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Suppose that X ~ N(1,2)


$E(X-1)^4 and E(X^4)$

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.

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Is there anyone who can help please? – dreamer Mar 19 '13 at 17:00
up vote 3 down vote accepted

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

$$ \mathbb{E}[Y] = \int_{-\infty}^\infty y f_Y(y) dy = 0 \mathbb{E}[Y^3] = \int_{-\infty}^\infty y^3 f_Y(y) dy = 0 $$


$$\mathbb{E}[Y^2] = var(Y) = 2^2 = 4.$$

Now we are looking for $$\begin{split} \mathbb{E}[(X-1)^4-X^4] &= \mathbb{E}[Y^4-(Y+1)^4] \\ &= \mathbb{E}[-4Y^3+6Y^2-4Y+1] \\ &= 1 -4 \mathbb{E}[Y] + 6 \mathbb{E}[Y^2] -4 \mathbb{E}[Y^3]\\ &= 1 -4\cdot 0 + 6\cdot 2^2 - 4 \cdot 0 \\ &= 25. \end{split} $$

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Thanks for your help! Could you please show me the full answer though because I tried, but I still do not understand it fully. Thanks a lot in advance. – dreamer Mar 18 '13 at 19:50
@mause I wrote up a full solution for you. – gt6989b Mar 19 '13 at 21:31
Thanks a lot for your help! – dreamer Mar 21 '13 at 15:59
No problem, glad to be of help. – gt6989b Mar 21 '13 at 16:01

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