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==Solution 2== | ==Solution 2== | ||
| − | + | <math>\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}</math> | ||
| − | <math> | + | <math>\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}</math> | ||
| − | + | Therefore, | |
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| + | <math>Var(X)=E(X^2)-E(X)^2=\frac{1}{\lambda^2}</math> | ||
<font color="red"><u>'''Critique on Solution 2:'''</u> | <font color="red"><u>'''Critique on Solution 2:'''</u> | ||
| − | + | Solution 2 is correct. In addition, calculating <math>E(X)</math> first is better since the result can be used in calculating <math>E(X^2)</math>. | |
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</font> | </font> | ||
Revision as of 07:35, 4 November 2014
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) $.
