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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. :D

Do you know a better way for this problem?

StubbornAtom
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Chuong Nguyen
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  • Oh actually, right after I typed the question, I solved this problem. To be precise, integrating out $y_2,...,y_n$ is not that bad. But I am still interested in another approach if there is one. – Chuong Nguyen Sep 01 '19 at 08:01
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    $X_i$ are iid $\mathrm{Exp}(\theta) \Rightarrow$ $Z_i = \theta X_i$ are iid $\mathrm{Exp}(1)$. But $X_1/\overline X = Z_1/\overline Z$, so the distribution of this statistic is independent of $\theta$, qed – zhoraster Sep 01 '19 at 10:46
  • Thank you so much, zhoraster. This is a much better way. – Chuong Nguyen Sep 01 '19 at 15:53

1 Answers1

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@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 $\theta$.

StubbornAtom
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