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Let X have the probability density function given by:

f(x)=$\frac{4}{\beta^3\sqrt{\pi}}x^2$exp{$\frac{-1}{\beta^2}x^2$} from 0< x < $\infty$

a)Verify that f(x) is a probability density function.

b) Find E(x) and Var(x).

My first impression was to use change of variables u=$x^2$ and integrate the function. The answer should be equal to 1. I started and the whole thing got messy.

And E(x)=$\int$ xf(x)dx. I got stuck with this too.

J.R.
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1 Answers1

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Hint

Your life would be easier if, for the integration process, you change variable $x=\beta y$. So, $$I=\int \frac{4 x^2 e^{-\frac{x^2}{\beta ^2}}}{\sqrt{\pi } \beta ^3}dx=\int\frac{4 e^{-y^2} y^2}{\sqrt{\pi }}dy$$ Do one integration by parts and find $$I=\text{erf}(y)-\frac{2 y e^{-y^2} }{\sqrt{\pi }}$$

Claude Leibovici
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  • This definitely helps. I tried it out and got 1. The interesting thing about the E(x) is the cubic function and how to integrate. Does it follow from integration by parts too. And E($x^2$) would have $x^4$ in it. – J.R. Oct 02 '14 at 13:51