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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!

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  • $\begingroup$ The problem doesn't mention if X and Y are continuous, I'm also assuming that. $\endgroup$ Jun 7 '20 at 21:05
  • $\begingroup$ your pdf is negative for $y<-x$, how is that possible? $\endgroup$
    – Exodd
    Jun 7 '20 at 21:07
  • $\begingroup$ @Exodd how so? Do you think there's an error in the problem? $\endgroup$ Jun 7 '20 at 21:11
  • $\begingroup$ That is not a proper pdf so i believe the Intended constraint was $y\ge 1$ $\endgroup$ Jun 7 '20 at 21:50
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    $\begingroup$ @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 $\mathcal{R}$ $\endgroup$ Jun 7 '20 at 22:38
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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?

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    $\begingroup$ What do you mean by $\mathbb{I}(xy \le z)$? Can you please elaborate? thanks! $\endgroup$ Jun 7 '20 at 21:15
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    $\begingroup$ @JoseLuisPacheco added clarification $\endgroup$
    – gt6989b
    Jun 7 '20 at 21:22
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    $\begingroup$ @JoseLuisPacheco please see the update $\endgroup$
    – gt6989b
    Jun 7 '20 at 22:38
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    $\begingroup$ @JoseLuisPacheco figured it out, i think; please see update #2 $\endgroup$
    – gt6989b
    Jun 7 '20 at 23:56
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    $\begingroup$ @JoseLuisPacheco $z/x$ does not have to be bigger or smaller than one. For example, if $z = 0.5,x=0.4$ then $z/x>1$... $\endgroup$
    – gt6989b
    Jun 8 '20 at 2:22

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