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.


1 Answer 1


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):

  1. $n$ times the minimum of $n$ independent Exponential variables has an Exponential distribution.

  2. The Exponential is "memoryless": the distributions of the $Y_i-Y_1$, conditional on $Y_1$, are all Exponential and independent of $Y_1$.

  3. 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.