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Find the density function of $Z=XY$.

I was given $f(x,y)$ for $x>0$, $y>0$.

I know $F_{z}(z)= \iint f(x,y)dxdy$. But I need help finding the bounds for the double integral. I think the first one goes from $0$ to infinity and the second one goes from $0$ to z, but I am unsure.

Thank you!

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    $\begingroup$ The second goes from $0$ to $z/y$ $\endgroup$ – user384138 Nov 11 '16 at 19:36
  • $\begingroup$ @OpenBall would the first one then go from $0$ to $z$? $\endgroup$ – Silvia Rossi Nov 11 '16 at 19:38
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    $\begingroup$ The first goes from $0$ to $\infty$ $\endgroup$ – user384138 Nov 11 '16 at 19:46
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We have

$$P(XY<Z)=P(X<\frac{Z}{Y})=F(\frac{Z}{Y})$$

Now you simply differentiate

$$\frac{d}{dz}F=\int_0^{\infty}\frac{d}{dz}\int_0^{\frac{Z}{Y}}f(x,y)dxdy$$

By Leibnitz integral rule we have $$\frac{d}{dZ}F=\int_0^{\infty}\frac{f(\frac{Z}{y},y)}{y}dy$$

Where $$\frac{d}{dZ}F=p_z(z)$$

Note that the absolute value in the term $\frac{1}{y}$ is not required, as our domain is $x>0,y>0$.

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  • $\begingroup$ Oops, correction, the identity $$P(XY<Z)=P(X<\frac{Z}{Y})$$ is true, yes, but it is absurd too since $Z$ is defined as $Z=XY$ hence $$P(XY<Z)=0=P(X<Y/Z)$$ a fact which does not help to solve the question. $\endgroup$ – Did Dec 6 '16 at 23:53

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