Partial sum of order statistics of exponential r.v.'s and $\chi^2$ Suppose $X_i \sim Exp(\frac{1}{\lambda}), i = 1,\cdots,n$, where
$f(x) = I_{(0,\infty)}\frac{1}{\lambda}e^{-\frac{x}{\lambda}}$
is the p.d.f. of $X_i$'s. 
And we have a positive integer $r$, and the order statistics
$$X_{(1)}\leq X_{(2)} \leq \cdots \leq X_{(r)}$$ where $1<r \leq n$. 
Then, denote $$T = \sum_{i=1}^{r} X_{(i)} + (n-r)X_{(r)}$$
The problem is to prove that $\frac{2T}{\lambda} \sim \chi^2_{2r}$.
I'm quite at loss here. It seems to have a lot to do with Gamma distribution, but it doesn't seem the sum of first r- order statistics follows it. 
Even if the sum of the first r items does follow Gamma, I don't know how to handle the following $(n-r)X_{(r)}$.
It doesn't seem right to directly compute the p.d.f. of $T$, which I've tried and failed. 
I'd appreciate it enormously if anyone can give me any hint or solution !
 A: Your question was actually answered eight decades ago by P. V. Sukhatme. I'll explain how to prove the problem assuming that you are familiar with the fundamentals of probability theory and the relationships between the exponential, gamma and $\chi^2$ distributions. First, if you reviewed any textbook that addresses order statistics, then you will observed that the joint probability density function of the order statistics $X_{(1)}, \ldots, X_{(n)}$ is defined to be
$$
f(x_{(1)}, \ldots, x_{(n)}) = n! \prod^n_{i = 1} f(x_{(i)}).
$$
By integrating the above joint probability density function with respect to $x_{(n)}, \ldots, x_{(r + 1)}$, respectively, one will obtain
$$
g(x_{(1)}, \ldots, x_{(r)}) = \frac{n!}{(n - r)!} \prod^r_{i = 1} f(x_{(i)}) \left[1 - F(x_{(r)})\right]^{n - r} = (n - r + 1)! \prod^r_{i = 1} f(x_{(i)}) \left[1 - F(x_{(r)})\right]^{n - r},
$$
where $F(\cdot)$ is the corresponding cumulative distribution function. In the case of the exponential distribution with scale parameter $\lambda$, the above density is given by
$$
g(x_{(1)}, \ldots, x_{(r)}) = \dfrac{(n - r + 1)!}{\lambda^r} \exp\left(\lambda^{-1} T\right),
$$
where $T = \sum^{r}_{i = 1} X_{(i)} + (n - r) X_{(r)}$. Second, define the following transformations: 
$$
\begin{array}{l}
S_1 = n X_{(1)} \\
S_2 = (n - 1) \left[X_{(2)} - X_{(1)}\right] \\
\vdots \\
S_{r - 1} = [n - (r - 1) + 1] \left[X_{(r - 1)} - X_{(r - 2)}\right] \\
S_{r} = [n - r + 1] \left[X_{(r)} - X_{(r - 1)}\right] \\
\end{array}
$$
Clearly, 
$$
\sum^r_{i = 1} S_i = \sum^{r - 1}_{i = 1} X_{(i)} + (n - r + 1) X_{(r)} = \sum^{r}_{i = 1} X_{(i)} + (n - r) X_{(r)} = T,
$$
and
$$
\begin{array}{l}
X_{(1)} = \frac{S_1}{n} \\
X_{(2)} = \frac{S_2}{n - 1} + \frac{S_1}{n} \\
\vdots \\
X_{(r - 1)} = \frac{S_{r - 1}}{n - r + 2} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n} \\
X_{(r)} = \frac{S_{r}}{n - r + 1} + \frac{S_{r - 1}}{n - r + 2} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n} \\
\end{array}
$$
Note that $S_1, \ldots, S_r$ are called spacings in statistical literature. Third, acquire the joint probability density function of the transformations $S_1, \ldots, S_r$, which is simply
$$
h(s_1, \ldots, s_r) = g\left(\frac{S_1}{n}, \frac{S_2}{n - 1} + \frac{S_1}{n}, \ldots, \frac{S_{r}}{n - r + 1} + \cdots + \frac{S_2}{n - 1} + \frac{S_1}{n}\right) \mathbf{J}
$$
where $\mathbf{J}$ is the Jacobian of transformation. Clearly, since
$$
\mathbf{J} = \left|
\begin{matrix}
\frac{1}{n} & 0 & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n - 1} & 0 & \cdots & 0 \\
\frac{1}{n} & \frac{1}{n - 1} & \frac{1}{n - 2} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{n} & \frac{1}{n - 1} & \frac{1}{n - 2} & \cdots & \frac{1}{n - r + 1} \\
\end{matrix}
\right| = \frac{1}{(n - r + 1)!},
$$
then the above joint probability density function of the transformations $S_1, \ldots, S_r$ is reduced to
$$
h(s_1, \ldots, s_r) = \frac{1}{\lambda^r} e^{-\lambda^{-1} T} = \left(\dfrac{1}{\lambda} e^{-\lambda^{-1} s_1}\right) \cdots \left(\dfrac{1}{\lambda} e^{-\lambda^{-1} s_r}\right),
$$
i.e. $S_1, \ldots, S_r$ are independent and identically-distributed random variables that follow $\mathrm{Exp}(\lambda^{-1})$. Which means $T = \sum^r_{i = 1} S_i$ follows $\mathrm{Gamma}(r, \lambda)$. Hence, $\frac{2T}{\lambda} \sim \chi^2_{2r}$.
References: P. V. Sukhatme (1937), Tests of significance for samples of the $\chi^2$ population with two degrees of freedom, Ann. Eugenics 8, 52-56.
