# Given the normal distribution X=N(1,2), what is E[X^3]? [duplicate]

Given $X \sim N(1,2)$, what is $E[X^3]$?

I've been told I should use the moment generating function, but I'm not sure how to apply it here in this instance.

Using the given mean and variance, I subbed the values into the normal density function $f(x)$, then tried to compute $E[X^3]$ by integrating $\int_{-\infty}^{\infty} f(x) \, x^3 \mathrm{d}x$ . At this point, I have no idea how to compute the integral, which leaves me pondering if I've veered way off track. I'd appreciate any help!

• One can do this computation with or without using the MGF, but I assume you've been given this problem to help you understand what an MGF is. Do you know what the MGF is for a normal distribution? Do you know how MGFs work in general? – Gordon Smyth Feb 6 '17 at 6:46

You can use the fact that $E[Z^3] = 0$ where $Z\sim\mathcal{N}(0,1)$. That is, normal distribution has zero skewness(or symmetric). You can relate your $X$ and $Z$ by $X=\sqrt{2}Z+1$ and you have $$E[X^3]=E[(\sqrt{2}Z+1)^3] = E[2\sqrt{2}Z^3+6Z^2+3\sqrt{2}Z+1]=7$$ since $E[Z]=0$ and $E[Z^2]=1$.
If a moment-generating function exists for a random variable $X$, then the $r^{th}$ moment of $X$ can be found by evaluating the $r^{th}$ derivative of the moment-generating function at $t = 0$.
$E(X^3)=\dfrac{d^3}{dt^3}M_X(0)$
For the Normal Distribution, $M_X(t)=\exp(\mu t + \dfrac{\sigma^2 t^2}{2})$