# The difference of two order statistics of exponential distribution

If we have a random sample $X_1,X_2, \ldots, X_n \stackrel{\text{iid}}\sim f(x\mid\theta)=e^{-(x-\theta)}I(x >\theta)$. We want to prove $$2\sum X_i-2n X_{(1)} \sim \chi^2_{n-2}$$ where $X_{(1)}$ is the smallest order statistic.

I tried: $$2\sum X_i-2n X_{(1)} =2\left[\sum X_i-n X_{(1)}\right]=2\left[\sum X_{(i)}-n X_{(1)}\right]=2\left[\sum \left(X_{(i)}- X_{(1)}\right)\right]$$ And I was trying to find the distribution of $$X_{(i)}- X_{(1)}$$

And I searched that $$X_{(i)}-X_{(i-1)} \sim \operatorname{Exp}\left(\frac{1}{n+1-i}\right) \text{ if } X_i \stackrel{\text{iid}}\sim \operatorname{Exp}(1)$$ Any ideas? Thank you~

• It should be $\chi^2_{2n-2},$ right? Nov 29, 2017 at 3:42
• @spaceisdarkgreen I am not quite sure, it's just something my professor mentioned in class and he wrote this. Nov 29, 2017 at 3:44
• First, by translation invariance of the quantity you're calculating, WLOG you can set $\theta =0.$ You're told the minimum $m = X_{(1)}$... so you know that you have $n-1$ others that larger than this minimum. By memoryless property, conditional on this information the others are exponential with location $m,$ so their difference with $X_{(1)}$ is standard exponential. So you have the sum of $n-1$ independent standard exponentials ($\Gamma(n-1,1)$). By the relationship of Gamma and chi-squared, two times this is $\chi^2_{2n-2}.$ (I comment rather than answer since this is far from rigorous.) Nov 29, 2017 at 3:49
• (Hopefully it's clear that I'm not saying that $X_{(j)} -X_{(1)}$ are independent in any sense... that's certainly not true.) Nov 29, 2017 at 4:22
• Definitely it should be $\chi^2_{2(n-1)},$ not $\chi^2_{n-2}. \qquad$ Nov 29, 2017 at 6:23

Let $X_1,X_2,\ldots,X_n$ be a random sample from exponential distribution with mean $1.$ Then joint probability density of order statistics $X_{(1)},X_{(2)},\ldots,X_{(n)}$ is $$f_{X_{(1)},X_{(2)},\ldots,X_{(n)}}(x_1,x_2,\ldots,x_n)= n! e^{-\sum_{i=1}^{n}x_i}, 0\leq x_1\leq x_2\leq \cdots \leq x_n \leq \infty$$ Let us consider transformation

$$Y_1=nX_{(1)}, Y_2=(n-1)(X_{(2)}-X_{(1)}), Y_3=(n-2)(X_{(3)}-X_{(2)}),\ldots,Y_n= X_{(n)}-X_{(n-1)}$$

$$\Rightarrow X_{(1)}=\frac{Y_1}{n}, X_{(2)}=\frac{Y_1}{n}+\frac{Y_2}{n-1},\ldots, X_{(n)}=\frac{Y_1}{n}+\frac{Y_2}{n-1}+\frac{Y_3}{n-2}+\cdots+Y_n$$

Jacobian of above transformation is $\frac{1}{n!}$.

So joint probability density function of $Y_1,Y_2,\ldots,Y_n$ is given by

$f_{Y_1,Y_2,\ldots,Y_n}(y_1,y_2,\ldots,y_n)= e^{-\sum_{i=1}^n y_i}; 0\leq y_1,y_2,\ldots,y_n\leq \infty$.

This follows, using factorization theorem, $Y_1,Y_2,Y_3,\ldots,Y_n$ are identically and independently distributed as exponential variate with mean $1.$

$\Rightarrow Y_i=(n-i+1)(X_{(i)}-X_{(i-1)}) \stackrel{\text{iid}}{\sim} \operatorname{exp}(1)$; $i=2,3,\ldots,n$.

Hence $\sum_{i=2}^{n} Y_i= \sum_{i=1}^n(X_i-X_{(1)})$ is sum of $(n-1)$ independent $exp(1)$ variates, so $\sum_{i=1}^n(X_i-X_{(1)})\sim \operatorname{gamma}(n-1)$.

Ref: "Order Statistics & Inference" by Balakrishnan & Cohen. https://www.amazon.com/Order-Statistics-Inference-Estimation-Methods/dp/149330738X

Comment: @spaceisdarkgreen, I simulated the case $n = 10,\, \theta = 0$ to see if this seems to work at all and to confirm your correction of the degrees of freedom. The red curve is for $\mathsf{Chisq}(n-2)$ and the green for $\mathsf{Chisq}(2n-2).$ Of course, this doesn't prove anything, but (to me anyhow) it offers hope your argument might be made rigorous.

R code in case it is of any use:

m = 10^5;  n = 10
x = rexp(m*n);  MAT = matrix(x, nrow=m)
t = rowSums(MAT);  v = apply(MAT, 1, min)
y = 2*t - 2*n*v
hist(y, prob=T, br= 25, col="skyblue2", ylim=c(0,.12))
curve(dchisq(x, n-2), 0, 50, lwd=2, col="red", add=T)
• Let $J = \text{the index$j\in\{1,\ldots,n\}$for which } X_j = X_{(1)}.$ Then $J$ is uniformly distributed in $\{1,\ldots,n\}$. Then we have $$\sum_{i=1}^n \left(X_{(i)}- X_{(1)} \right) = \sum_{i=1}^n (X_i - X_J) \quad (\text{Note that one of the n terms in this last sum is 0.})$$ And $$\Pr\left( \sum_{i=1}^n (X_i-X_J) \in A \right) = \operatorname E\left(\Pr\left( \sum_{i=1}^n (X_i-X_J) \in A \right) \mid J \right)$$ and this last probability does not depend on the value of $J.$ Since it does not depend on $J,$ this expected value is equal to $\qquad$ Nov 29, 2017 at 6:25
• its conditional probability given that $J=1,$ i.e. to $$\Pr\left( \sum_{i=2}^n (X_i - X_1) \in A \mid J=1 \right).$$ By memorylessness of the exponential distribution, the distribution of $X_i-X_1$ given that $X_1<X_i$ is the same exponential distribution as that of $X_i.$ Nov 29, 2017 at 6:25
First try to show that $$X_i - X_{(1)} \sim \begin{cases} \delta_0 & \text{if } X_i = X_{(1)}, \\ \chi^2_2 & \text{otherwise}. \end{cases}$$ Then think about how to show that $X_i-X_{(1)},\, X_j-X_{(1)}$ are independent. Then use standard properties of the chi-square distribution.