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I need to show that, given a random sample of independent variables $X_1, ... , X_n$, each following a distribution EXP($\theta$,$\eta$), that is, $f(x_i)=\frac{1}{\theta}\exp(-\frac{x_i-\eta}{\theta})$, the statistics $X_{1:n}$ (the first order statistic) and $\overline{X}=(1/n)(X_1+...+X_n)$ are jointly sufficient for $\theta$ and $\eta$.

I already did this, by using a so-called ''factorization criterion'' (I'm following the terminology of Bein and Engelhardt's ''Introduction to Probability and Mathematical Statistics'') and the indicator function. See this related post.

I'd also like to do it using the formula $f_{\vec{X}|\vec{S}}(x_1,...,x_n)=\begin{cases} \frac{f(x_1,...,x_n;\vec{\theta})}{f_{\vec{S}}(\vec{s};\vec{\theta})} & \text{if } \mathit{s}(x_1,...,x_n)=s \\ 0 & \text{otherwise}\end{cases}$ (one has to show that this is independent of $\vec{\theta}$, the parameters; in this case, $(\theta_1,\theta_2)=(\theta,\eta)$).

(To be honest, I don't understand what that formula really means or where it comes from.)

My trouble is in finding the joint distribution of $X_{1:n}$ and $\overline{X}$. How does one calculate this?

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What do you mean by $X_{1:n}$? – Robert Israel May 23 '11 at 1:21
@Robert: $X_{1:n}=\min\{X_1,...,X_n\}$. – Weltschmerz May 23 '11 at 1:28
I changed the title to better reflect what it sounds like you were actually looking for. If you don't like the new title feel free to change it back. – Mike Spivey May 25 '11 at 21:34
up vote 4 down vote accepted

It looks like your question is really about finding the joint distribution of $X_{1:n}$ and $\overline{X}$, so that's the one I'm going to address. We'll do this by finding the marginal distribution of $X_{1:n}$ and the conditional distribution of $\overline{X}$ given $X_{1:n}$. Multiplying them together will give the joint distribution.

First, the marginal of $X_{1:n}$. The distribution of the minimum $X_{1:n}$ of $n$ independent exponential random variables is again exponential, and it's not too hard to prove that a similar result holds for the exp$(\theta, \eta)$ distribution. We have that $X_{1:n}$ is exp$(\theta/n,\eta)$; i.e., has pdf $$f_{X_{1:n}}(x) = \frac{n}{\theta} e^{n(\eta - x)/\theta} {\bf 1}_{(x \geq \eta)}.$$

Then, the conditional distribution of $\overline{X}$ given $X_{1:n}$. Given that $X_{1:n} = a$, each of the other $n-1$ variables from the sample has the cdf $P(X < x| X \geq a)$, where $X$ is exp$(\theta, \eta)$. As with the memoryless property of the usual exponential distribution, it is not too hard to show that each of these other variables is exp$(\theta,a)$; i.e., has pdf $$\frac{1}{\theta} e^{(a-x)/\theta} {\bf 1}_{(x \geq a)}.$$

The distribution of $\overline{X}$ given $X_{1:n} = a$, then, is the distribution of $\frac{1}{n}(X_1 + X_2 + \cdots + X_{n-1} + a)$, where $X_i$ is exp$(\theta,a)$. The moment generating function of $\overline{X}$ given $X_{1:n} = a$ can then be found (using standard mgf techniques) to be $$\frac{e^{at}}{(1- \theta t/n)^{n-1}}.$$

Using the time shift property of the Laplace transform (or by direct calculation) we can see that this is the mgf of what we might reasonably call a gamma$(n-1,\theta/n,a)$ distribution. (Since the mean of exponentials is gamma, it is reasonable that the mean of shifted exponentials is a shifted gamma.) Thus the pdf of $\overline{X}$ given $X_{1:n} = a$ is $$\left(\frac{n}{\theta}\right)^{n-1} \frac{1}{\Gamma(n-1)} (\bar{x} - a)^{n-2} e^{n(a - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a)}.$$

Finally, compute the joint distribution. Since $f_{X_{1:n},\overline{X}}(a,\bar{x}) = f_{X_{1:n}}(a) f_{\overline{X}|a}(\bar{x}|a)$ we have that

$$f_{X_{1:n},\overline{X}}(a,\bar{x}) = \frac{n}{\theta} e^{n(\eta - a)/\theta} \left(\frac{n}{\theta}\right)^{n-1} \frac{1}{\Gamma(n-1)} (\bar{x} - a)^{n-2} e^{n(a - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a)} {\bf 1}_{(a \geq \eta)}.$$ Simplifying, the joint pdf of $X_{1:n}$ and $\overline{X}$ that you're looking for is $$f_{X_{1:n},\overline{X}}(a,\bar{x}) = \left(\frac{n}{\theta}\right)^n \frac{(\bar{x} - a)^{n-2}}{\Gamma(n-1)} e^{n(\eta - \bar{x})/\theta} {\bf 1}_{(\bar{x} \geq a \geq \eta)}.$$

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