So I have a practice question on an example exam, and I am a bit stumped by it:
Suppose that $X \sim N(1,2)$.
Find: $$ E((X−1)^4) $$
and $$ E(X^4) $$
I am a bit confused as to how to proceed. Now, I know it is possible to get these using the moment generating function, but for the normal that function is awfully complicated and it's a pain to take all those derivatives. I've been trying to find shortcuts.I found the following post here on the Math exchange:
Calculate expected values for a normal distribution
Which purports to answer the question. However, I am confused by the last step. This is what they wrote for the solution:
$$ [(X−1)^4−X^4]=[Y^4−(Y+1)^4]=[−4Y^3+6Y^2−4Y+1]=1−4[Y]+6[Y^2]−4[Y^3]=1−4⋅0+6⋅22−4⋅0=25 $$
But how is this actually a solution? This doesn't give you either E(X−1)^4 or E(X^4) individually, it only evaluates the expected value of the difference between them. I understand the algebra involved, but it seems to me this doesn't actually answer the question. Further, searching around I actually can't find ANY proofs online regarding the 4th moment of the normal distribution.
Is it really not possible to evaluate these without the MGF? If not, how do I get E(X−1)^4 or E(X^4) from the "solution" offered at the other thread?
Anyone have any help? Or even any shortcuts for evaluating the MGF derivatives?
Hint: I am not sure whether $2$ is the variance or the standard deviation. (Some use $\sigma^2$ as the second parameter, and some use $\sigma$.)
So to cover all bases, we assume that $X$ has standard deviation $\sigma$. Let $Y=X-1$. Then $Y$ has mean $0$ and standard deviation $\sigma$. If we know $E(Y)$, $E(Y^2)$, $E(Y^3)$, and $E(Y^4)$, we can find anything we may need by using the linearity of expectation.
The calculation of $E(Y)$ and $E(Y^3)$ is no problem, by symmetry they are both $0$.
The calculation of $E(Y^2)$ is no problem either, it is $\text{Var}(Y)+(E(Y))^2$, so it is $\sigma^2$.
For $E(Y^4)$, we need to do some work. Note first that $Y=\sigma Z$, where $Z$ is standard normal. So $E(Y^4)=\sigma^4 E(Z^4)$. We show how to calculate $E(Z^4)$. $$E(z^4)=\int_{-\infty}^\infty z^4 \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\,dz.$$
We integrate by parts, letting $u=z^3$ and $dv=\frac{1}{\sigma\sqrt{2\pi}}ze^{-z^2/2}\,dz$. The integral turns out to be $\int_{-\infty}^\infty kz^2e^{-z^2/2}\,dz$, where $k$ is a constant. We can now find an explicit answer, since $E(Z^2)=1$.