# Normal distribution theoretical moments

how we can show that the following equality holds

$E[(x-\mu)/\sigma]=0$

$E[(x-\mu)^2/\sigma^2-1]=0$

$E[(x-\mu)^3/\sigma^3]=0$

$E[(x-\mu)^4/\sigma^4-3]=0$

-
Convert each of those into someting about an integral. Then examine the integrals. – GEdgar Feb 6 '13 at 13:05

If $X\sim\mathcal{N}(\mu,\sigma^2)$ then $Z=\frac{X-\mu}{\sigma}\sim\mathcal{N}(0,1)$, and so you're asked to show $$E[Z], \quad E[Z^2]-1,\quad E[Z^3],\quad E[Z^4]-3$$ are all equal to $0$. This can be done by calculating the approriate integrals using the law of the unconscious statistician. You can compare your expressions with this list

We know that $Z$ has density $f_Z$ given by $$f_Z(x)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}z^2\right),\quad z\in\mathbb{R},$$ and consequently \begin{align} E[Z]&=\int_{-\infty}^\infty z\cdot f_Z(z)\,\mathrm dz=\frac{1}{2\pi}\int_{-\infty}^\infty z\cdot \exp\left(-\frac{1}{2}z^2\right)\,\mathrm dz. \end{align} Now an anti-derivative of $z\cdot \exp\left(-\frac{1}{2}z^2\right)$ is the function $$g(z):=-\exp\left(-\frac{1}{2}z^2\right),\quad z\in\mathbb{R}.$$ Thus $$E[Z]=\frac{1}{2\pi}\left[\lim_{z\to\infty}g(z)-\lim_{z\to -\infty}g(z)\right]=\frac{1}{2\pi}\left[0-0\right]=0.$$

-
colud you please show it for the first moment $E[Z]$ – user45689 Feb 6 '13 at 13:22
@user45689: See the edit. – Stefan Hansen Feb 6 '13 at 16:01
@user45689: Here is a detailed explanation. But if you already know that the variance $\mathrm{Var}(Z)$ is $1$, then you can use the formula $\mathrm{Var}(Z)=E[Z^2]-E[Z]^2$. – Stefan Hansen Feb 6 '13 at 16:54
No, that's not what I'm saying. I'm saying that if you already know that $\mathrm{Var}(Z)=1$ and from earlier you know that $E[Z]=0$, then the formula $E[Z^2]=\mathrm{Var}(Z)+E[Z]^2=1+0^2=1$ yields the result. – Stefan Hansen Feb 6 '13 at 19:57
The third moment is $0$, because you integrate $z^3\cdot f_Z(z)$ which is an odd function. For the fourth moment I would recommend to follows Sasha's answer. – Stefan Hansen Feb 7 '13 at 6:14