Calculating Density of Y for a Given Function

[sneaky666]

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...

 

[vela]

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))

You're pretty much done at this point (though you need to check your algebra/calculus). The idea is that

$$f_X(x)\,dx = f_Y(y)\,dy$$

so that

$$f_Y(y) = \frac{f_X(x)}{|dy/dx|}$$

Then it's just a matter of expressing the righthand side in terms of y. You can simplify your result a bit by evaluating what f_x(arcsin Y) is equal to.

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...

You're probably thinking of an alternate approach where you find the cumulative distribution function F_Y(y) first and then differentiate it to find f_Y(y). To find F_Y(y), you need to identify what Y≤y means in terms of X. In this case, you'd have

$$F_Y(y) = P(Y\le y) = P(X \le \arcsin y)$$

Since you know f_X(x), you can work out P(X≤arcsin y) to find F_Y(y). Once you have that, differentiate it to find f_Y(y).

If you can't tell what the error I'm alluding to in your first attempt, you might try using this method and seeing what answer you get. The difference might help you identify where you went wrong in the first attempt.