Gaussian distribution

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November 28th 2006, 10:23 AM
iljitj

Gaussian distribution

Hi!

I'm having a hard time solving this problem:

http://s10.imagehosting.us/uploadpoi...-1_1778949.JPG

I hope somebody can give me a little help. I've been strugling with this for quite a while now.

/Iljtitj
November 28th 2006, 10:50 AM
ThePerfectHacker

You want to find,
$\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}}x^2\exp \left( - \frac{(x-\mu)^2}{2\sigma^2} \right) dx$
Let us use the hint,
$u=\frac{x-\mu}{\sqrt{2} \sigma}$
Also,
$u'=\frac{1}{\sigma \sqrt{2}}$
Thus,
$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} x^2 \exp (-u^2) u' dx$
Since,
$x=\sqrt{2} \sigma u+\mu$
Thus, by substitution rule,
$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} (\sqrt{2} \sigma u+\mu)^2 \exp (-u^2) du$
Thus, open,
$\frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} (2\sigma^2u^2 +u\sqrt{2} \sigma \mu+\mu^2) \exp(-u^2)du$
Maybe, I will return to it later.
But that is the idea.

Also note that,
$\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}$
This is why $\pi$ will cancel out in the end.
Hope that helps.
November 28th 2006, 11:19 AM
iljitj

Thanks

Thanks alot, I'll give it another try with this.
November 28th 2006, 12:01 PM
CaptainBlack

Quote:

Originally Posted by iljitj

Hi!

I'm having a hard time solving this problem:

http://s10.imagehosting.us/uploadpoi...-1_1778949.JPG

I hope somebody can give me a little help. I've been strugling with this for quite a while now.

/Iljtitj

$ I=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{\infty} x^2 \exp \left( -\frac{(x-\mu)^2}{2\sigma ^2}\right) dx $

Now let $u=\frac{x-\mu}{\sqrt{2}\sigma}$ , so $x=\sqrt{2}\sigma u+\mu$ . Hence:

$ I=\frac{1}{ \sqrt{ \pi}} \int_{-\infty}^{\infty} (\sqrt{2}\sigma u+\mu)^2 \exp (-u^2) du=$ $ \frac{1}{\sqrt{ \pi}} \int_{-\infty}^{\infty} [2\sigma^2 u^2+2 \sqrt{2}\sigma u \mu+\mu^2] \exp (-u^2) du $

as the middle term in the integral is odd it integrates up to $0$ so:

$ I=\frac{1}{\sqrt{ \pi}} \int_{-\infty}^{\infty} [2\sigma^2 u^2+\mu^2] \exp (-u^2) du $

The last term integrates up to $\mu^2$ as

$ \frac{1}{ \sqrt{ \pi}} \int_{-\infty}^{\infty} \exp(-u^2) du=1 $

and the first term integrates up to $\sigma^2$ as:

$ \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{\infty}x^2 \exp(-x^2/2) dx=1 $

so:

$ I=\sigma^2+\mu^2$

RonL
November 28th 2006, 12:30 PM
iljitj

Cool, thanks..

Many thanks. I see now :) It's very much appreciated.

/Iljitj