Moment Generating Function for $\bar{Y}^4$ where $Y_i$ is i.i.d. Normal [closed]

Question

I am trying to find $\mathrm{E}[\bar{Y}^4]$ where $Y_1, \dots, Y_n$ constitute a random sample from a normal distribution with mean $\mu$ and variance 1. I was hoping to use the mgf of $\bar{Y}$ to solve this, but I am having a hard time finding it.

What is the mgf of $\bar{Y}$ and why?

Answer 1

The mgf of $Y_i\sim\mathscr N(\mu, 1)$ is $\psi_i(t)=\exp \left(\mu t+\frac 1 2 t^2\right)$. The mgf of $Y_1+Y_2+...+Y_n$ is the product of their individual moment generating functions, $\psi(t)=\prod\exp\left(\mu t+\frac 1 2 t^2\right)=\exp\left(n(\mu t+\frac 1 2 t^2)\right)$. The mgf of $\bar Y$ is $\psi(t/n)=\exp\left(\mu t+\frac 1 2\frac {t^2} n\right)$. (Alternatively, note that $\bar Y\sim \mathscr N(\mu, 1/n)$ so that the same mgf is achieved.)

$$\begin{split}\psi'(t)&=\left(\mu+\frac t n\right)\exp\left(\mu t+\frac {t^2}{2n}\right)\\ \psi''(t)&=\left(\mu+\frac {t}{n}\right)^2\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 1 n\exp\left(\mu t+\frac {t^2} {2n}\right)\\ \psi'''(t)&=\left(\mu+ \frac t n\right)^3\exp\left(\mu t+\frac {t^2} {2n}\right)+2\left(\mu+\frac t n\right)\frac 1 n\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 1 n\left(\mu+\frac t n\right)\exp\left(\mu t+\frac {t^2} {2n}\right)\\ &=\left(\mu+\frac t n\right)^3\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 3 n\left(\mu+\frac t n\right)\exp\left(\mu t+\frac {t^2} {2n}\right)\\ \psi^{(4)}(t)&=\left(\mu+\frac t n\right)^4\exp\left(\mu t+\frac {t^2} {2n}\right)+3\left(\mu+\frac t n\right)^2\frac 1 n\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 3 n\frac 1 n\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 3 n\left(\mu+\frac t n\right)^2\exp\left(\mu t+\frac {t^2} {2n}\right)\\ &=\left(\mu+\frac t n\right)^4\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 6 n\left(\mu+\frac t n\right)^2\exp\left(\mu t+\frac {t^2} {2n}\right)+\frac 3 {n^2}\exp\left(\mu t+\frac {t^2} {2n}\right)\end{split}$$

where we just applied the product rule a few times and grouped some like terms together. Therefore $$E(\bar Y^4)=\psi^{(4)}(0)=\mu^4+\frac {6\mu^2}n+\frac 3 {n^2}$$