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I need help with the following problem. Suppose $Z=N(0,s)$ i.e. normally distributed random variable with standard deviation $\sqrt{s}$. I need to calculate $E[Z^2]$. My attempt is to do something like \begin{align} E[Z^2]=&\int_0^{+\infty} y \cdot Pr(Z^2=y)dy\\ =& \int_0^{+\infty}y\frac{1}{\sqrt{2\pi s}}e^{-\frac y{2s}}dy\\ =&\frac{1}{\sqrt{2\pi s}}\int_0^{\infty}ye^{-\frac y{2s}}dy. \end{align}

By using integration by parts we get

$\int_0^{\infty}ye^{-\frac y{2s}}dy=\int_0^{+\infty}2se^{-\frac y{2s}}dy=4s^2.$

Hence $E[Z^2]=\frac{2s\sqrt{2s}}{\sqrt{\pi}},$ which does not coincide with the answer in the text. Can someone point the mistake?

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    Any luck with my suggestion above?2013-03-29

2 Answers 2

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This is old, but I feel like an easy derivation is in order.

The variance of any random variable $X$ can be written as $ V[X] = E[X^2] - (E[X])^2 $

Solving for the needed quantity gives $ E[X^2] = V[X] + (E[X])^2 $

But for our case, $E[X] = 0$, so the answer of $\sigma^2$ is immediate.

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The answer is $s = \sigma^2$. The integral you want to evaluate is

$E[Z^2] = \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{\infty} dz \: z^2 \exp{(-\frac{z^2}{2 \sigma^2})}$

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    Your method is wrong because you misunderstand the definition of expected value. The expected value of a random variable $Z$ having probability distribution $f(z)$ is $\int_{-\infty}^{\infty} dz \: z f(z) $. For the expected value of a function $g(Z)$, the result is $\int_{-\infty}^{\infty} dz \: g(z)f(z) $.2012-12-23