Science:Math Exam Resources/Courses/MATH103/April 2015/Question 07 (d)

MATH103 April 2015

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Question 07 (d)
Consider the Normal (or Gaussian) probability density function (pdf ) given by $f(x)={\sqrt {\frac {2}{\pi }}}e^{-2x^{2}}$ for $-\infty <x<\infty$. (d) Find the variance $V$. (Hint: what is the anti-derivative of $xe^{-2x^{2}}$?)

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that if $f(x)$ is the probability density function on the domain $[a,b]$, then the variance is given by $V=\int _{a}^{b}(x-{\bar {x}})^{2}\,f(x)\,dx$

Hint 2
To calculate the integral, use integration by parts.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Using Hint and part (a), we have $V=\int _{-\infty }^{\infty }x^{2}f(x)dx={\sqrt {\frac {2}{\pi }}}\int _{-\infty }^{\infty }x^{2}e^{-2x^{2}}dx.$ Note that the integrand is an even function, we have $V={\frac {2{\sqrt {2}}}{\sqrt {\pi }}}\int _{0}^{\infty }x^{2}e^{-2x^{2}}dx.$ To calculate the integral, we first check the anti-derivative of $xe^{-2x^{2}}$. By the change of variable $t=2x^{2}$, we have ${\frac {1}{4}}dt=xdx$ and hence $\int xe^{-2x^{2}}dx={\frac {1}{4}}\int e^{-t}dt=-{\frac {1}{4}}e^{-2x^{2}}+C.$ Then, using the integration by parts with $u=x$ and $v'=xe^{-2x^{2}}$. we have $\int _{0}^{\infty }x^{2}e^{-2x^{2}}dx=\int _{0}^{\infty }uv'dx=-{\frac {1}{4}}xe^{-2x^{2}}{\bigg |}_{0}^{\infty }+{\frac {1}{4}}\int _{0}^{\infty }e^{-2x^{2}}dx={\frac {1}{4}}\int _{0}^{\infty }e^{-2x^{2}}dx$ Since $e^{-2x^{2}}$ is an even function and $I=\int _{-\infty }^{\infty }e^{-2x^{2}}dx={\sqrt {\frac {\pi }{2}}}$ from part (c), we have $\int _{0}^{\infty }x^{2}e^{-2x^{2}}dx={\frac {1}{4}}\int _{0}^{\infty }e^{-2x^{2}}dx={\frac {1}{8}}I={\frac {1}{8}}\cdot {\sqrt {\frac {\pi }{2}}}$ Collecting the terms, we have ${\begin{aligned}V={\frac {2{\sqrt {2}}}{\sqrt {\pi }}}\int _{0}^{\infty }x^{2}e^{-2x^{2}}dx={\frac {2{\sqrt {2}}}{\sqrt {\pi }}}\cdot {\frac {1}{8}}\cdot {\sqrt {\frac {\pi }{2}}}=\color {blue}{\frac {1}{4}}\end{aligned}}$