Mathematically finding the bounds of a CDF for a random variable

Question

I am trying to solve this question that asks for the pdf of a new variable V such that V = X*Y. The given information is $\begin{align*} f(x,y)&=\frac{2x+y}{36} & 0 \leq y \leq x, \hspace{5mm} x + 2y \leq 6 \\ \end{align*}$ . My Issue is I can't figure out the bounds/cases for the new variable V, algebraically. When using desmos to graph the bounds and cases, I can clearly see that the cases are $0 \leq v \leq 4$ and $4 \leq v \leq 4.5$. However I am struggling to find these bounds without the use of a graphing aid. My method of solving is to find the CDF of each bound and then take the derivative of each component to find the pdf.

Answer 1

First we figure out for which $(x,y)$ the above pdf takes nonzero values. Notice that $0 \leq x,y$ by the first inequality and that $x \leq 6$ and $y \leq 3$ by the second. Therefore, for any given value of $x \in [0,6]$, the density is nonzero precisely for $y \in [0, \min(x, 3 - \frac{x}{2})]$. Now let us look at the cdf of $V$. Clearly $0 \leq V$ so that it is enough to consider $v \geq 0$: \begin{align} \mathbb{P}(V \leq v) &= \mathbb{P}(XY \leq v)\\ &= \int_\mathbb{R} \int_{\mathbb{R}} f(x,y)\, \mathbb{I}(xy \leq v)\, dy \,dx \\ &= \int_\mathbb{R} \int_\mathbb{R} \frac{2x + y}{36} \mathbb{I}(0 \leq y \leq x,\, x + 2y \leq 6,\, xy \leq v)\, dy\,dx \\ &= \int_0^6 \int_0^{\min(3 - \frac{x}{2}, x, \frac{v}{x})} \frac{2x+y}{36} dy\,dx. \end{align} Now you can try to look at the $3$ possible cases. We have \begin{align} x \leq 3 - \frac{x}{2} \quad&\iff\quad x \leq 2 \\ x \leq \frac{v}{x} \quad&\iff\quad x^2 \leq v \\ \frac{v}{x} \leq 3 - \frac{x}{2} \quad&\iff\quad x \not\in [3 - \sqrt{9-2v}, 3 + \sqrt{9-2v}]. \end{align} From the above you start seeing the different behaviour depending on the value of $v$ that you observed graphically. For example: if $v > 4.5$ then $\frac{v}{x} > 3 - \frac{x}{2}$ for every $x$.