Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been stuck on this problem for some days. I'm hoping someone would help by chipping in a few comments. I have two i.i.d. r.v.:

$$ f_X(x)=\frac{\left(1-e^{-\frac{x}{\alpha }}\right)^{\tilde{r}-1} exp({-\frac{x \left(K-\tilde{r}+1\right)}{\alpha })}}{\alpha } $$ and $$ f_Y(y)=\frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} exp({-\frac{y (K-r+1)}{\beta })}}{\beta } $$ Since they are i.i.d, the joint pdf is given as: $$ f_{XY}(x,y)=\frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} \left(1-e^{-\frac{x}{\alpha }}\right)^{\tilde{r}-1} e^{-\frac{x \left(K-\tilde{r}+1\right)}{\alpha }-\frac{y (K-r+1)}{\beta }}}{\alpha \beta } $$

Goal is to find the the PDF of Z = XY. Suppose z>0, I'm using the product relation given in Henry Stark's, (page 137, also here on wiki http://en.wikipedia.org/wiki/Product_distribution)

$$ f_Z(z)=\int_{-\infty }^{\infty } \frac{1}{\left| y\right| } f_{XY}\left(\frac{z}{y}, y\right) \, dy $$ Subtituting, I have,

$$ f_Z(z)=\int_{-\infty }^{\infty } \frac{\left(1-e^{-\frac{y}{\beta }}\right)^{r-1} \left(1-e^{-\frac{z}{\alpha y}}\right)^{\tilde{r}-1} \exp \left(-\frac{z \left(K-\tilde{r}+1\right)}{\alpha y}-\frac{y (K-r+1)}{\beta }\right)}{\alpha \beta y} \, dy $$

From here, I don't seem to know how to proceed to obtain a closed form equation. Almost at my wits end but hoping I might get some guidance here.

share|improve this question
Neither $f_X$ nor $f_Y$ is a PDF in general. –  Did May 13 '13 at 14:00
Thanks Did. Yes they are. I removed the constants to reduce the form. –  Afloz May 13 '13 at 14:12

1 Answer 1

As explained in comments, neither $f_X$ nor $f_Y$ is a PDF in general. For example, using the change of variable $t=\mathrm e^{-x/\alpha}$ in $f_X$ yields $\mathrm dt=-t\mathrm dx/\alpha$, hence $$ \int_0^{+\infty}f_X(x)\mathrm dx=\int_0^1(1-t)^{\tilde{r}-1} t^{K-\tilde{r}}\mathrm dt=\mathrm{Beta}(K-\tilde{r}+1,\tilde{r}). $$ For each $\bar r\gt0$, there is exactly one value of $K$ such that $f_X$ is a PDF (for example, if $\bar r=2$ then $K=\frac12(1+\sqrt5)$).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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