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Math Help - Gaussian distribution

  1. #1
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    Gaussian distribution

    Hi!

    I'm having a hard time solving this problem:



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

    /Iljtitj
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  2. #2
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    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.
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  3. #3
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    Thanks

    Thanks alot, I'll give it another try with this.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by iljitj View Post
    Hi!

    I'm having a hard time solving this problem:



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

    /Iljtitj
    <br />
I=\frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{\infty} x^2 \exp \left( -\frac{(x-\mu)^2}{2\sigma ^2}\right) dx<br />

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

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

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

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

    The last term integrates up to \mu^2 as

    <br />
\frac{1}{ \sqrt{ \pi}} \int_{-\infty}^{\infty} \exp(-u^2) du=1<br />

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

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

    so:

    <br />
I=\sigma^2+\mu^2

    RonL
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  5. #5
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    Cool, thanks..

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


    /Iljitj
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