Let a random sample of size $n$ from an exponential distribution $X_i \sim EXP(1)$. Give the pdf of

(1) The sample range, $R = Y_n - Y_1$

(2) The first r order statistics

The answers are supposed to be

(1) $f_R(r) = (n-1)e^{-r}(1 - e^{-r})^{n-2}$ and (2) $g(y_1, \dots, y_r) = \frac{n}{(n-r)!}\exp\left ( -\sum_{i=1}^{r} y_i -(n-r)y_r\right)$

I have absolute no idea how they got -2 in the first answer and I don't know how in the second answer, they got a sum

EDIT: I figured out the sum part. They basically just put everything together into the exponent

EDIT#2. The largest order statistic $Y_n$ is $n e^{-y_n}(1 - e^{y_n})^{n-1}$.


Thanks to the memoryless property of the exponential distribution, the difference between $Y_{(n)}$ and $Y_{(1)}$ is independent of the actual value of $Y_{(1)}$. So to find the distribution of $R = Y_{(n)} - Y_{(1)}$ we can operate under the assumption that $Y_{(1)} = 0$.

Then $P(R < r)$ is the probability that the remaining $n-1$ sample observations all fall in the range $(0,r)$. This is because (under the assumption that the smallest observation is $0$), $R < r$ means that the largest observation must be smaller than $r$. But in order for that to happen, all $n-1$ of the remaining sample observations must be smaller than $r$. (In fact, the two statements are equivalent.)

Thus $$ \begin{align} P(R < r) &= P(n-1 \text{ independent sample observations are all smaller than } r) \\ &= \left(\int_0^r e^{-x} \, dx\right)^{n-1} \\ &= (1-e^{-r})^{n-1}.\end{align}$$ Differentiating to obtain the pdf of $R$, we get $$f_R(r) = (n-1)(1-e^{-r})^{n-2} e^{-r}.$$

  • $\begingroup$ Sorry if i am asking a stupid question, but did you mean $P(R > r)$ $\endgroup$ – Hawk Nov 28 '12 at 6:55
  • $\begingroup$ @sizz: No, I do mean $P(R < r)$. Let me update my answer. $\endgroup$ – Mike Spivey Nov 28 '12 at 16:14

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