Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2013

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

\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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