find E($\bar{Y^4})$ by using moment generating function for a normal distribution with mean μ and variance 1.

Let $Y_1, Y_2, . . . , Y_n$be a random sample from a normal distribution with mean μ and variance 1.

I would like to find E($\bar{Y^4})$ by using moment generating function.

The setup I have right now is the following:

E($\bar{Y^4}) = \int_{-\infty}^\infty \frac{\bar{Y}^4}{2\pi^{n/2}} e^{-\sum_{i=1}^n y_i^2 - 2\mu n\bar{Y} + n\mu^2}$

First of all, is the setup correct? If so, I am wondering how to get rid of $y_i$?

• Assuming that $Y_1,\ldots,Y_n$ are independent, what distribution does $\bar{Y}$ follow? Do you know its moment-generating function? And how can you extract ${\rm E}[\bar{Y}^4]$ using only the moment-generating function of $\bar{Y}$? – Stefan Hansen Apr 21 '14 at 13:11
• $\bar{Y}$ follows normal distribution right? If so, do we apply $$e^{(\mu t + t^2\sigma^2/2)}?$$ – afsdf dfsaf Apr 21 '14 at 13:54
• Yes. And the moment-generating function of a normal distributed random variable with mean $\mu$ and variance $\sigma^2$ is as you state. So find the mean and variance of $\bar{Y}$ and plug them in. Then you can calculate moments by differentiating the MGF, see e.g. here. – Stefan Hansen Apr 21 '14 at 14:00
• what should I plug in for t? – afsdf dfsaf Apr 21 '14 at 14:50
• The MGF is a function of $t$. You shouldn't plug anything in for $t$. You do know what an MGF is, right? – Stefan Hansen Apr 21 '14 at 14:58