2
$\begingroup$

Let $X_i ~ \sim \operatorname{Exp}(1), \text{ i.i.d. } i=1,2,\ldots,n$, and let $X_{(i)}$ denote the $i$-th order statistic of $(X_i)$. In other words, $(X_{(i)} : i\text{-th smallest one among } X_1, \ldots, X_n)$

Then how can I find the distribution of $Y = X_1-X_{(1)}$?

I know that if $X_1$ is the minimum (which happens with probability $1/n$), $Y$ is $0$, but I don't know how to deal with the other cases.

$\endgroup$
  • $\begingroup$ I am not sure. At $n=1$ it is always 0 as you write. Perhaps it would help to limit $n=2$ and try to compute there and see what happens when $n=3$ and look for a pattern... $\endgroup$ – gt6989b Sep 13 '17 at 18:29
  • $\begingroup$ Mixed discrete-continuous model with a discrete mass at zero i.e. in a sample of size $n$, $P(Y=0) = \frac{1}{n}$, .... and then presumably a weighted standard Exponential for $y>0$: the latter might follow neatly via the memoryless property of Exponentials or some similar result. $\endgroup$ – wolfies Sep 13 '17 at 19:11
1
$\begingroup$

$\newcommand{\e}{\operatorname{E}}$ Suppose $x>0.$ Then \begin{align} \Pr( X_1 - X_{(1)} > x) & = \Pr(X_1 > X_{(1)}) \Pr(X_1 -X_{(1)} >x \mid X_1 > X_{(1)}) \\[10pt] & = \frac{n-1} n \Pr(X_1 - X_{(1)} > x \mid X_1 > X_{(1)}). \tag 0 \end{align} Let $J \in\{1,\ldots,n\}$ be the index $j$ for which $X_j = X_{(1)}.$ Then $J$ is uniformly distributed in $\{1,\ldots,n\}.$ So \begin{align} \Pr(X_1 - X_{(1)} > x\mid X_1 \ne X_{(1)}) & = \Pr(X_1-X_J> x\mid J\ne 1) \\[10pt] & = \e( \Pr( X_1 - X_J > x \mid J ) \mid J\ne 1 ). \tag 1 \end{align} But $\Pr(X_1-X_J>x\mid J=j)$ is the same for all $j\in\{2,\ldots,n\};$ thus as a function of $J$ it is constant on the set $J\ne 1.$ Thus the event $X_1-X_J>x$ is conditionally independent of $J$ given the event $J\ne1.$ Therefore the expected value in $(1)$ is the expected value of a constant random variable, and so the expected value is equal to that constant. And the constant is $\Pr(X_1-X_j>x\mid J=j),$ for any $j\in\{2,\ldots,n\}.$ So it is the same as $\Pr(X_1-X_2 > x\mid X_1>X_2).$ Now the memorylessness of the exponential distribution tells us that that is $\Pr(X_1> X_2+x\mid X_1>X_2) = e^{-x}.$ So line $(0)$ above becomes $\dfrac{n-1} n \cdot e^{-x}.$

The bottom line is that for $x\ge0$ we have $$ \Pr(X_1-X_{(1)} \le x) = \frac 1 n + \frac{n-1} n ( 1 - e^{-x}) = 1 - \frac{n-1} n e^{-x}. $$

$\endgroup$
  • $\begingroup$ Hello, I was trying to solve this problem, but in the final answer you gave, we should have $P(X_1-X_{(1)}\leq 0)=0$, but from your answer, it is $1/n$. So, I am confused. Can you please tell me why is this happening? $\endgroup$ – Stat_prob_001 Sep 3 '18 at 9:40
  • $\begingroup$ ok, Remind me of this about five hours from now and I'll look at it again..... $\endgroup$ – Michael Hardy Sep 3 '18 at 18:19
  • $\begingroup$ Thanks${}{}{}{}$ $\endgroup$ – Stat_prob_001 Sep 3 '18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.