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.

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- January 25th 2013, 04:57 PM #1

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- January 25th 2013, 06:03 PM #2

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## 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, 03:46 AM #3

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## 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, 04:57 AM #4

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- January 26th 2013, 05:08 AM #5

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## 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, 05:53 PM #6

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- January 27th 2013, 05:33 AM #7

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## 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, 05:49 AM #8

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- January 27th 2013, 06:59 AM #9

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- January 28th 2013, 03:50 AM #10

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