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 $\theta$.

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

1 Answer 1


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 $XY$


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