Here is what I have done so far: Consider the random vector $X=(X_1,\ldots,X_n)$ which has pdf $$f(x_1,\ldots,x_n; \theta)=\theta^n e^{-\theta(x_1+\cdots+x_n)}.$$
Let $Y=T(X)$ where $T$ be the transformation that sends $(x_1,…,x_n)$ to $(nx_1/(x_1+\cdots+x_n),x_2,x_3,\cdots,x_n)$. The the pdf of $Y$ is
$$\theta^n\frac{n(y_2+\cdots+y_n)}{(n-y_1)^2}\exp\left\{ \frac{n\theta(y_2+\cdots+y_n)}{n-y_1}\right\}.$$
Thus, to get the pdf of $X_1/\bar{X}$ I just need to integrate out the $y_2,…,y_n$ to show that the pdf is independent of $\theta$. And this is where I got stuck. I have tried it with $n=2$ and it worked, but integrating out this $y_2,…,y_n$, I got….lazy. 😀
Do you know a better way for this problem?
Best Answer
@zhoraster's comment is a perfect answer to your question.
Alternatively, the exact distribution of $X_1/\overline X$ can be obtained with ease using well-known relations between Exponential, Gamma and Beta distributions. We have
$$\frac{X_1}{\overline X}=n\cdot\frac{X_1}{X_1+Y}\,\,,$$
where $Y=\sum\limits_{i=2}^n X_i$ is independent of $X_1$.
Since $X_1,\ldots,X_n$ are i.i.d $\mathsf{Exp}$ with mean $1/\theta$ (or equivalently $\Gamma(1,\theta)$), it is known that $Y\sim \Gamma(n-1,\theta)$. Hence it is also known (see this or this) that $$\frac{X_1}{X_1+Y}\sim \mathsf{Beta}(1,n-1)$$
So if you are familiar with these results, there is nothing to calculate to conclude that $X_1/\overline X$ is ancillary for $\theta$.