3.4 Problem 84 chapter 8
Problem 90 chapter 4
by Nasser Abbasi 5:20 PM, oct 26, 2007
Problem:
Assuming that \(X \sim N(0,\sigma ^{2})\) use the mgf to show that the odd moments are zero and the even moments are
\(\frac {\left ( 2n\right ) !\ \sigma ^{2n}}{2^{n}\left ( n!\right ) }\)
Answer:
\[ M_{X}\left ( t\right ) =e^{\frac {t^{2}\sigma ^{2}}{2}}\]
First obtain a recursive formula for the moment generation function
\begin{align*} M_{X}^{\left ( 1\right ) }\left ( t\right ) & =t\sigma ^{2}e^{\frac {t^{2}\sigma ^{2}}{2}}\\ M_{X}^{\left ( 2\right ) }\left ( t\right ) & =\sigma ^{2}e^{\frac {t^{2}\sigma ^{2}}{2}}+\left ( t\sigma ^{2}\right ) ^{2}e^{\frac {t^{2}\sigma ^{2}}{2}}\\ & =\sigma ^{2}M_{X}\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( 1\right ) }\left ( t\right ) \\ M_{X}^{\left ( 3\right ) }\left ( t\right ) & =\sigma ^{2}M_{X}^{\left ( 1\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( 2\right ) }\left ( t\right ) +\sigma ^{2}M_{X}^{\left ( 1\right ) }\left ( t\right ) \\ & =2\sigma ^{2}M_{X}^{\left ( 1\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( 2\right ) }\left ( t\right ) \\ M_{X}^{\left ( 4\right ) }\left ( t\right ) & =2\sigma ^{2}M_{X}^{\left ( 2\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( 3\right ) }\left ( t\right ) +\sigma ^{2}M_{X}^{\left ( 2\right ) }\left ( t\right ) \\ & =3\sigma ^{2}M_{X}^{\left ( 2\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( 3\right ) }\left ( t\right ) \end{align*}
Hence
\[ \fbox {$M_{X}^{\left ( r\right ) }\left ( t\right ) =\left ( r-1\right ) \sigma ^{2}M_{X}^{\left ( r-2\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( r-1\right ) }\left ( t\right ) $}\]
Using the above, we generate odd and even moments.
\begin{align*} E\left ( x^{1}\right ) & =M_{X}^{\left ( 1\right ) }\left ( 0\right ) =0\\ E\left ( x^{2}\right ) & =M_{X}^{\left ( 2\right ) }\left ( 0\right ) =\sigma ^{2}M_{X}\left ( 0\right ) =\sigma ^{2}\\ & etc... \end{align*}
odd moments: Proof by induction., See class notes, Math 502 lecture 10/24/07
even moments:
proof by induction.
