# Does the distribution of $XY$ depend on $\theta$, when $X\sim\text{Exp}(\theta)$, $Y\sim\text{Exp}(1/\theta)$ and $X$ independent with $Y$?

Does the distribution of $$XY$$ depend on $$\theta$$, when $$X\sim\text{Exp}(\theta)$$, $$Y\sim\text{Exp}(1/\theta)$$ and $$X$$ independent with $$Y$$?

I understand that from Wiki Parametrization 1 we have $$XY$$ does not depend on $$\theta$$ and intuitively this is correct, but I don't know how to show it. Here is what I've done.

\begin{align*} \mathbb{P}(XY\leq t)&=\int_{0}^{\infty}\mathbb{P}(XY\leq t|Y=s)f_{Y}(s)ds\\ &=\int_{0}^{\infty}\mathbb{P}(X\leq t/s)f_{Y}(s)ds \text{ -- by X independent with Y}\\ &=\int_{0}^{\infty}\left[1-e^{\theta t/s}\right]\frac{1}{\theta}e^{-\frac{s}{\theta}}ds\\ &=\frac{1}{\theta}\int_{0}^{\infty}\left[1-e^{\theta t/s}\right]e^{-\frac{s}{\theta}}ds\\ &=\text{a function of }\theta \end{align*}

I don't know why the distribution of $$XY$$ is independent with $$\theta$$.

• It is because $X=\frac1{\theta} U$ and $Y=\theta V$ where $U,V\sim \text{Exp}(1)$. Jan 9, 2021 at 19:40
• @StubbornAtom ha, thanks a lot. I was trying to find a general way of finding an ancillary statistic and missed this smart solution
– Tan
Jan 9, 2021 at 19:47

The distribution of $$X\theta$$ does not depend on $$\theta$$; it is Exp(1).
The distribution of $$Y/\theta$$ does not depend on $$\theta$$; it is also Exp(1).
So the distribution of $$(X\theta)(Y/\theta)$$ does not depend on $$\theta$$. But that is just the distribution of $$XY$$