Calculate the mean of the normal distribution function $\frac1 {2\pi \sigma^2}exp[-\frac {(x-\mu)^2} {2\sigma^2}]$ by integration.

Question

I know that it must be $\mu$ but I cannot get the answer. This is my attempt so far:

Normal distribution function = $N(x)=\frac1 {2\pi \sigma^2}exp[-\frac {(x-\mu)^2} {2\sigma^2}]$

$$\langle x\rangle=\int_{-\infty}^\infty xN(x)dx$$ Where $N(x)$ is the normal distribution function.

Substitute $y=\frac{x-\mu} {\sqrt2 \sigma} $. The resulting integral is:

$$\frac 1{2\sigma^2 \sqrt\pi}\int_{-\infty}^\infty e^{-y^2}$$

Since the integral can be evaluated as $\sqrt\pi$ the mean that I calculate is $\frac 1{2\sigma^2}$ not $\mu$ but I don't know what I have done wrong. Please help.

Answer 1

Hint. The change of variable $\displaystyle y=\frac{x-\mu} {\sqrt{2 \pi}\sigma} $ rather gives you $$\langle x\rangle=\int_{-\infty}^\infty xN(x)dx=\int_{-\infty}^{+\infty} (\sqrt{2 \pi}\sigma \:y +\mu /\sqrt{\pi})e^{-y^2}dy=\mu$$ since the first integral vanishes, the second one being standard.

Answer 2

$N\left(x\right)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}$. Use substitution $u=\frac{x-\mu}{\sigma}$.

Note that $\frac{1}{\sigma\sqrt{2\pi}}\int e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}dx=\frac{1}{\sqrt{2\pi}}\int e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}d\left(\frac{x-\mu}{\sigma}\right)$.