1. change in variable question

Y=sin(x)
let X~uniform[0,pi/2]
compute density of fy(y) for Y.

my attempt
------------
arcsin(Y) = X

fx(arcsinY)
-------------- = sqrt(1-y^2) * fx(arcsinY)
(1/sqrt(1-y^2))

then

/\ arcsin(pi/2) sqrt(1-y^2) * fx(arcsinY) dy
\
.\
\/ arcsin(0)

/\ infinity sqrt(1-y^2) * fx(arcsinY) dy
\
.\
\/ 0

and i am stuck here as i cannot evaluate this, i probably did something wrong earlier...

2. Originally Posted by Sneaky
Y=sin(x)
let X~uniform[0,pi/2]
compute density of fy(y) for Y.

my attempt
------------
arcsin(Y) = X

fx(arcsinY)
-------------- = sqrt(1-y^2) * fx(arcsinY)
(1/sqrt(1-y^2))

then

/\ arcsin(pi/2) sqrt(1-y^2) * fx(arcsinY) dy
\
.\
\/ arcsin(0)

/\ infinity sqrt(1-y^2) * fx(arcsinY) dy
\
.\
\/ 0

and i am stuck here as i cannot evaluate this, i probably did something wrong earlier...
Just like I have previously suggested: The cdf is $\displaystyle \displaystyle G(y) = \frac{2}{\pi} \int_0^{\sin^{-1} y} dy$ and the pdf is $\displaystyle \displaystyle \frac{dG}{dy}$. The support is $\displaystyle 0 \leq y \leq 1$.

fY(y) = 2/(pi*sqrt(1-y^2)) for 0<=y<=1 and equals 0 otherwise.

I don't understand how to get the answer like that.

4. how did u get G(x) and where did u get 2/pi from?

5. Originally Posted by Sneaky
how did u get G(x) and where did u get 2/pi from?
Have you understood any of my posts on this topic?

My expression for G(y) in this thread comes from using exactly the same approach that I explained in more detail here: http://www.mathhelpforum.com/math-he...ge-159842.html. Have you read it? What don't you understand?

As for where the 2/pi comes from, you said that X~uniform[0,pi/2]. Are you familiar with the pdf of a continuous uniform distribution? (I assumed that you were).

Originally Posted by Sneaky
fY(y) = 2/(pi*sqrt(1-y^2)) for 0<=y<=1 and equals 0 otherwise.

I don't understand how to get the answer like that.
Simply differentiate the cdf I gave in my previous reply.

6. Can you explain your steps in your second post as I am very lost in this topic.
edit: i get where 2/pi came from now

so now i don't understand how to evaluate the integral part

you have integral from 0 to arcsin y of dy, how do you evaluate that?

7. Originally Posted by Sneaky
Can you explain your steps in your second post as I am very lost in this topic.
edit: i get where 2/pi came from now
I intended the answer I gave here:

http://www.mathhelpforum.com/math-he...ge-159842.html

to be an example you could follow when trying to answer other similar questions. Please try to apply each step of that answer to the current problem. Show all your work, say where you get stuck.

Originally Posted by Sneaky
[snip]
so now i don't understand how to evaluate the integral part

you have integral from 0 to arcsin y of dy, how do you evaluate that?
You do exactly the same as if you were finding $\displaystyle \displaystyle \int_a^b dy$.

8. thanks i understand now