Density Functions

janvdl · #1 $Old$ May 15th, 2008, 06:54 AM

Quote:

A particle of mass $M$ has a random velocity, $V$ , which is normally distributed with parameters $\mu = 0$ and $\sigma$ . Find the density function of the kinetic energy.

I've asked this before in a previous post, however, while not quite necessary then, the answer did not include full workings, and I think I sort of need that just now to make sure I understand this.
I'm writing another semester test tomorrow, so I would appreciate if someone could show me how this is done.

(Not sure whether it would be classified as "bumping" to bring up the old topic again, I decided to just as well post it in a new thread. Please show full workings.)

Thanks in advance guys

mr fantastic · #2 $Old$ May 15th, 2008, 07:23 AM

Quote:

Originally Posted by janvdl $View Post$

I've asked this before in a previous post, however, while not quite necessary then, the answer did not include full workings, and I think I sort of need that just now to make sure I understand this.
I'm writing another semester test tomorrow, so I would appreciate if someone could show me how this is done.

(Not sure whether it would be classified as "bumping" to bring up the old topic again, I decided to just as well post it in a new thread. Please show full workings.)

Thanks in advance guys

Let $F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)$

$\Rightarrow F(x) = \Pr \left( -\sqrt{\frac{2x}{M}} < V < \sqrt{\frac{2x}{M}} \right)$

$= \frac{1}{\sigma \sqrt{2 \pi}} \, \int_{-\sqrt{\frac{2x}{M}} }^{\sqrt{\frac{2x}{M}} } e^{-u^2/(2\sigma^2)} \, du$

$= \frac{2}{\sigma \sqrt{2 \pi}} \, \int_{0}^{\sqrt{\frac{2x}{M}} } e^{-u^2/(2 \sigma^2)} \, du$ .

Therefore $f(x) = \frac{dF}{dx} = \frac{2}{\sigma \sqrt{2 \pi}} \, e^{-x/(\sigma^2 M)} \, \left (\frac{1}{\sqrt{2Mx}} \right) \,$ , $x > 0$

where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative

$= \frac{e^{-x/(\sigma^2 M)}}{\sigma \sqrt{\pi M x}} \,$ , $x > 0$

and f(x) = 0 for x < 0.

$\int_{-\infty}^{+\infty} f(x) \, dx = 1$ so I don't think I've slipped up anywhere here .......

janvdl · #3 $Old$ May 15th, 2008, 07:25 AM

Quote:

Originally Posted by mr fantastic $View Post$

Let $F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)$

$\Rightarrow F(x) = \Pr \left( -\sqrt{\frac{2x}{M}} < V < \sqrt{\frac{2x}{M}} \right)$

$= \frac{1}{\sqrt{2 \pi}} \, \int_{-\sqrt{\frac{2x}{M}} }^{\sqrt{\frac{2x}{M}} } e^{-u^2/2} \, du$

$= \frac{2}{\sqrt{2 \pi}} \, \int_{0}^{\sqrt{\frac{2x}{M}} } e^{-u^2/2} \, du$ .

Therefore $f(x) = \frac{dF}{dx} = \frac{2}{\sqrt{2 \pi}} \, e^{-x/M} \, \left (\frac{1}{\sqrt{2Mx}} \right) \,$ , $x > 0$
where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative

$= e^{-x/M} \, \frac{1}{\sqrt{\pi M x}} \,$ , $x > 0$

and f(x) = 0 for x < 0.

Thank you Mr F, I will work through this now and try to understand