Let $X_1, \ldots, X_n$ be i.i.d. exponentially distributed random variables with density
$$\eqalign{\theta^{-1} e^{-x/\theta}, &x \ge 0 \\ 0, &x \lt 0} $$
and let $Y_i = X_{(i)}$ denote the order statistics such that $Y_1 \leq \cdots \leq Y_n$.
How to show that $$ 2\frac{\left(\sum_{i=1}^{r}Y_i\right) + (n-r)Y_r}{\theta} $$ has a chi-square distribution with $2r$ degrees of freedom?
I wrote the joint density of $(Y_1,Y_2,...,Y_r)$ but nothing became apparent.
Upon reparameterizing and rescaling ($\chi^2$ distributions are special Gamma distributions), the question is equivalent to showing that
$$Z_r = \sum_{i=1}^r Y_i + (n-r)Y_r$$
has a Gamma$(r)$ distribution. Let's rewrite this suggestively as
$$Z_r = nY_1 + (n-1)(Y_2-Y_1) + \cdots + (n-r+1)(Y_r - Y_{r-1}).$$
Exploit these basic (and easily proven) properties of Exponential distributions (with unit scale):
$n$ times the minimum of $n$ independent Exponential variables has an Exponential distribution.
The Exponential is "memoryless": the distributions of the $Y_i-Y_1$, conditional on $Y_1$, are all Exponential and independent of $Y_1$.
The sum of $r$ iid Exponential variables has a Gamma$(r)$ distribution.
These imply (with a simple inductive proof) that $Z_r$ has the same distribution as the sum of $r$ iid Exponential variables, QED. After all, the first term $nY_1$ has an Exponential distribution and the remaining terms are independent of it. Thus the next term, $(n-1)(Y_2-Y_1)$, is the smallest of $n-1$ iid Exponential variates, and also has an Exponential distribution, etc.
