3.1 Important examples of expectation

We get Moments of a pdf by setting $g(X)=X^n$ in this previous equation ,

• $n=1$ : 1st order moment, $(x)$ = Mean value
• $n=2$ : 2nd order moment, $(x^2)$ = Mean-squared value (Power or energy)
• $n> 2$ : Higher order moments, $(x^n)$ , give more detail about $f_X(x)$ .

Central moments

Central moments are moments about the centre or mean of a distribution,

• Variance, $n=2$ :
  $\sigma ^2=((X-\langle X\rangle )^2)=\int (x-\langle X\rangle )^2{f}_X(x)\,d x=\int x^2{f}_X(x)\,d x-2\langle X\rangle \int x{f}_X(x)\,d x+\langle X\rangle ^2\int f_X(x)\,d x=(X^2)-2\langle X\rangle ^2+\langle X\rangle ^2=(X^2)-\langle X\rangle ^2$
• Standard deviation, $\sigma =\sqrt{\mathrm{variance}}$ .
• Skewness, $n=3$ :
  $\gamma =\frac{((X-\langle X\rangle )^3)}{\sigma ^3}$
  $\gamma =0$ if the pdf of $X$ is symmetric about $\langle X\rangle$ , and becomes more positive if the tail of the distribution is heavier when $X> \langle X\rangle$ .
• Kurtosis (or excess), $n=4$ :
  $\kappa =\frac{((X-\langle X\rangle )^4)}{\sigma ^4}-3$
  $\kappa =0$ for a Gaussian pdf and becomes more positive for distributions with heavier tails.

Example: central moments of a normal distribution

The normal (or Gaussian) pdf with zero mean is given by:

Since the mean is zero, the $n$ th order central moment is given by

• Valid pdf, $n=0$ :
  $(X^0)=\frac{1}{\sqrt{\pi }}\Gamma (\frac{1}{2})=1$
  as required for a valid pdf.
  The normalization factor $\frac{1}{\sqrt{2\pi \sigma ^2}}$ in the expression for the pdf of a unit variance Gaussian (e.g. ) arises directly from the above result.
• Mean, $n=1$ :
  $(X)=0$
  so the mean is zero.
• Variance, $n=2$ :
  $((X-\langle X\rangle )^2)=(X^2)=\frac{1}{\sqrt{\pi }}2\sigma ^2\Gamma (\frac{3}{2})=\frac{1}{\sqrt{\pi }}2\sigma ^2\frac{\sqrt{\pi }}{2}=\sigma ^2$
  Therefore standard deviation = $\sqrt{\mathrm{variance}}=\sigma$ .
• Skewness, $n=3$ :
  $(X^3)=0$
  so the skewness is zero.
• Kurtosis, $n=4$ :
  $((X-\langle X\rangle )^4)=(X^4)=\frac{1}{\sqrt{\pi }}(2\sigma ^2)^2\Gamma (\frac{5}{2})=\frac{1}{\sqrt{\pi }}(2\sigma ^2)^2\frac{3\sqrt{\pi }}{4}=3\sigma ^4$
  Hence
  $\kappa =\frac{((X-\langle X\rangle )^4)}{\sigma ^4}-3=3-3=0$

Evaluation of the gamma function

From the definition of $\Gamma$ and substituting $u=x^2$ :