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Suppose $X$ has a uniform $[-1,2]$ distribution. Find he density of $X^2$.... What I did was the change of variables, (and I know I'm supposed to do it because it's that section of the textbook) and I first found the range of $Y$ which is $[1,4]$ because were assuming $Y=X^2$ and $x$'s domain is from $-1$ to $4$ and the range would just be the end points. Next I found $f_Y(y)=\frac{f_X(x)}{|\frac{dy}{dx}|}$ which is pretty simple, it came out to be$$f_Y(y)=\frac{1}{6\sqrt y}\quad \forall \quad 1\le y \le 4$$ I know that this is correct because in the back of the book it lists this as the brief answer. However, there is also another part to it that says $$f_Y(y)=\frac{1}{3\sqrt y}\quad \forall \quad 0\lt y \lt 1$$ and I'm not sure where that second part came from. The interval [0,1] isn't even in the range of $Y$ so why would I be looking for it?

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    What is the value of $Y$ if $X = 0.1$?2012-11-05
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    Sketch the plane with coordinate axes $x$ and $y$. Mark on it the points $(−1,0)$ and $(2,0)$, and then sketch the parabola$ y=x^2$ between these points. The point $(X,Y)$ always lies on this segment of the parabola. Then ask yourself: for fixed $y in [0,1]$, what values of the random variable $X$ result in the occurrence of the event $\{Y\leq y\}$? Can you find the probability $P\{Y\leq y\}$ which is $F_Y(y)$ for $y \in [0,1]$? Repeat for fixed $y \in (1,4]$ and note that the values of $X$ that result in $\{Y\leq y\}$ have a different form... Do not rely on magical formulas....2012-11-05
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    I see, thank you for that. I just remembered my professor mentioned this in lecture.2012-11-05

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