# Whats the joint PDF of Z=XY given the a joint pdf f(x,y)?

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!

• The problem doesn't mention if X and Y are continuous, I'm also assuming that. Jun 7 '20 at 21:05
• your pdf is negative for $y<-x$, how is that possible? Jun 7 '20 at 21:07
• @Exodd how so? Do you think there's an error in the problem? Jun 7 '20 at 21:11
• That is not a proper pdf so i believe the Intended constraint was $y\ge 1$ Jun 7 '20 at 21:50
• @JoseLuisPacheco What happens when x=1000, and y=1 does it lie in (0,1). Under your constraints then the range of $XY$ is any number in $\mathcal{R}$ Jun 7 '20 at 22:38

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?

• What do you mean by $\mathbb{I}(xy \le z)$? Can you please elaborate? thanks! Jun 7 '20 at 21:15
• @JoseLuisPacheco added clarification Jun 7 '20 at 21:22
• @JoseLuisPacheco please see the update Jun 7 '20 at 22:38
• @JoseLuisPacheco figured it out, i think; please see update #2 Jun 7 '20 at 23:56
• @JoseLuisPacheco $z/x$ does not have to be bigger or smaller than one. For example, if $z = 0.5,x=0.4$ then $z/x>1$... Jun 8 '20 at 2:22