Calculate expected values for a normal distribution Suppose that X ~ N(1,2)
Find:
$E(X-1)^4
and  E(X^4)$
I have no idea how to get started. Do I just need to integrate the pdf of the normal distribution multiplied by what is between the brackets? If so, how is this done (these seem very complex integrals). Thanks in advance.
 A: Note that $X^4$ cancels in the expansion.
Convert to 0-mean normals $Y \sim N(0,\sigma=2)$ and use the symmetry of pdf of $Y$ around $0$ to conclude that $\mathbb{E} \left[Y^k\right]=0$ for all odd $k$. That would only leave you with $\mathbb{E} \left[Y^2\right]$ term, but $\mathbb{E} \left[Y^2\right] = \sigma^2$.
Further hint To convert to $0$-mean. note that if $\mathbb{E}[X]=\mu$, then $\mathbb{E}[X-\mu]=0$.
Full Solution Let $Y = X-1$ and note that $$\mathbb{E}[Y] = \mathbb{E}[X-1] = \mathbb{E}[X]-1 = 1-1 = 0,$$ and the variance of $Y$ is unaffected by translation, so it still is $2$. Thus, $Y \sim \mathcal{N}(0,2)$ and pdf of $Y$, $f_Y(y)$ is symmetric around $y=0$. Note by symmetry that
$$
\mathbb{E}[Y] = \int_{-\infty}^\infty y f_Y(y) dy = 0
\mathbb{E}[Y^3] = \int_{-\infty}^\infty y^3 f_Y(y) dy = 0
$$
and 
$$\mathbb{E}[Y^2] = var(Y) = 2^2 = 4.$$
Now we are looking for
$$\begin{split}
\mathbb{E}[(X-1)^4-X^4]
 &= \mathbb{E}[Y^4-(Y+1)^4] \\
 &= \mathbb{E}[-4Y^3+6Y^2-4Y+1] \\
 &= 1 -4 \mathbb{E}[Y] + 6 \mathbb{E}[Y^2] -4 \mathbb{E}[Y^3]\\
 &= 1 -4\cdot 0 + 6\cdot 2^2 - 4 \cdot 0 \\
 &= 25.
\end{split}
$$
