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Math Help - Gaussian PDF - Expectation Value

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    Gaussian PDF - Expectation Value

    <x> = \int_{-\infty}^{\infty} x \frac{1}{\sqrt{2\pi}{\sigma}}  e^{-\frac{(x-m)^2} {2{\sigma}^{2}}} dx=m

    Why?

    Can this be shown without doing the integration in polar form or using the error function?
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  2. #2
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    Quote Originally Posted by _jonny_ View Post
    <x> = \int_{-\infty}^{\infty} x \frac{1}{\sqrt{2\pi}{\sigma}} e^{-\frac{(x-m)^2} {2{\sigma}^{2}}} dx=m

    Why?

    Can this be shown without doing the integration in polar form or using the error function?
    Start by proving that the mean of the standard normal distribution is equal to 0:

    E(Z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z \, e^{-\frac{z^2}{2}} \, dz = 0.

    Hint: z \, e^{-\frac{z^2}{2}} is an odd function.

    Then note that Z = \frac{X - \mu}{\sigma} \Rightarrow X = \, ......

    And you know that if X = a Z + b then E(X) = a E(Z) + b \, ....
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