Variance in Univariate Normal

**ynotidas** · August 31st 2009, 10:17 AM

In Univariate Normal, Show
$Var(x)=\sigma^2$

**pedrosorio** · August 31st 2009, 10:54 AM

What have you tried?

**Chris L T521** · August 31st 2009, 11:25 AM

Originally Posted by ynotidas

In Univariate Normal, Show
$Var(x)=\sigma^2$

Well, we need to use the fact that $E\left[X\right]=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}xe^{-\tfrac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu$

Thus, $E\left[X^2\right]=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx$

Let $u=\frac{x-\mu}{\sigma}\implies x=\sigma u+\mu\implies \,dx=\sigma\,du$

Therefore, $\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx\xrightarrow{u=(x-\mu)/\sigma}{}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\sigma u+\mu\right)^2e^{-\tfrac{1}{2}u^2}\,du$ $=\frac{1}{\sqrt{2\pi}}\left[\sigma^2\int_{-\infty}^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du-2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du+\mu^2\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du\right]$

Analyze each integral individually:

$\sigma^2\int_{-\infty}^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du=2\sigma^2\int_0^{\infty}u^2e^ {-\tfrac{1}{2}u^2}\,du$

Let $z=\tfrac{1}{2}u^2\implies \,dz=u\,du$

Thus, $2\sigma^2\int_0^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2 }=2\sqrt{2}\sigma^2\int_{0}^{\infty}z^{\frac{1}{2} }e^{-z}\,dz=2\sqrt{2}\sigma^2\Gamma\!\left(\tfrac{3}{2} \right)$ $=\sqrt{2}\sigma^2\Gamma\!\left(\tfrac{1}{2}\right) =\sqrt{2\pi}\sigma^2$

$2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du$

Let $z=\tfrac{1}{2}u^2\implies \,dz=u\,du$

Thus,
$2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}2\mu\sigma\int_{\infty}^{\infty}e^{-z}\,dz=0$

$\mu^2\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=\mu^2\sqrt{2\pi}$

Thus, $E\left[X^2\right]=\frac{1}{\sqrt{2\pi}}\cdot\left[\sigma^2\sqrt{2\pi}+\mu^2\sqrt{2\pi}\right]=\sigma^2+\mu^2$

Thus, $\text{Var}\left(X\right)=E\left[X^2\right]-\left(E\left[X\right]\right)^2=\sigma^2+\mu^2-\mu^2=\sigma^2$

Does this make sense?

**matheagle** · August 31st 2009, 08:24 PM

Chris, I mean anteater (my bad)
I thought you did this last week.

**ynotidas** · September 1st 2009, 12:46 AM

last week Chris did the expected value proof

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