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Thread: [SOLVED] integral

  1. #1
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    [SOLVED] integral

    $\displaystyle \int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

    $\displaystyle = e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du} $

    what now?

    sorry, I didn't mean to put this thread here...
    Last edited by akane; May 30th 2010 at 08:01 AM.
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by akane View Post
    $\displaystyle \int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

    $\displaystyle = e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du} $

    what now?

    sorry, I didn't mean to put this thread here...
    You want to use the fact that $\displaystyle \int f_X\text{ dx }= 1.$

    Turn your integrand into a normal pdf.
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  3. #3
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    Quote Originally Posted by Anonymous1 View Post
    You want to use the fact that $\displaystyle \int f_X\text{ dx }= 1.$

    Turn your integrand into a normal pdf.
    I have this:

    $\displaystyle g_{Z}(z) = \int_{-\infty}^{\infty} f(u,u-z) du = \int_{-\infty}^{\infty} \frac{1}{2 \pi \sigma^2} e^{-\frac{u^2+(u-z)^2}{2 \sigma^2}} du$

    Z should also be a normal variable, I don't know how to get a function of only one variable under the integral
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  4. #4
    Super Member Anonymous1's Avatar
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    Have you ever heard of the error function?

    $\displaystyle \int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} =\left(\frac{1}{2}\sqrt{\pi\sigma^2} e^{-\frac{z^2}{4\sigma^2}}\right)\cdot\left[\text{erf}(-\frac{z-2u}{2\sigma})|_{-\infty}^{+\infty}\right]$

    Also, $\displaystyle \text{erf}(\infty)=1$ and $\displaystyle \text{erf}(-\infty)=-1$
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  5. #5
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    Quote Originally Posted by Anonymous1 View Post
    Have you ever heard of the error function?

    $\displaystyle \int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} =\left(\frac{1}{2}\sqrt{\pi\sigma^2} e^{-\frac{z^2}{4\sigma^2}}\right)\cdot\left[\text{erf}(-\frac{z-2u}{2\sigma})|_{-\infty}^{+\infty}\right]$

    Also, $\displaystyle \text{erf}(\infty)=1$ and $\displaystyle \text{erf}(-\infty)=-1$
    thanks for help

    I've heard of it, but this is actually the first time it appeared in my assignment so I wasn't able to recognize it
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by akane View Post
    $\displaystyle \int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

    $\displaystyle = e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du} $

    what now?

    sorry, I didn't mean to put this thread here...
    You have:

    $\displaystyle \int_{-\infty}^{\infty} e^{-\frac{u^2+(u-z)^2}{2\sigma^2}}\;du$

    To do this we complete the square of the integrand:

    $\displaystyle u^2+(u-z)^2=\left(\sqrt{2}u-\frac{z}{\sqrt{2}}\right)^2+\frac{z^2}{2}$

    So:

    $\displaystyle \int_{-\infty}^{\infty} e^{-\frac{u^2+(u-z)^2}{2\sigma^2}}\;du
    =e^{\frac{z^2}{4\sigma^2}} \int_{-\infty}^{\infty}$$\displaystyle

    e^{-\frac{\left(\sqrt{2}u-z/\sqrt{2}\right)^2}{2\sigma^2}}\;du
    $

    Now make the change of variable:

    $\displaystyle v=\frac{\sqrt{2}u-z/\sqrt{2}}{\sigma}$

    and see what happens.

    CB
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