# Thread: Gaussian PDF - Expectation Value

1. ## Gaussian PDF - Expectation Value

$ = \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?

2. Originally Posted by _jonny_
$ = \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 $E(X) = a E(Z) + b \, ....$