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 $d$?
Best Answer
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 $\vert Z\vert \geq \left(\frac{1}{2t}^{\frac{1}{d-2}}\right)$.