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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?

  • 1
    Note that $\Pr(Z^2=y)=0$, identically. In fact you might want to debunk thoroughly the misconceptions which led you to write the identity where this quantity appears.2013-03-23
  • 0
    Any luck with my suggestion above?2013-03-29

2 Answers 2