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I have the following problem: Let the random variables X and Y have the probability density function (pdf) $$f(x, y) = \begin{cases} 1 & \text{for } 0<x<1, \,0<y<1\\ &\\ 0 & \text{elsewhere} \end{cases}$$

Find the probability distribution function (PDF) of $Z = XY$. I got as far as $$Fz(z)=P(Z\le z)=P(XY\le z)=P(X\le z/Y)\ldots$$ but I'm not sure where to go from here. I was thinking that the next time might be: $P(X\le z/Y)=F_X(z/Y)$ where $F_X$ is the probability distribution function for $X$. But, doing this, I end up with $F_Z(z,y)$ and not $F_Z(z)$. The PDF for Z should just be a function of $z$.

Any help is appreciated.

Jimmy R.
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user130398
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2 Answers2

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Draw the $1\times 1$ square on which the joint density function "lives."

Fix a number $z$ between $0$ and $1$. Draw the curve $xy=z$. (For the drawing, use say $z=0.4$.)

Let $D_z$ be the part of the square which is below the curve $xy=z$. Then $XY\le z$ happens precisely if the pair $(X,Y)$ lands in $D_z$. Note that our curve has equation $y=\frac{z}{x}$.

The probability that $Z\le z$ is the integral over $D_z$ of the joint density. Our joint density is $1$ on the square, so $\Pr(Z\le z)$ is the area of $D_z$.

To find the area, integrate. One way to do the calculation is to break up the region into $2$ parts, the part $0\le x\le z$, which is a rectangle of area $z$, and the part $z\le x\le 1$, whose area is $\int_z^1 \frac{z}{x}\,dx$.

André Nicolas
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$$P(XY\leqslant z)=\iint_{\mathbb R^2}\mathbf 1_{xy\leqslant z}f_X(x)f_Y(y)\mathrm dx\mathrm dy$$

Did
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