I have a probability issue that i am dealing with now. Maybe you could help me a bit :-) .

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 found that the domain of Y is $-3 < Y < 0$.

I found that distribution of $Y$ is: $0$ when $y < -3$ and $1$ when $y >0$.

At $-3<Y<0$ is :

$$\int_{-\sqrt{1-y}}^{\sqrt{1-y}}f(x) dx = \cdots = [2(1-y)^{\frac{3}{2}} + 6(1-y)^{\frac{1}{2}}]27$$

So, finally the density function of $Y$ is the derivative of $[2(1-y)^{\frac{3}{2}} + 6(1-y)^{\frac{1}{2}}]27 = \cdots = \frac{x-2}{9\sqrt{1-x}}$ at $-3<y<0$, $0$ else.

I think that the general idea is correct, but I am not sure at all for the results, for example maybe my domain is wrong or I might miss a calculation.

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