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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~

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    $\begingroup$ It should be $\chi^2_{2n-2},$ right? $\endgroup$ Nov 29, 2017 at 3:42
  • $\begingroup$ @spaceisdarkgreen I am not quite sure, it's just something my professor mentioned in class and he wrote this. $\endgroup$
    – Matata
    Nov 29, 2017 at 3:44
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    $\begingroup$ 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.) $\endgroup$ Nov 29, 2017 at 3:49
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    $\begingroup$ (Hopefully it's clear that I'm not saying that $X_{(j)} -X_{(1)}$ are independent in any sense... that's certainly not true.) $\endgroup$ Nov 29, 2017 at 4:22
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    $\begingroup$ Definitely it should be $\chi^2_{2(n-1)},$ not $\chi^2_{n-2}. \qquad$ $\endgroup$ Nov 29, 2017 at 6:23

3 Answers 3

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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

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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.

enter image description here

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)
 curve(dchisq(x, 2*n-2), lwd=2, col="darkgreen", add=T)
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    $\begingroup$ 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 $\qquad$ $\endgroup$ Nov 29, 2017 at 6:25
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    $\begingroup$ 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 $X_i.$ $\endgroup$ Nov 29, 2017 at 6:25
  • $\begingroup$ @michaelhardy, yes this is the part I was stuck on too (see my comments above). Seems "obviously" true since all you know is that they're all greater than some value... knowing one doesn't give info about the others. Yet couldn't see how to make this rigorous. $\endgroup$ Nov 29, 2017 at 16:53
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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.

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