Finding the mean of a Maxwell Distribution

Question

Let the velocity of gas particles be modeled by the Maxwell distribution. The probability density function is

$$ f(v) = 4\pi \cdot\left( \frac{m}{2\pi K T} \right) ^ {\frac{3}{2}}v^2\cdot e^{-v^2(m/[2KT])}$$

I found that the mean is $2a \sqrt{\frac{2}{\pi}}$ where $a=\sqrt{\frac{kT}{m}}$ from Wikipedia.

Could you please explain how is the mean obtained?

A source indicates that this is equivalent to a chi distribution with df = 3. But I only understand chi-square and beta distributions. Is there a way to transform the Maxwell distribution to another type of distribution? — user12488, Oct 24 '14 at 04:01
@kolonel Yes, $v$ is the random variable. Sorry - I should have clarified that. — user12488, Oct 24 '14 at 04:27

Satish Ramanathan · Accepted Answer · 2014-10-24T08:55:59.923

1

I have given the hand-written solution of the problem. Let me know if it is legible and clear

enter image description here

edited Oct 24 '14 at 08:55

answered Oct 24 '14 at 08:28

Satish Ramanathan

Thank you very much indeed. I went over your solution twice and got it finally (: – user12488 Oct 24 '14 at 18:58
You are welcome! – Satish Ramanathan Oct 24 '14 at 19:24

1 Answers1