MGF of standard normal raised to a power

Question

Let $Z \sim \mathcal N(0,1)$ be a standard normal variable. What is the moment generating function $M_{Z^d}(t)$ for the variable $Z^d$ where $d$ is a positive integer?

For $d = 1$, we have $M_Z(t) = e^{t^2/2}$, and for $d = 2$, thinking of the variable as Chi-Squared with $1$ degree of freedom, we have that $M_{Z^2}(t) = \frac{1}{\sqrt{1 - 2t}}$. Is computation of $M_{Z^d}(t)$ tractable for general $d$?

Answer 1

Here's one way to construct a general MGF for $M_{Z^d}(t)$.

Since the moments of the random variable $X$ don't really change, what changes are the closed form solutions of the coefficients in the expansion of the MGF.

For example
\begin{align*} M_Z(t) &= e^{t^2/2}\\ &= 1+\frac{t^2}{2\times1!}+\frac{t^4}{4\times2!}+\frac{t^6}{8\times3!}+\dots++\frac{t^{2i}}{2^i\times{i}!}+\dots\\ \end{align*}

Now of course we can read off the moments of the normal distribution from here. For example, $$\mathbb E [X^{2i}] = \frac{(2i)!}{2^i\cdot i!}$$

For example, $\mathbb E [X^{4}]=3$. Note that $\mathbb E [X^{2i+1}] = 0\quad \forall i$

Then $M_Z^d(t)$ are the moments of $Z^d$ \begin{align*} M_{Z^d}(t) &= 1+ \mathbb E [X^{d}]\cdot\frac{t}{1!}+ \mathbb E [X^{2d}]\cdot\frac{t^2}{2!}+\dots+\mathbb E [X^{id}]\cdot\frac{t^i}{i!}+\dots\\ \end{align*} Now we have two cases, when d is even and d is odd. First consider the case when d is even, ie d = 2p. \begin{align*} M_{Z^d}(t) &= 1+ \mathbb E [X^{2p}]\cdot\frac{t}{1!}+ \mathbb E [X^{4p}]\cdot\frac{t^2}{2!}+\dots+\mathbb E [X^{2ip}]\cdot\frac{t^i}{i!}+\dots\\ &= 1+ \frac{(2p)!}{2^p\cdot p!}\cdot\frac{t}{1!}+ \frac{(4p)!}{2^{2p}\cdot (2p)!}\cdot\frac{t^2}{2!}+\dots+\frac{(2ip)!}{2^{ip}\cdot (ip)!}\cdot\frac{t^i}{i!}+\dots \end{align*} Now consider that $d$ is odd, ie $d = 2p+1$. In this case, we will get only the even terms. \begin{align*} M_{Z^d}(t) &= 1+ \mathbb E [X^{2p+1}]\cdot\frac{t}{1!}+ \mathbb E [X^{4p+2}]\cdot\frac{t^2}{2!}+\dots+\mathbb E [X^{2ip+i}]\cdot\frac{t^i}{i!}+\dots\\ &= 1+ 0+ \frac{(4p+2)!}{2^{2p+1}\cdot (2p+1)!}\cdot\frac{t^2}{2!}+0+\frac{(8p+4)!}{2^{4p+2}\cdot (4p+2)!}\cdot\frac{t^4}{4!}+\dots\\ \end{align*} Check: One can substitute $p=0$ to get back the MGF of the normal distribution.

I'm not sure how to get a general closed form for these (or whether such closed forms exist for all d).

Answer 2

It is worth noting that the moment-generating function $M_{Z^d}(t)$ of $Z^d$ corresponds to the moment-generating function of the Chi-square distribution for $d=2$. Furthermore, for all other $d>2$ where $d$ is even, it is only finite if $t \leq 0$, whereas it is infinite for $t>0$. This is because \begin{align} \mathbb{E} \left[e^{Z^dt} \right] &= \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}} e^{Z^d t -\frac{1}{2}Z^2} dZ \\ &\ge \frac{1}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} e^{Z^d t -\frac{1}{2}Z^2}dZ+ \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{-\frac{1}{2t}^{\frac{1}{d-2}}} e^{Z^d t -\frac{1}{2}Z^2}dZ \\ & \ge \frac{1}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} 1 dZ+ \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{-\frac{1}{2t}^{\frac{1}{d-2}}} 1 dZ \\ &= \frac{2}{\sqrt{2\pi}} \int\limits_{\frac{1}{2t}^{\frac{1}{d-2}}}^{\infty} 1 dZ = \infty, \end{align} which holds because $Z^d t -\frac{1}{2}Z^2 \geq 0$ for all $t>0$ and $Z$ with $\vert Z\vert \geq \left(\frac{1}{2t}^{\frac{1}{d-2}}\right)$.