Thread: Y=1-x^2

The random variable X has density given with
p(x)={
(x+1)/2 , -1<x<1
0, in other cases
}
find the density of the random variable Y=1-X^2

I'm stuck with this problem
any help would be appreciated

Hey dejandenib.

Are you aware of the transformation theorem?

Transformations of Variables

Since f(x) = 1 - x^2 has an inverse in the appropriate domain, you can use this.

yes, I'm aware of that. I used it, and don't know how to solve it completely.
After I integrate it, I have: (1/4)x^2 + x/2 + C, -1<x<1,
and because in the main formula : y=1-x^2, i have x squared, when I calculate the root, I'll have both signs, plus and minus, in front of the square root.
I don't know how to handle these two cases in one.

Originally Posted by dejandenib

The random variable X has density given with
p(x)={
(x+1)/2 , -1<x<1
0, in other cases
}
find the density of the random variable Y=1-X^2

I'm stuck with this problem
any help would be appreciated

Do you have the right answer?my answer comes out to be .If it matches with your answer, post it,i will post the solution.I might have done something wrong.

So I'm given the density (x+1)/2
To find out the function of distribution, I integrate it
integrate &#40;x&#43;1&#41;&#47;2 dx - Wolfram|Alpha
what do I do next?
I dont know what exactly is your answer you are showing me.
I need hint on what to do next after this step.
I know that eventually I'll use Py(x)=|Px(f[-1](x))*|f[-1](x)|
but I'm having problems with the domain, because I have a square in the equation, and the domain of x can be negative.
Thanx for the help so far

Given a r.v. X with pdf f(x) with a r.v. Y = g(X) then:

g(y) = f(g^-1(y)) * |dg^-1(y)/dy|

Now g(x) = 1 - X^2 so g^-1(x) = SQRT(1 - x) and dg^-1(x)/dx = -1*1/2*1/SQRT(1-x)

This information will give you the PDF for Y.

Two questions:
1.
why is g^-1(x) = SQRT(1 - x)
but not = + - SQRT(1-x)
?
2.
after i substitute the values in the formula for the PDF of Y, do I get:
g(y) =[(sqrt(1-x)+1)/2 ]*[1/2*sqrt(1-x)]
which is

Originally Posted by dejandenib

Two questions:
1.
why is g^-1(x) = SQRT(1 - x)
but not = + - SQRT(1-x)
?
2.
after i substitute the values in the formula for the PDF of Y, do I get:
g(y) =[(sqrt(1-x)+1)/2 ]*[1/2*sqrt(1-x)]
which is

AS you can see, I got the same answer too(excuse the typo...) but with a (-) sign in front of it.Do you have the correct answer??

i believe the minus should be ignored because we calculate the ABSOLUTE value of the derivative.

can anyone explain how to handle the case with both signs, plus and minus in front of the sqrt(1-x) ?