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

Results 1 to 10 of 10

- January 25th 2013, 05:57 PM #1

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6

- January 25th 2013, 07:03 PM #2

- Joined
- Sep 2012
- From
- Australia
- Posts
- 4,506
- Thanks
- 871

## 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.

- January 26th 2013, 04:46 AM #3

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6

## 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.

- January 26th 2013, 05:57 AM #4

- Joined
- Feb 2010
- From
- in the 4th dimension....
- Posts
- 122
- Thanks
- 9

- January 26th 2013, 06:08 AM #5

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6

## 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 (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

- January 26th 2013, 06:53 PM #6

- Joined
- Sep 2012
- From
- Australia
- Posts
- 4,506
- Thanks
- 871

- January 27th 2013, 06:33 AM #7

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6

## 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

- January 27th 2013, 06:49 AM #8

- Joined
- Feb 2010
- From
- in the 4th dimension....
- Posts
- 122
- Thanks
- 9

- January 27th 2013, 07:59 AM #9

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6

- January 28th 2013, 04:50 AM #10

- Joined
- Jan 2013
- From
- skopje
- Posts
- 6