The question I have to do is essentially this:

A distribution, X, is modelled by $\displaystyle f(x)= \frac{x}{\sigma^2}e^{-x^2/2\sigma^2},\ x\ge0. $ Show that $\displaystyle E(X)=\sigma \sqrt{\frac{\pi}{2}}$ using the gamma function and its properties.

My attempt:

Using $\displaystyle E(X)=\int_a^bxf(x)\ dx: $

$ \displaystyle =\int_0^\infty x(\frac{x}{\sigma^2}e^{-x^2/2\sigma^2})\ dx$

$ \displaystyle =\int_0^\infty \frac{x^2}{\sigma^2}e^{-x^2/2\sigma^2}\ dx$

Let $\displaystyle u=\frac{x^2}{2\sigma^2}.$ When $x=\infty,u=\infty$ and when $x=0, u=0$ so our limits of integration are the same.

Also $\displaystyle \frac{du}{dx}=\frac{x}{\sigma^2}$

$\displaystyle \implies du=\frac{x}{\sigma^2}dx$

So $ \displaystyle \int_0^\infty \frac{x^2}{\sigma^2}e^{-x^2/2\sigma^2}\ dx = \displaystyle \int_0^\infty xe^{-u}\ du$

which isn't very helpful and not in the gamma function form so I can't substitute anything. Different change of variables maybe?


\begin{align} \int_0^\infty \frac{x^2}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}} \, dx &= \int_0^\infty (2u) e^{-u} \frac{\sigma^2}{x} \, du \\ &= \int_0^\infty\frac{2u\sigma^2}{x}e^{-u}\, du \\ &= \int_0^\infty\frac{2x\sigma^2}{2\sigma^2}e^{-u}\, du \\ &= \int_0^\infty xe^{-u}\, du \\ &= \sqrt2 \int_0^\infty \sigma u^\frac12 e^{-u}\, du \\ &= \sqrt2 \sigma \Gamma(1+0.5) \\ &=\sqrt2 \sigma \cdot\frac{2!}{4}\cdot\sqrt{\pi} \\ &=\sigma \sqrt{\frac{\pi}{2}} \end{align}

  • $\begingroup$ How did you get from the line with $xe^{-u}$ to the $\sqrt2$ line? $\endgroup$ – Sonjov Aug 23 '18 at 6:51
  • $\begingroup$ $u = \frac{x^2}{2\sigma^2}$, take square root on both sides to express $x$ in terms of $u$. $\endgroup$ – Siong Thye Goh Aug 23 '18 at 6:53
  • $\begingroup$ Ah yep, thank you $\endgroup$ – Sonjov Aug 23 '18 at 6:54

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