# Y=1-x^2

• Jan 25th 2013, 04:57 PM
dejandenib
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
• Jan 25th 2013, 06:03 PM
chiro
Re: Y=1-x^2
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.
• Jan 26th 2013, 03:46 AM
dejandenib
Re: Y=1-x^2
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.
• Jan 26th 2013, 04:57 AM
earthboy
Re: Y=1-x^2
Quote:

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 $-(\frac{1}{4}+\frac{1}{4\sqrt{1-y}})$ .If it matches with your answer, post it,i will post the solution.I might have done something wrong.
• Jan 26th 2013, 05:08 AM
dejandenib
Re: Y=1-x^2
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
• Jan 26th 2013, 05:53 PM
chiro
Re: Y=1-x^2
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.
• Jan 27th 2013, 05:33 AM
dejandenib
Re: Y=1-x^2
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
$(\frac{1}{4}+\frac{1}{4\sqrt{1-x}})$
• Jan 27th 2013, 05:49 AM
earthboy
Re: Y=1-x^2
Quote:

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
$(\frac{1}{4}+\frac{1}{4\sqrt{1-x}})$

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??(Thinking)
• Jan 27th 2013, 06:59 AM
dejandenib
Re: Y=1-x^2
i believe the minus should be ignored because we calculate the ABSOLUTE value of the derivative.
• Jan 28th 2013, 03:50 AM
dejandenib
Re: Y=1-x^2
can anyone explain how to handle the case with both signs, plus and minus in front of the sqrt(1-x) ?