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Since I know the gamma function

$$ \Gamma(s) = \int_{0}^{\infty} t^{s-1} {\rm e}^{-t}dt\,. $$

Making the change of variables $y=\frac{x^2}{b^2} $ casts the integral to the gamma function

$$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;= \frac{2b}{\pi}\int _{0}^{\infty }\sqrt {y}\,{{\rm e}^{-y}}{dy} = \frac{2b}{\pi} \Gamma(\frac{3}{2}) = \frac{b}{\pi}\Gamma(\frac{1}{2})= \frac{b}{\sqrt{\pi}}\,. $$

But how do I show that

$$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx$$ is a pdf? Since I clearly did not get $1$. Then, how do I find the mean and variance? When I find E(X) do I compute the following? $$\int_0^\infty\frac4{\pi b^2}x^3e^{-x^2/b^2}dx.$$ If so, how do I do this using the gamma distribution? Is there a short cut or do I have to do integration by parts? Please explain.

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$f(x)$ is a p.d.f. if $\int_{-\infty}^{\infty}f(x)=1$ and $f(x) \ge 0$ for all $x$ –  Daniel Littlewood Sep 30 '12 at 18:03
yes this i know, but i integrated it and got b/pi^-1/2 not 1. –  user43126 Sep 30 '12 at 19:00
Therefore it is not a pdf and the problem is solved. What made you think that it would be a pdf in the first place? Did you copy the question correctly? –  Byron Schmuland Sep 30 '12 at 19:06
Then $f$ is not a PDF. You might want to show that $4x^2e^{-x^2/b^2}/(b^3\sqrt\pi)$ is a PDF. –  Did Sep 30 '12 at 19:06
agh! no i didn't copy it down correctly. sorry i am changing it now! –  user43126 Sep 30 '12 at 19:54

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