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Please correct my answer (Probability)

I have to calculate the density function of the random variable $Y= 1-X^2$, given that: $f(x) = \frac{1}{9}(x+1)^2$, where $-1 < x < 2$.

So I finally understood that the domain of Y is $-3 < Y < 1$.

But it seems that i cant continue further... My method is to find the distribution of Y by using the density function of X , and finally ill find the density function of Y by taking the derivative of Y's distribution func.

Its not working at all, i am taking wrong integral, wrong spaces, everything wrong...

Could someone please explain me what is the correct way?

Thanks

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    Draw the graph and you see that $Y$ is a non-one-to-one function of $X$, and that's where the difficulty lies, and that's why the thing comes out as a piecewise defined density function.2012-05-24

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It is easiest to start with CDF: $ F_Y(y) = \mathbb{P}(Y \leqslant y) = \mathbb{P}(1-X^2 \leqslant y) = \mathbb{P}(X^2 \geqslant 1 - y) = \mathbb{P}(X \geqslant \sqrt{1 - y}) + \mathbb{P}(X \leqslant -\sqrt{1 - y}) $ Since $F_X(x) = \begin{cases} 0 & x < -1 \\ 1 &x > 2 \\ \left(\frac{x+1}{3}\right)^3 & -1 \leqslant x \leqslant 2 \end{cases}$ one easily arrives as $F_Y(y)$ $ F_Y(y) = F_X(-\sqrt{1-y}) + 1 - F_X(\sqrt{1-y}) = \begin{cases} 1 & y \geqslant 1 \\ 0 & y \leqslant -3 \\ \frac{23 + 3 y}{27} + \frac{y-4}{27} \sqrt{1-y} & y \leqslant 0 \\ 1 + 2 \sqrt{1-y} \frac{y-4}{27} & 0 < y < 1 \end{cases} $ Now, differentiating with respect to $y$: $ f_Y(y) = \begin{cases} \frac{\left(1 + \sqrt{1-y}\right)^2}{18 \sqrt{1-y}} & -3 < y \leqslant 0 \\ \frac{2-y}{9 \sqrt{1-y}} & 0 < y< 1 \\ 0 & \text{otherwise}\end{cases} $

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    bgins has corrected his answer and now it matches yours.2012-05-24
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Used wrong density function.${}{}{}{}{}$

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    @Dilip Nice to know.2012-05-24