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#1
09-16-2016, 09:01 PM
 jchaney Member CAS Join Date: Oct 2014 College: Univ. of Washington Posts: 96
Finan 38.5

Gas molecules move about with varying velocity which has, according to the
Maxwell- Boltzmann law, a probability density given by

$f(v) = cv^2e^{\beta v^2}$

The kinetic energy is given by $Y=E= \frac{1}{2}mv^2$ where m is the mass. What is the density function of Y ?

I simplify the Y equation as $\frac{2y}{m} = v^2$ and then just sub it in with and multiple by the abs d/dx. But, I get a different solution.

Is there something wrong with the method? In a pdf transformation, do I need to simplify $v^2$ to just $v$?
Now that I think about I bet the square factor changes the derivation.

Correct solution is:

$\frac{c}{m}{(\frac{2y}{m})}^.5e^{\frac{2\beta y}{m}}$

Last edited by jchaney; 09-16-2016 at 09:34 PM..
#2
09-16-2016, 09:51 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,528

It one to one so write v in terms of y and substitute. Then calculate |dv/dy| in terms on y and multiply.
#3
09-16-2016, 10:20 PM
 jchaney Member CAS Join Date: Oct 2014 College: Univ. of Washington Posts: 96

Quote:
 Originally Posted by Academic Actuary It one to one so write v in terms of y and substitute. Then calculate |dv/dy| in terms on y and multiply.
Are you saying simplify further to ${(\frac{2y}{m})^.5}=v$ ? I understand this method. I just wondered if there was a way to solve f(y) to f(v^2). But, it looks like it's probably just too messy. Thank you for the reply.