0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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 $$ \int_{0}^z dw\int_{w}^1 dx\ \frac{1}{|x|}=\int_0^z dw [-\log w]=z-z \log (z)\ . $$

$\endgroup$
  • $\begingroup$ Awesome, thank you. I understand that the key step which I was doing wrong was that I wasn't truly ensuring that $f_Y$ was 1 over the integration bounds. $\endgroup$ – vancan1ty Mar 1 '16 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.