Revision as of 07:24, 4 November 2014

Part 3

Let $ X $ be an exponential random variable with parameter $ \lambda $, so that $ f_X(x)=\lambda{exp}(-\lambda{x})u(x) $. Find the variance of $ X $. You must show all of your work.

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

Assume

$ Y=\left(\begin{array}{c}Y_i \\ Y_j\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}a_{11}X_i+a_{12}X_j \\ a_{21}X_i+a_{22}X_j\end{array} \right) $.

Then

$ \begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\ =a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j) \end{array} $

For $ |i-j|\geq1 $, $ E(X_i,X_j)=0 $. Therefore, $ a_{11}a_{21}+a_{12}a_{22}=0 $.

One solution can be

$ A=\left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array} \right) $.

Critique on Solution 2:

1. $ E(Y_iY_j)=0 $ is not the condition for the two random variables to be independent.

2. "For $ |i-j|\geq1 $, $ E(X_i,X_j)=0 $" is not supported by the given conditions.

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