how we can show that the following equality holds
$E[(x-\mu)/\sigma]=0$
$E[(x-\mu)^2/\sigma^2-1]=0$
$E[(x-\mu)^3/\sigma^3]=0$
$E[(x-\mu)^4/\sigma^4-3]=0$
Asked
Active
Viewed 484 times
1
-
1Convert each of those into someting about an integral. Then examine the integrals. – GEdgar Feb 06 '13 at 13:05
1 Answers
2
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 $$ E[Z]=\frac{1}{2\pi}\left[\lim_{z\to\infty}g(z)-\lim_{z\to -\infty}g(z)\right]=\frac{1}{2\pi}\left[0-0\right]=0. $$
Stefan Hansen
- 26,160
- 7
- 62
- 95
-
-
-
thanks I calculted it already. but I'm stuck at the second one $E[Z^2]$ any hint – user45689 Feb 06 '13 at 16:29
-
1@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 06 '13 at 16:54
-
i'm afraid i'm stuck !! how can we use the formula $Var(Z)=E[Z^2]-E[Z]^2$ – user45689 Feb 06 '13 at 19:43
-
If you know that $\mathrm{Var}(Z)=1$ and that $E[Z]=0$, then you can calculate $E[Z^2]$. – Stefan Hansen Feb 06 '13 at 19:44
-
-
1No, 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 06 '13 at 19:57
-
thanks for your guidance, I have managed to calculate it by taking integral of $E[Z^2]$ . but now I'm stuck at 3rd moment can i do it in a similar fashion ? – user45689 Feb 06 '13 at 23:41
-
1The 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 07 '13 at 06:14
-
thank you. I have calculated the first three momnents. but 4th $E[Z^4]-3=0$ i'm not getting help me taking the deravitive of \begin{align} E[Z^4]&=\int_{-\infty}^\infty z^4\cdot f_Z(z),\mathrm dz=\frac{1}{2\pi}\int_{-\infty}^\infty z^4\cdot \exp\left(-\frac{1}{2}z^2\right),\mathrm dz. \end{align} – user45689 Feb 07 '13 at 14:46
-
Again, please take a look at this answer. I don't think it can be done any easier than this. Just do the substitution $z=\sqrt{2u}$. – Stefan Hansen Feb 07 '13 at 15:14
-
thanks, but what is the intuition behind this substitution does that mean $E[Z^4]=E[(\sqrt2u)^4]$ – user45689 Feb 07 '13 at 15:21
-
I don't think that there is any intuition behind it. It's just an application of integration by substitution. No it certainly doesn't mean that $E[Z^4]=E[(\sqrt{2}u)^4]$. – Stefan Hansen Feb 07 '13 at 15:57