Let $X_1,\dots,X_n$ be a sample of iid exponential random variables with mean $\beta$, and let $X_{(1)},\dots,X_{(n)}$ be the order statistics from this sample. Let $\bar X = \frac{1}{n}\sum_{i=1}^n X_i$.

Define spacings $$W_i=X_{(i+1)}-X_{(i)}\ \forall\ 1 \leq i \leq n-1\,.$$ It can be shown that each $W_i$ is also exponential, with mean $\beta_i=\frac{\beta}{n-i}$.

Question: How would I go about finding $\mathbb{P}\left( \frac{W_i}{\bar X} > t \right)$, where $t$ is known and non-negative?

Attempt: I know that this is equal to $1 - F_{W_i}\left(t \bar X\right)$. So I used the law of total probability like so: $$ \mathbb{P}\left( W_i > t \bar X \right) = 1 - F_{W_i}\left( t \bar X \right) = 1 - \int_0^\infty F_{W_i}(ts)f_{\bar X}(s) \mathrm{d}s \,, $$

which turns into a messy but I think tractable integral.

Am I on the right track here? Is this a valid use of the Law of Total Probability?

Another approach might be to look at the difference distribution: $$ \mathbb{P}\left( W_i - t \bar X > 0\right) $$

Or even break apart the sums: $$ \mathbb{P}\left( W_i - t \bar X > 0 \right) = P \left( \left(X_{(i+1)} - X_{(i)}\right) + \frac{t}{n}\left(X_{(1)} + \dots + X_{(n)} \right) \right) $$

A solution to the exponential case would be great, but even better would be some kind of general constraints on the distribution. Or at the very least, its moments, which would be enough to give me Chebyshev and Markov inequalities.

Update: here's the integral from the first method: $$\begin{align} 1 - \int_0^\infty \left( 1 - \exp \left( -\frac{ts}{\beta_i} \right) \right) \left( \frac{1}{\Gamma(n)\beta^n} s^{n-1} \exp \left( -\beta s \right) \right) \mathrm{d}s \\ 1 - \int_0^\infty \left( 1 - \exp \left( -\frac{(n-i)ts}{\beta} \right) \right) \left( \frac{1}{\Gamma(n)\beta^n} s^{n-1} \exp \left( -\beta s \right) \right) \mathrm{d}s \end{align}$$

I've been playing around with it for a little while and I'm not sure where to go with it.

  • 1
    $\begingroup$ The integral you get looks relatively straightforward after you distribute the the parentheses terms. After a change of variables, it looks like you'll get some gamma functions. $\endgroup$
    – Alex R.
    Commented Jan 6, 2016 at 20:33
  • $\begingroup$ @AlexR indeed it does, but after getting halfway through it I started to suspect that it wouldn't be bounded between 0 and 1. I'm more looking for confirmation that I set up the problem correctly. If I get stuck with the integral itself I'll ask on Math.SE $\endgroup$ Commented Jan 7, 2016 at 14:08

1 Answer 1


The difficulty you have here is that you have an event relating non-independent random variables. The problem can be simplified and solved by manipulating the event so that it compares the independent increments. To do this, we first note that for $X_1, ..., X_N \sim \text{IID Exp}(\beta)$, each of the order statistics can be written as:

$$X_{(k)} = \beta \sum_{i=1}^{k} \frac{Z_i}{n-i+1},$$

where $Z_1, Z_2, ..., Z_n \sim \text{IID Exp} (1)$ (see e.g., Renyi 1953, David and Nagaraja 2003). This allows us to write $W_k = \beta Z_{k+1} / (n-k)$ and we can write the sample mean as:

$$\begin{equation} \begin{aligned} \bar{X} \equiv \frac{\beta }{n} \sum_{k=1}^n X_{(k)} &= \frac{\beta }{n} \sum_{k=1}^n \sum_{i=1}^k \frac{Z_i}{n-i+1} \\ &= \frac{\beta }{n} \sum_{i=1}^n \sum_{k=i}^n \frac{Z_i}{n-i+1} \\ &= \frac{\beta }{n} \sum_{i=1}^n Z_i. \end{aligned} \end{equation}$$

To facilitate our analysis we define the quantity:

$$a \equiv \frac{t(n-k)}{n-t(n-k)}.$$

For $a > 0$ we then have:

$$\begin{equation} \begin{aligned} \mathbb{P}(W_k \geqslant t \bar{X}) &= \mathbb{P} \left( \frac{Z_{k+1}}{n-k} \geqslant \frac{t}{n} \sum_{i=1}^n Z_i \right) \\ &= \mathbb{P} \left( \frac{n}{n-k} \cdot Z_{k+1} \geqslant t \sum_{i = 1}^k Z_i \right) \\ &= \mathbb{P} \left( \left( \frac{n}{n-k} - t \right) Z_{k+1} \geqslant t \sum_{i \neq k} Z_i \right) \\ &= \mathbb{P} \left( \left( \frac{n}{n-k} - t \right) Z \geqslant t G \right) = \mathbb{P} \left( Z \geqslant a G \right), \end{aligned} \end{equation}$$

where $Z \sim \text{Exp} (1)$ and $G \sim \text{Ga}(n-1, 1)$ are independent random variables. For the trivial case where $t \geqslant n / (n-k)$ we have $\mathbb{P}(W_k \geqslant t \bar{X}) = 0$. For the non-trivial case where $t < n / (n-k)$ we have $a>0$, and the probability of interest is:

$$\begin{equation} \begin{aligned} \mathbb{P}(W_k \geqslant t \bar{X}) &= \int\limits_0^\infty \text{Ga} (g| n-1, 1 ) \int\limits_{ag}^\infty \text{Exp}(z|1) dz dg \\ &= \int\limits_0^\infty \frac{1}{\Gamma{(n-1)}} g^{n-2} \exp{(-g)} \int\limits_{ag}^\infty \exp{(-z)} dz dg \\ &= \int\limits_0^\infty \frac{1}{\Gamma{(n-1)}} g^{n-2} \exp{(-g)} \left( 1 - \exp (ag) \right) dg \\ &= \int\limits_0^\infty \frac{1}{\Gamma{(n-1)}} g^{n-2} \exp{(-g)} dg - \int\limits_0^\infty \frac{1}{\Gamma{(n-1)}} g^{n-2} \exp{(-(a+1)g)} dg \\ &= 1 - (a+1)^{-(n-1)} \\ &= 1 - \left( 1- \frac{n-k}{n} \cdot t \right)^{n-1}. \end{aligned} \end{equation}$$

This answer is intuitively reasonable. This probability is strictly decreasing in $t$, with unit probability when $t=0$ and zero probability when $t = \frac{n}{n-k}$.


Your Answer

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

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