First show it is true for the base case \(n=1.\)(base case)
From the above we see this is indeed the case because \(E\left ( x^{2}\right ) =\sigma ^{2}\), which is the same as saying \(E\left ( x^{2n}\right ) =\frac {\left ( 2n\right ) !\ \sigma ^{2n}}{2^{n}\left ( n!\right ) }\) when
\(n=1\)
Now assume it is true for some \(n\geq 1\), i.e. assume that
\begin{equation} E\left ( x^{2n}\right ) =\frac {\left ( 2n\right ) !\ \sigma ^{2n}}{2^{n}\left ( n!\right ) }\tag {1}\end{equation}
Then we need to show that the relation is true for \(n+1\) (the next even number), i.e. we need to show
that
\begin{equation} \fbox {$E\left ( x^{2\left ( n+1\right ) }\right ) =\frac {\left ( 2\left ( n+1\right ) \right ) !\ \sigma ^{2\left ( n+1\right ) }}{2^{\left ( n+1\right ) }\left ( \left ( n+1\right ) !\right ) }$}\tag {*}\end{equation}
Use the moment generation recursive formula to show the above. since from deļ¬nition we know
that
\[ E\left ( x^{2\left ( n+1\right ) }\right ) =M_{X}^{2\left ( n+1\right ) }\left ( 0\right ) \]
But we showed that
\begin{equation} M_{X}^{\left ( r\right ) }\left ( t\right ) =\left ( r-1\right ) \sigma ^{2}M_{X}^{\left ( r-2\right ) }\left ( t\right ) +t\sigma ^{2}M_{X}^{\left ( r-1\right ) }\left ( t\right ) \tag {2}\end{equation}
Then replace \(r\) in (2) with \(2\left ( n+1\right ) \) we obtain (and setting \(t=0\))
\begin{align*} M_{X}^{\left ( 2\left ( n+1\right ) \right ) }\left ( 0\right ) & =E\left ( x^{2\left ( n+1\right ) }\right ) \\ & =\left ( 2\left ( n+1\right ) -1\right ) \sigma ^{2}M_{X}^{\left ( 2\left ( n+1\right ) -2\right ) }\left ( 0\right ) \\ & =\left ( 2n+2-1\right ) \sigma ^{2}M_{X}^{\left ( 2n+2-2\right ) }\left ( 0\right ) \\ & =\left ( 2n+1\right ) \sigma ^{2}M_{X}^{\left ( 2n\right ) }\left ( 0\right ) \end{align*}
But \(M_{X}^{\left ( 2n\right ) }\left ( 0\right ) \) in the above is just \(E\left ( x^{2n}\right ) \) which we assumed in (1) to be \(\frac {\left ( 2n\right ) !\ \sigma ^{2n}}{2^{n}\left ( n!\right ) }\), hence the above can be written
as
\[ E\left ( x^{2\left ( n+1\right ) }\right ) =\left ( 2n+1\right ) \sigma ^{2}\frac {\left ( 2n\right ) !\ \sigma ^{2n}}{2^{n}\left ( n!\right ) }\]
But \(\sigma ^{2}\sigma ^{2n}=\sigma ^{2(n+1)}\) and \(\left ( 2n+1\right ) \left ( 2n\right ) !=\left ( 2n+1\right ) !\) hence the above becomes
\begin{equation} E\left ( x^{2\left ( n+1\right ) }\right ) =\frac {\left ( 2n+1\right ) !\ \sigma ^{2(n+1)}}{2^{n}\left ( n!\right ) }\tag {3}\end{equation}
But \(\left ( 2n+1\right ) !=\frac {\left ( 2n+2\right ) !}{\left ( 2n+2\right ) }\)
Hence (3) becomes
\begin{align*} E\left ( x^{2\left ( n+1\right ) }\right ) & =\frac {\left ( 2n+2\right ) !\ \sigma ^{2(n+1)}}{\left ( 2n+2\right ) 2^{n}\left ( n!\right ) }\\ & \\ & =\frac {\left ( 2\left ( n+1\right ) \right ) !\ \sigma ^{2(n+1)}}{2n\times 2^{n}\left ( n!\right ) +2\times 2^{n}\left ( n!\right ) }\\ & \\ & =\frac {\left ( 2\left ( n+1\right ) \right ) !\ \sigma ^{2(n+1)}}{2^{n+1}n\left ( n!\right ) +2^{n+1}\left ( n!\right ) }\\ & \\ & =\frac {\left ( 2\left ( n+1\right ) \right ) !\ \sigma ^{2(n+1)}}{2^{n+1}\left ( n!\left ( n+1\right ) \right ) }\end{align*}
But \(n!\left ( n+1\right ) =\left ( n+1\right ) !\) hence the above becomes
\[ E\left ( x^{2\left ( n+1\right ) }\right ) =\frac {\left ( 2\left ( n+1\right ) \right ) !\ \sigma ^{2(n+1)}}{2^{n+1}\left ( n+1\right ) !}\]
Compare to (*) we see it is the same. QED