# Transformation of f(x,y) = 1 to f(z) where Z=XY

1. Oct 9, 2013

### rayge

This is from a chapter on distributions of two random variables. Let X and Y have the pdf f(x,y) = 1, 0<x<1 and 0<y<1, zero elsewhere. Find the cdf and pdf of the product Z=XY.

My current approach has been to plug in X=Z/Y in the cdf P(X<=x) , thus P(Z/Y<=x), and integrate over all values of Y. The integral I've tried to solve is ∫(0 to 1)∫(0 to Z/Y) 1 dxdy (sorry for the formatting).

This is somehow wrong, because from that integral I get z*(ln(1) - ln(0)), which is undefined (unless I'm solving the integral incorrectly, which would actually be a relief).

The correct answer is -ln(z), 0<z<1, though it's amiguous as to whether that's the cdf or pdf.

Thanks for any solutions or suggestions about where my approach is going wrong.

2. Oct 9, 2013

### mathman

-ln(z) is a density function.

When doing the integral note that the upper limit on x is min(1,z/y).

3. Oct 9, 2013

### economicsnerd

Another way of seeing the same thing... [It looks a little less direct, but there's value in thinking about sums of exponential random variables.]

Let $\tilde X = -ln(X)$, $\tilde Y = -ln(Y)$, and $\tilde Z = -ln(Z) = \tilde X+\tilde Y$, all taking values on $(0,\infty)$. Define the CDF $G(\tilde x)= P(\tilde X\leq \tilde x)= P(X\geq e^{-\tilde x})=1-e^{-\tilde x}$, which is of course also the CDF of $\tilde Y$. Differentiating, we get the PDF $g(\tilde x)=e^{-\tilde x}$. Now we can compute $$P(\tilde Z> \tilde z\geq \tilde X) = \int_0^{\tilde z} P(\tilde x+\tilde Y> \tilde z) g(\tilde x) \text{d}\tilde x = \int_0^{\tilde z} [1- G( \tilde z-\tilde x)] g(\tilde x) \text{d}\tilde x = \int_0^{\tilde z} e^{\tilde x-\tilde z} e^{-\tilde x} \text{d}\tilde x = \int_0^{\tilde z} e^{-\tilde z} \text{d}\tilde x = \tilde z e^{-\tilde z}.$$ This yields $$P(Z<z\leq X) = P(\tilde Z> -ln(z)\geq \tilde X) = [-ln(z)]z.$$ Finally, we notice that (as $Z\leq X$) $$P(Z<z)=P(X<z)+P(Z<z\leq X) = z - [-ln(z)]z = [1-ln(z)]z.$$ If your end goal is a density, you can differentiate for a PDF of $-ln(z)$.

4. Oct 10, 2013

### rayge

Thanks for the responses. Before I dive into economicsnerd's solution (wow), I'm a bit confused by min(1,Z/Y). When Z<Y, min(1,Z/Y) is Z/Y. When Z>Y, it's 1. So...

∫(0 to 1)∫(0 to Z/Y)dxdy where 0<z<y<1
∫(0 to z)∫(0 to 1)dxdy where 0<y<z<1

Isn't the first integral still undefined? Or am I applying this incorrectly?

5. Oct 10, 2013

### economicsnerd

The CDF of $Z$ is $$F(z)= P(Z\leq z) = 1- P(Z>z) = 1-P(X\geq Z > z) = 1-\int_z^1 \int_{\frac{z}{x}}^1 1\enspace \text{d}y \enspace \text{d}x = 1-\int_z^1 \left(1-\frac{z}{x}\right) \enspace \text{d}x.$$ What do you get when you compute that integral?

6. Oct 10, 2013