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Consider random variable $X$ and its distribution $f = \dfrac{3x^2}{2}$ where $-1 < x <1$

What is the distribution of $Y= |X|$? What about $V=X^2$? And $W=X^3$?

I basically did the following $F(y) = P(Y \leq y) = P(|X|\leq y) = P(-y \leq X \leq y) = \int_{-y}^{y}3x^2/2 dx = y^3$

Is the first one even right? I am worried about the $-y$ since the bounds on $x$ is $-1 < x <1$, do I need to substitute the end points? How would I do it if I were to use transformations? Example I would try $Y = X$ and $Y=-X$. Split them up and find their distribution and then add them. I eventually got to $f_x(y)(y)' = \dfrac{3y^2}{2}$. So multiply by 2 and I get $3y^2$, which contradicts my other answer...

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\mid is a binary operator symbol, used e.g. to denote divisibility. The symbol for a norm bar is simply |. The two lead to different spacing and shouldn't be used in place of each other. – joriki Dec 4 '12 at 2:37
up vote 1 down vote accepted

Your first procedure is correct. The function F you have found is the CDF of Y, so to get the pdf you'll have to take the derivative, which gives you: $$ f_{Y}(y)=3y^{2} $$ which is the same result that you have found with the second method.

You'll have to also determine where the new distribution is defined. This is given by the range of the function you are applying to the X, e.g. for |X| this becomes $$ x \epsilon [0, 1) $$

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How did you get the range from $0$ to $1$? – Hawk Dec 4 '12 at 18:25
@sizz, that's the range of the function modulus when applied to the domain of the original variable X. In other words the function modulus transforms (-1, 1) into [0, 1) – darioses Dec 4 '12 at 22:58

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