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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 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, $0$ 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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    >So I found that the domain of $Y$ is $−3. Check your work. First, the _range_ of $Y$ (i.e., the set of values of $Y$ that can be observed) is what you want, not the _domain_ of $Y$. But even after correcting the nomenclature, consider what is the value of $Y$ when $X = 0$? Shouldn't that be included in the range of $Y$?2012-05-23
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    For starters, $F_X(x)=\frac{1}{9}(x+1)^2$ would be the **CDF**, not the PDF, of $X$ on the interval $(-1,2)$, since it is strictly increasing from $0$ to $1$ on this interval. $Y$ would fall in the interval $(-3,1]$. Try graphing $Y$ versus $X$, as will as their CDFs and PDFs.2012-05-23
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    @bgins Curiously enough, $\frac{1}{9}(x+1)^2\mathbf 1_{[-1,2]}$ is also a valid probability density function.2012-05-23
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    @DilipSarwate: Ahh, of course (not so curious really), since the average height under a parabola from its vertex out to some extent is one third the maximum height, and the extent is three.2012-05-23

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