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Motivation:
I am a graduate student in the Department of Statistics at Kansas State University. Everyday I create a "question of the day" for myself, and it has been going well for the past year. These are often little distribution puzzles (sum of iid r.v. from Chi-Square is also Chi-Square, etc). Today, I came up with a question that seems simple (and perhaps is); however, I am quite stumped. The question is as follows:

Question:

Let $\{X_n\} \stackrel{iid}{\sim}$ Exp($\theta$), where $\theta > 0$. What is the distribution of $\{X_{(i)} - X_{(i-1)}\}$, $i = 2, \dots, n$?

(Note: $X_{(i)}$ denotes the i-th order statistic)

Thought Process:

I began by looking at the pdf for order statistics, both marginally and jointly; however, this did not lead anywhere because I am interested in the difference between two order statistics, and not the joint pdf of two order statistics. My other thought was to examine the mgf of each, but I cannot find an explicit method for determining the mgf of an order statistic.

aidan kerns
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1 Answers1

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This is simple given:

  • With $n$ i.i.d. $Exp(\lambda)$-distributed random variables, $X_{(1)} \sim Exp(n\lambda)$ and here $\lambda=1$
  • The exponential distribution is memoryless

So when $i-1$ events have occurred by time $X_{(i-1)}$, there are $n-i+1$ left, and the additional time to the next event (i.e. $X_{(i)}$-$X_{(i-1)}$) is distributed as $Exp(n-i+1)$ with expectation $\dfrac{1}{n-i+1}$.

Adding these up then implies $\mathbb E[X_{(n)}]=H_n=\sum\limits_{j=1}^n \frac1j$ and $\mathbb E[X_{(j)}]=H_n-H_{n-j}=\sum\limits_{j=n-i+1}^n \frac1j$.

Henry
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