# Change in variable question

#### 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...

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#### vela

Staff Emeritus
Homework Helper
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 fx(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 FY(y) first and then differentiate it to find fY(y). To find FY(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 fX(x), you can work out P(X≤arcsin y) to find FY(y). Once you have that, differentiate it to find fY(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.

Last edited:

"Change in variable question"

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