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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.

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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. –  Richard May 13 '13 at 14:12
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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 $K=\frac12(1+\sqrt5)$).

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