Pdf of a normal distribution with mean\(\mu \)and variance\({\sigma ^2}\)is
\(f\left( {x|\mu ,{\sigma ^2}} \right) = \frac{1}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}\exp \left[ { - \frac{1}{2}{{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}} \right]\) for \( - \infty < x < \infty \)
Hence
\(\begin{aligned}{}E\left[ {{{\left( {X - \mu } \right)}^{2n}}} \right] &= \int\limits_{ - \infty }^\infty {{{\left( {X - \mu } \right)}^{2n}}} \frac{1}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}\exp \left[ { - {{\left( {\frac{{x - \mu }}{{2\sigma }}} \right)}^2}} \right]dx\\ &= \frac{2}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}\int\limits_\mu ^\infty {{{\left( {X - \mu } \right)}^{2n}}} \exp \left[ { - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}} \right]dx\end{aligned}\)
Let \(y = {\left( {x - \mu } \right)^2}\)
Hence \(x = \frac{{\left( {x - 2\mu } \right) + {\mu ^2}}}{y}\)
\(\frac{{dx}}{{dy}} = \frac{{d\left( {\frac{{\left( {x - 2\mu } \right) + {\mu ^2}}}{y}} \right)}}{{dy}}\)
\(\frac{{dx}}{{dy}} = \frac{1}{{\left( {2{y^{\frac{1}{2}}}} \right)}}\)
Hence \(dx = \frac{{dy}}{{\left( {2{y^{\frac{1}{2}}}} \right)}}\)
Hence expectations can be written as follows,
\(\begin{aligned}{}E\left[ {{{\left( {X - \mu } \right)}^{2n}}} \right] &= \frac{2}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}\int\limits_0^\infty {{y^n}} \exp \left\{ { - \frac{y}{{2{\sigma ^2}}}} \right\}\frac{1}{{2{y^{\frac{1}{2}}}}}dy\\& = \frac{1}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}\int\limits_0^\infty {{y^{n - \frac{1}{2}}}} \exp \left\{ { - \frac{y}{{2{\sigma ^2}}}} \right\}dy\end{aligned}\)
The integrand in this integral is the pdf of a gamma distribution with parameters \(\alpha = n + \frac{1}{2}\) and \(\beta = \frac{1}{{\left( {2{\sigma ^2}} \right)}}\) ,except for the constant factor
\(\frac{{{\beta ^\alpha }}}{{\left| \!{\overline {\, {\left( \alpha \right)} \,}} \right. }}\frac{1}{{{{\left( {2{\sigma ^2}} \right)}^{n + \frac{1}{2}\left| \!{\overline {\, {\left( {n + \frac{1}{2}} \right)} \,}} \right. }}}}\)
Since the integral of the pdf of the gamma distribution must be equal to 1, it follows that
\(\begin{array}{} = \int\limits_0^\infty {{y^{n - \frac{1}{2}}}} \exp \left\{ { - \frac{y}{{2{\sigma ^2}}}} \right\}dy\\ = {\left( {2{\sigma ^2}} \right)^{n + 1}}\left| \!{\overline {\, {\left( {n + \frac{1}{2}} \right)} \,}} \right. \end{array}\)
We know
\(\left| \!{\overline {\, {\left( {n + \frac{1}{2}} \right)} \,}} \right. = \left( {n - \frac{1}{2}} \right)\left( {n - \frac{1}{2}} \right)...\left( {\frac{1}{2}} \right){\pi ^{\frac{1}{2}}}\)
Therefore
\(\begin{aligned}{}E\left[ {{{\left( {X - \mu } \right)}^{2n}}} \right] &= \frac{1}{{{{\left( {2\pi } \right)}^{\frac{1}{{2\sigma }}}}}}{\left( {2{\sigma ^2}} \right)^{n + \frac{1}{2}}}\left( {n - \frac{1}{2}} \right)\left( {n - \frac{1}{2}} \right)...\left( {\frac{1}{2}} \right){\pi ^{\frac{1}{2}}}\\ &= {2^n}\left( {n - \frac{1}{2}} \right)\left( {n - \frac{3}{2}} \right)...\left( {\frac{1}{2}} \right){\sigma ^{2n}}\\ &= \left( {2n - 1} \right)\left( {2n - 3} \right)...\left( 1 \right){\sigma ^{2n}}\end{aligned}\)
Hence the value of \(E\left[ {{{\left( {X - \mu } \right)}^{2n}}} \right]\) is \(\left( {2n - 1} \right)\left( {2n - 3} \right)...\left( 1 \right){\sigma ^{2n}}\)
