Let $X$ and $Y$ be random variables with joint pdf:
$$f(x,y)=x + y \quad \text{if } x \ge 0, y \le 1$$
Let $Z=XY$. Calculate the pdf of $Z$.
I'm a bit confused about solving this problem, I'm trying to get to the pdf by calculating the cdf to derive it afterwards, so I know that the cdf of $Z$ would be something like this:
$$F(XY \le z) = \iint(x+y) \,dydx.$$
But I'm not so sure how to would the limits of the definite would be...Im guessing it's:
$$F(XY \le z)= \int_0^\infty\int_{-\infty}^{z/x}(x+y) \,dy dx.$$
But this integral's result is divergent, so I know something is wrong but I'm a bit lost there. Is there a better approach on solving it? Any thoughts?
I appreciate any help!
So indeed, if $Z = XY$, let $\mathbb{I}(A)$ denote the indicator of the event $A$ (i.e. $\mathbb{I}(A)=1$ if $A$ is true and is $0$ otherwise). You have $$ \begin{split} F_Z(z) &= \mathbb{P}[Z \le z] \\ &= \mathbb{P}[XY \le z] \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \mathbb{I}(xy \le z) f_{X,Y}(x,y)\ dxdy \\ &= \int_{x=-\infty}^{x=0} \int_{y=z/x}^{y=\infty} f(x,y) dy dx + \int_{x=0}^{x=\infty} \int_{y=-\infty}^{y=z/x} f(x,y) dy dx \end{split} $$
UPDATE
Sorry, I missed that you gave the definition for $f(x,y) = x+y$ for $x \ge 0$ and $y \le 1$. I don't understand how this is a valid pdf -- you must have $$ 1 = \int_{x=0}^{x=\infty} \int_{y = -\infty}^{y=1} (x+y)dxdy $$ but the RHS integral diverges...
UPDATE 2
I think the intent is to have $f(x,y)=x+y$ for $0 \le x,y \le 1$, which means $0 \le x \le 1$ and $0 \le y \le 1$. Indeed, $$ \begin{split} \int_0^1 \int_0^1 (x+y)dxdy &= \int_0^1 \left[y + \left(\int_0^1 x dx\right) \right]dy\\ &= \int_0^1 \left[y + \frac12 \right]dy \\ &= \int_0^1 y dy + \frac12 \\ &= \frac12 + \frac12 \\ &= 1. \end{split} $$ Then, $$ \begin{split} F_Z(z) &= \int_{x=0}^{x=1} \int_{y=0}^{y=\min\{z/x,1\}} (x+y) dy dx \end{split} $$ Can you now finish?