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

Printable View

- Jan 25th 2013, 04:57 PMdejandenibY=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 PMchiroRe: 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 AMdejandenibRe: 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 AMearthboyRe: Y=1-x^2
- Jan 26th 2013, 05:08 AMdejandenibRe: Y=1-x^2
So I'm given the density (x+1)/2

To find out the function of distribution, I integrate it

integrate (x+1)/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 PMchiroRe: 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 AMdejandenibRe: 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

$\displaystyle (\frac{1}{4}+\frac{1}{4\sqrt{1-x}})$ - Jan 27th 2013, 05:49 AMearthboyRe: Y=1-x^2
- Jan 27th 2013, 06:59 AMdejandenibRe: 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 AMdejandenibRe: 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) ?