What am I doing wrong on this integral for CDF/PDF of product of random variables?

Question

I am trying to use the integral expression given here Wikipedia: Product Distribution to determine the CDF of the product $Z=XY$ of two independent Uniform(0,1) random variables $X$ and $Y$. I already know the answer to this question, and also know several ways to get to the answer (see, for example this question).

What I am asking for is help evaluating a certain integral -- I am trying to learn more about solving these types of problems in general.

The general integral for the CDF of the product of two independent random variables is given as: $$\int_{-\infty}^\infty\int_{-\infty}^z f_X(x)f_Y(w/x){1 \over |x|}\ dw\ dx$$ according to the Wikipedia link above.

I reasoned that since $X,Z$ are only defined on the interval $[0,1]$, that we can change the integration bounds for this problem to $[0,1]$,$[0,z]$ respectively like so:

$$\int_{0}^1\int_{0}^z f_X(x)f_Y(w/x){1 \over |x|}\ dw\ dx$$

Is that step wrong?

Next, I reasoned that since the pdfs $f_X$ and $f_Y$ are always $1$ over their support, and our new integration bounds limit the integration to the support of $f_X$ and $f_Y$, that we can simply replace the pdfs $f_X$ and $f_Y$ by $1$ under the integral symbol, like so:

$$\int_{0}^1\int_{0}^z 1\cdot 1{1 \over |x|}\ dw\ dx$$

Is that step wrong?

Next, I swap the bounds of integration, in an attempt to obtain the integral for the pdf.

$$\int_{0}^z\int_{0}^1 {1 \over |x|}\ dx\ dw$$

I am pretty sure that step is legal.

Next, I attempt to evaluate the inner integral to obtain the pdf.

$$\int_{0}^z \ln(x)]_{0}^{x=1}\ dw = \int_{0}^z \ln(1)-\ln(0) \ dw$$

But I can't evaluate that integral since $ln(0)$ is undefined.

I think I followed reasonable steps to work this integral, but my final result is obviously wrong. What am I doing incorrectly? Thank you for any help!

Answer 1

This is all fine $$ \int_{-\infty}^z dw\int_{-\infty}^\infty dx\ f_X(x)f_Y(w/x){1 \over |x|}= \int_{0}^z dw\int_{0}^1 dx\ f_X(x)f_Y(w/x){1 \over |x|}\ . $$ Now, $f_Y(t)=1$ for $0\leq t\leq 1$. So $f_Y(w/x)=1$ for $w/x\geq 0$ and $w/x\leq 1$. The first inequality is always verified, but the second one is not: you need to impose $x\geq w$. Therefore your $x$-integral gets truncated and the whole thing reads $$ \int_{0}^z dw\int_{w}^1 dx\ \frac{1}{|x|}=\int_0^z dw [-\log w]=z-z \log (z)\ . $$