1. Finding pdf question

I have this random variable X that has a uniform distribution on [-pi, pi]. I want to find the pdf for Y = cosX.

Here is my work so far:

Fy(y)= P( Y < y ) = P ( cos X < y) =*P ( X < cos-1(y) ) + **P ( X > cos-1(y) ) (note: I got * since cos(x) is increasing on [-pi , 0] and ** since cos(x) is decreasing on [0 , pi]).

$=\displaystyle{\int_{-pi}^{cos^{-1}(y) } {\left( {1/2pi} \right)} dx + \int_{cos^{-1}(y)}^{pi} {\left( {1/2pi} \right)} dx}$

I am sure that I can't write this as one integral as the two cos-1(y) are not the same (and if I did it looks like I'll get 1)

Since cos(x) is even I know that somehow these two integrals are equal but I just can't show it.

Can someone please give me a hint?

2. Re: Finding pdf question

Hey JaguarXJS.

You will have to restrict the domain so that the inverse function exists - and note that the inverse function should only go from 0 to +pi for inverse cosine.

3. Re: Finding pdf question

Originally Posted by chiro
Hey JaguarXJS.

You will have to restrict the domain so that the inverse function exists - and note that the inverse function should only go from 0 to +pi for inverse cosine.

chiro, thank you for your reply. It's late here in New York and I'm turning in now. I will look at your response closely tomorrow.

4. Re: Finding pdf question

Originally Posted by chiro
Hey JaguarXJS.

You will have to restrict the domain so that the inverse function exists - and note that the inverse function should only go from 0 to +pi for inverse cosine.

$F y (y)$ $=\displaystyle{\int_{-cos^{-1}(y)}^{pi } {\left( {1/2pi} \right)} du + \int_{cos^{-1}(y)}^{pi} {\left( {1/2pi} \right)} dx}$.
$f y (y)$= [-(d/dy(-arccos(y))-(d/dy(arccos(y))]/(2pi) which I think should be 0. Where did I go wrong?