Joint distribution of $X_{(1)}$ and $\sum_{i=1}^n X_i - X_{(1)}$ i've been stuck on this question for a while now. 
I am looking for the joint pdf of $(X_{(1)}, \sum_{i=1}^n (X_i - X_{(1)})$ where X is distributed Exp(1). My current thought is first try and compute:
$$f(X_{(1)}, \sum X_i) = \int f(\sum x_i | X_{(1)}) * f(X_{(1)}) $$
Where $f(\sum x_i | X_{(1)}) $ is defined using the memory less property. But i am unsure how to do this. 
 A: This question can be resolved using a useful representation originally due to the mathematician Alfréd Rényi (see e.g., David and Nagaraja 2003, p. 18).  If $X_1, ..., X_n \sim \text{IID Exp}(\lambda)$ then the corresponding order statistics $X_{(1)} \leqslant ... \leqslant X_{(n)}$ have the equivalent representation:
$$X_{(k)} = \frac{1}{\lambda} \sum_{i=1}^k \frac{Z_i}{n-i+1} \quad \quad \quad Z_1, ..., Z_n \sim \text{IID Exp}(1).$$
Using this representation you have $X_{(1)} = Z_1 / n \lambda$ and:
$$\begin{equation} \begin{aligned}
D_n \equiv \sum_{i=1}^n X_i - X_{(1)} 
&= \sum_{i=1}^n X_{(i)} - X_{(1)}  \\[6pt]
&= \frac{1}{\lambda} \Bigg[ \sum_{k=1}^n \sum_{i=1}^k \frac{Z_i}{n-i+1} - \frac{Z_1}{n} \Bigg] \\[6pt]
&= \frac{1}{\lambda} \Bigg[ \sum_{i=1}^n \sum_{k=i}^n \frac{Z_i}{n-i+1} - \frac{Z_1}{n} \Bigg] \\[6pt]
&= \frac{1}{\lambda} \Bigg[ \sum_{i=1}^n \frac{Z_i}{n-i+1} \cdot(n-i+1) - \frac{Z_1}{n} \Bigg] \\[6pt]
&= \frac{1}{\lambda} \Bigg[ \sum_{i=1}^n Z_i - \frac{Z_1}{n} \Bigg] \\[6pt]
&= \frac{1}{\lambda} \Bigg[ \sum_{i=2}^n Z_i + \frac{n-1}{n} Z_1 \Bigg] \\[6pt]
&= \frac{1}{\lambda} \sum_{i=2}^n Z_i + (n-1) X_{(1)}, \\[6pt]
\end{aligned} \end{equation}$$
where the components $X_{(1)}$ and $Z_2, ..., Z_n$ are independent.  This gives us the conditional distribution $D_n | x_{(1)} \sim (n-1) x_{(1)} + \text{Ga}(n-1,\lambda)$ so the joint distribution is:
$$\begin{equation} \begin{aligned}
f(d_n, x_{(1)}) 
&= f(d_n | x_{(1)}) \cdot f(x_{(1)}) \\[8pt]
&= \text{Ga}\Big( d_n - (n-1) x_{(1)} \Big| n-1,\lambda \Big) \cdot \text{Exp}\Big( x_{(1)} \Big| n \lambda \Big) \\[6pt]
&= \frac{\lambda^{n-1}}{\Gamma(n-1)} (d_n - (n-1) x_{(1)} )^{n-2} \exp \Big( - \lambda (d_n - (n-1) x_{(1)} ) \Big) \cdot n \lambda  \exp \Big( - n\lambda x_{(1)} \Big) \\[6pt]
&= \frac{n \lambda^{n}}{(n-2)!} (d_n - (n-1) x_{(1)} )^{n-2} \exp \Big( - \lambda (d_n - x_{(1)}) \Big), \\[6pt]
\end{aligned} \end{equation}$$
for all arguments over the support $x_{(1)} \geqslant 0$ and $d_n \geqslant (n-1) x_{(1)}$.
