TITLE: Derived PDF calculation - how do I do substitution between X and Y in this case?
QUESTION [0 upvotes]: Modify from this problem, if I have:
$X$ has a uniform distribution with PDF $f(x) = 1/2$, $x \in [0, 2]$, zero elsewhere.
Find the PDF of $Y = X^2 - 6X$, with $y\geq -9$.
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How do I do substitution between $X$ and $Y$ in this case?  Thanks.

REPLY [1 votes]: $P(Y\leq y)=P(X^2-6X\leq y)=P(X^2-6X-y\leq 0)$. Now the term on the left is an upward opening parabola. It is negative in between any possible roots. Looking at the roots, by quadratic formula,
$x^2-6x-y=0\implies x=\frac{6\pm\sqrt{36+4y}}{2}$. Real roots occur when $36+4y\geq 0$ or $y\geq -9$. Hence, based on your assusmptions, you can assume real roots always occur. There is a double root at $y=-9$ and two real roots when $y>-9$.
Going back to the original problem, $P(X^2-6-y\leq 0)=P(\frac{6-\sqrt{36+4y}}{2}\leq X\leq \frac{6+\sqrt{36+4y}}{2}).$
From here, you should be to solve this. Then just differentiate to get the PDF.

REPLY [0 votes]: Note that $\frac{d}{dx}(x^2-6x)=2x-6$, which is negative for $x\in [0,2]$ so the function is strictly monotone on this interval and the image is therefore $[-8,0]$. We can find it's inverse by solving
$$x^2-6x -y = 0   \Rightarrow x = \frac{6 \pm \sqrt{36 + 4y}}{2}=3\pm\sqrt{9+y}$$
We know that the range of $x$ values is $[0,2]$, and therefore we must have $X=3-\sqrt{9+Y}$
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