- #1

mbigras

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## Homework Statement

Show in detail that:

[tex]

\sigma_{x}^{2} = \int_{-\infty}^{\infty} (x -\bar{x})^{2} \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}} = \sigma^{2}

[/tex]

where,

[tex]

G_{X,\sigma}(x) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}}

[/tex]

## Homework Equations

[tex]

\int u dv = uv -\int v du

[/tex]

## The Attempt at a Solution

There are some hints that are given. Replace [itex]\bar{x}[/itex] with [itex]X[/itex] (according to the text, this is because they are equal after many trials). make the substitutions [itex]x-X=y[/itex] and [itex] y/\sigma=z[/itex]. Integrate by parts to obtain the result.

After following the hints I get the following integral:

[tex]

\frac{\sigma}{\sqrt{2\pi}} \int_{\infty}^{\infty} z^{2}e^{-\frac{z^{2}}{2}}dz

[/tex]

How would one go about using integration by parts for this integral?

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