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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>So I found that the domain of $Y$ is $−3<Y<0$. 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 $Y$? –  Dilip Sarwate May 23 '12 at 13:11
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. –  bgins May 23 '12 at 13:13
@bgins Curiously enough, $\frac{1}{9}(x+1)^2\mathbf 1_{[-1,2]}$ is also a valid probability density function. –  Dilip Sarwate May 23 '12 at 13:25
@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. –  bgins May 23 '12 at 13:46

From looking at the graph of $Y$ versus $X$ on the given $X$-domain, the open interval $(-1,2)$,

you can see that $Y\in(-3,1]$ (note that the interval is open at $-3$ but closed at $Y=1$ where $X=0$). The CDF of $Y$ can then be calculated from the CDF of $X$, after calculating that from the integral, as follows: \eqalign{ F_X(x) &=P(X\le x)= \int_{-1}^{x}\tfrac19(1+t)^2\,dt =\tfrac1{27}\left[(1+t)^3\right]_{-1}^{x} =\tfrac1{27}(1+x)^3 \\\\ F_Y(y) &= P(Y\le y)=P(1-X^2\le y)=P(X^2\ge1-y)=P(|X|\ge\sqrt{1-y}) \\\\ &=\left\{\array{ 0&y\le-3\\ P\left(X\ge\sqrt{1-y}\right)\qquad&-3\lt y\le0\\ 1-P\left(|X|\le\sqrt{1-y}\right)\qquad&0\lt y\le1 \\1&y\ge1 }\right. \\\\ &= \left\{\array{ 0&y\le-3\\ 1-F_X\left(\sqrt{1-y}\right)\qquad&y\in(-3,0]\\ 1-F_X\left(\sqrt{1-y}\right)+F_X\left(-\sqrt{1-y}\right)\qquad&y\in(0,1] \\1&y\ge1 }\right. \\\\ &= \left\{\array{ 0&y\le-3\\ 1-\tfrac1{27}\left(1+\sqrt{1-y}\right)^3\qquad&y\in(-3,0]\\ 1-\tfrac1{27}\left[ \left(1+\sqrt{1-y}\right)^3- \left(1-\sqrt{1-y}\right)^3 \right]\qquad&y\in(0,1] \\1&y\ge1 }\right. } which for $y\in(0,1]$ can also be simplified a bit further using $(1\pm r)^3=1\pm3r+3r^2\pm r^3$ thus: $$(1+r)^3-(1-r)^3=2\,(3r+r^3)=2r\,(3+r^2)\qquad\implies$$ $$F_Y(y)=1-\tfrac1{27}2\sqrt{1-y}~(3~+~1-y)=1-\frac{2\sqrt{1-y}\,(4-y)}{27} \quad\text{for}\quad y\in(0,1]$$ or differentiated to get the PDF: \eqalign{ f_Y(y) &=\left\{\array{ 0 & \qquad y\le-3\quad\text{or}\quad y\gt1\\\\ \frac19+\frac{2-y}{18\sqrt{1-y}} & \qquad y\in(-3,0]\\\\ \frac{2-y}{9\sqrt{1-y}} & \qquad y\in(0,1] }\right. } Here is the PDF $f_X(x)$, in red, and CDF $F_X(x)$, in blue, of $X$:

and likewise for $Y$ (the PDF $f_Y(y)$ has a vertical asymptote at $1$):

Note that both CDFs are in fact (not only right- but also left-) continuous, so that it doesn't matter to which case we assign the transition points $-3$, $0$ and $1$; as @Dilip has pointed out, it could be considered better pedagogy to use left-closed, right-open intervals to emphasize that (CDF) distributions must be right-continuous.

The key step in my method is being able to replace the probabilities with differences of the (cumulative) distribution function for $X$ in the middle bracketed RHS above during the derivation of $F_Y$. This technique is known as the distribution function method or method of distribution functions. Alternate methods exist, for example integrating the product of $f_X$ with the derivative of the transformation function (or for multivariate transformations, the Jacobian). There is also a method using moment generating functions.

Again starting with the graph $Y=1-X^2$ above, if we approach it from an integral involving the PDF, as you do, we can still start as I did above, up to the first bracketed RHS involving probabilities:

\eqalign{ F_Y(y)&=P\left(X\ge\sqrt{1-y}\right) \qquad\text{for}\qquad-3\lt y\le0\\ &=\int_{\sqrt{1-y}}^2f_X(x)\,dx =\int_{\sqrt{1-y}}^2\tfrac19(1+x)^2\,dx\\ &=\tfrac1{27}\left[(1+x)^3\right]_{\sqrt{1-y}}^2 =1-\tfrac1{27}\left(1+\sqrt{1-y}\right)^3 } as above, and similarly for $y\in(-3,0]$. The key insight is still that $$Y=1-X^2 \le y \iff X^2 \ge 1-y \iff |X| \ge \sqrt{1-y},$$ which then must be handled seperately, depending on the sign of $x$.

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You need some additional lines in your answer. The CDF must have value $0$ for $y \leq -3$ and value $1$ for $y > 1$ and your expressions don't meet this requirement. –  Dilip Sarwate May 23 '12 at 13:56
@DilipSarwate: thanks, although hopefully that is obvious once we know that the support of $Y$ is $(-3,1]$! –  bgins May 23 '12 at 14:11
Isn't your revised final answer for the CDF complex-valued for $y > 1$ since it works out to be $$\frac{1}{27}\left[ 1 -\left(1+\sqrt{1-y}\right)^3 \right]$$ for $y > 1$ and so $1-y < 0$, $\sqrt{1-y} = i\sqrt{y-1}$? –  Dilip Sarwate May 23 '12 at 14:54
Hey guys, alright, i got the domain part... obviously i forgot the case where X=0 therefore -3 < Y < 1 . But seriously i cant understand the poseter above. My answer is very simple. I will find first the distribution of Y using the the PDF of X. in this way: Integral -root(1-y) to root(1-y) f(x) dx. And what i find ill differate it to get the PDF of Y. I cant understand the above poster , to be honest. –  Jane Fon. May 23 '12 at 16:06
Bah, another question, this one's solution is part on Complex numbers ? –  Jane Fon. May 23 '12 at 16:57