Central moments versus moments of a normal random variable

Question

Is this statement correct:

$\mathbb{E}[(X_1-\mu)^4]$ of $X_1 \sim N(\mu,\sigma^2) $ is the same as $\mathbb{E}[X_2^4]$ of $X_2 \sim N(0,\sigma^2)$?

If so is there an easy way to show this?

I used the moment generating function for the normal random variable, but it is hugely "inefficient" (in that there are a lot of calculations and it is very easy to make a mistake; but I noticed a recurrence when I was differentiating, so there is probably a nice formula) when I consider the case when $\mu \neq 0$.

And is this generally true for other moments?

Answer 1

Let $Y = X_1-\mu$. Then using a one-to-one transformation we have $dy/dx = 1$, $X_1 = Y+\mu$, and $$f_Y(y) = \frac{f_X(y+\mu)}{|dy/dx|} = \frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{1}{2}\left(\frac{y+\mu-\mu}{\sigma}\right)^2\right\}=\frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac12\left(\frac{y}{\sigma}\right)^2\right\}.$$ Hence $Y\sim N(0,\sigma^2)$ and so $E[(X_1-\mu)^4] = E[Y^4] = E[X_2^4]$ since $Y\overset{d}= X_2$.

Further, $Z = (X_2/\sigma)\sim N(0,1)$, and so $$E[X_2^4] = E[(\sigma\cdot X_2/\sigma)^4]=\sigma^4E[Z^4]$$ which should be familiar.

Answer 2

You can also use the relationship between $X_1$ and $X_2$:

$$X_1 = \mu + X_2.$$