Probability density function

**shinn** · March 17th 2010, 04:55 AM

Hi, I have this question that I've been struggling with.

Given $D$ is a random distance with probability density function $f_d$ and expected value $\mu$ , then the sampled distance distribution has probability density function $g$ , where

$g(x)=\frac{xf_d(x)}{\mu}, x \geq 0$

Find the form of the distance probability density function $f_d$ and its expected value $\mu$ if the sample distances are fitted by the probability distance function $g(x)=\frac{x^2 e^{-\frac{x}{\alpha}}}{2 \alpha^3}, x \geq 0$ and $\alpha = 8.1$ .

Anyone help will be appreciated. thanks.

I have tried equating the two g(x) but I keep getting something ridiculous like showing $E_d(X) = \mu$

**Moo** · March 17th 2010, 10:42 AM

Hello,

Do you agree that we have $\frac{f_d(x)}{\mu}=\frac{xe^{-x/\alpha}}{2\alpha^3}$

Then $f_d(x)=\frac{\mu}{2\alpha^3}\cdot xe^{-x/\alpha}$ , for any positive x.

But we want $f_d(x)$ to be a pdf. So its integral over 0 to infinity should be 1. And this computation will give you $\mu$

Any further question ?

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