Sample distribution of some estimators Let $X_1, X_2, ...,X_n$ be random samples from a distribution with pdf $f(x) =\frac{1}{\sigma} e^{-(x-\mu)/\sigma}\mathrm I_{(\mu,\infty)}(x)$ where $\mathrm I_A$ is the indicator function of a set $A$.
Suppose that we cannot observe the whole random samples $X_1, X_2, ...,X_n$, but we can only observe $r$ smallest order statistics $X_{(1)}<...<X_{(r)}$ out of $n$ order statistics $X_{(1)}<...<X_{(r)}<...<X_{(n)}$ where $1\le r<n$.
We are going to estimate $\mu$ and $\sigma$ from the estimators below:
$\hat{\mu}=X_{(1)}$, $\hat{\sigma}=\frac{1}{r-1}\left( \sum_{k=1}^{r}(X_{(k)}-X_{(1)}) + (n-r)(X_{(r)}-X_{(1)}) \right)$.
Question: How can I find the joint pdf of $\hat{\mu}$ and $(r-1)\hat{\sigma}$?
What I know is that $\hat{\mu}$ and $(r-1)\hat{\sigma}$ are independent, so the joint pdf of $\hat{\mu}$ and $(r-1)\hat{\sigma}$ are the product of marginal pdf of $\hat{\mu}$ and $(r-1)\hat{\sigma}$, respectively.
 A: Rewrite $X_{(k)}-X_{(1)}$ in terms of spacings as 
$$
\left(X_{(k)}-X_{(k-1)}\right)+\left(X_{(k-1)}-X_{(k-2)}\right)\ldots+\left(X_{(2)}-X_{(1)}\right)
$$
and then you will get that
$$
(r-1)\hat{\sigma}=\sum_{k=1}^{r}(X_{(k)}-X_{(1)}) + (n-r)(X_{(r)}-X_{(1)}) $$ $$ = 
(n-1)\left(X_{(2)}-X_{(1)}\right) + (n-2)\left(X_{(3)}-X_{(2)}\right)+\ldots +(n-r+1)\left(X_{(r)}-X_{(r-1)}\right)
$$
Note that after subtraction the shift parameter $\mu$ disappears. This spacings distributed like the spacings from exponential distribution with mean $\sigma$. It is well-known that $X_{(i+1)}-X_{(i)}$ is also exponential distribution with mean $\frac{\sigma}{n-i}$ and all the spacings are independent and do not depend on $X_{(1)}$. See, for example, two answers here. 
Therefore all the summands 
$$(n-1)\left(X_{(2)}-X_{(1)}\right)$$ 
$$(n-2)\left(X_{(3)}-X_{(2)}\right)$$ 
$$\ldots$$ 
$$(n-r+1)\left(X_{(r)}-X_{(r-1)}\right)$$
are independent and exponentially distributed with the same mean $\sigma$. Their sum is Gamma-distributed $\Gamma(r-1,\sigma)$ with the PDF
$$
f_{(r-1)\hat{\sigma}}(x) = \frac{1}{\sigma^{r-1}(r-2)!}x^{r-2}e^{-x/\sigma}
$$
Sure, $\hat{\mu}=X_{(1)}$ has the PDF
$$
f_{\hat{\mu}}(x)=\frac{n}{\sigma} e^{-n(x-\mu)/\sigma}\mathrm I_{(\mu,\infty)}(x)
$$
and joint PDF of $\hat{\mu}$, $(r-1)\hat{\sigma}$ is the product. 
