e}^{-\lambda{x}}dx\\ =\int xde^(\lambda x)\\ =-xe^{-\lambda x}+\int e^{\lambda x}dx\\ =-xe^{-\lambda x}-\frac{1}{x}e^{\lambda x}\\ \end{array} $

Similar to the calculation of $ E(X^2) $,

$ E(X)=\frac{1}{\lambda} $.

Therefore,

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

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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