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I have been trying to find the general formula for the $k$th order statistics of $n$ i.i.d exponential distribution random variables with mean $1$. And how to calculate the expectation and the variance of the $k$th order statistics. Can someone give me some general formula? It would be nice if there is any proof.

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2 Answers 2

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The minimum $X_{(1)}$ of $n$ independent exponential random variables with parameter $1$ is exponential with parameter $n$. Conditionally on $X_{(1)}$, the second smallest value $X_{(2)}$ is distributed like the sum of $X_{(1)}$ and an independent exponential random variable with parameter $n-1$. And so on, until the $k$th smallest value $X_{(k)}$ which is distributed like the sum of $X_{(k-1)}$ and an independent exponential random variable with parameter $n-k+1$.

One sees that $X_{(k)}=Y_{n}+Y_{n-1}+\cdots+Y_{n-k+1}$ where the random variables $(Y_i)_i$ are independent and exponential with parameter $i$. Each $Y_i$ is distributed like $\frac1iY_1$, and $Y_1$ has expectation $1$ and variance $1$, hence $$ \mathrm E(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1i,\qquad \mbox{Var}(X_{(k)})=\sum\limits_{i=n-k+1}^n\frac1{i^2}. $$

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    $\begingroup$ Very nice, Didier. I'll remember this trick! $\endgroup$ Nov 10, 2011 at 7:45
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    $\begingroup$ Is there an easy way to generalize this to independent exponential with different rates? $\endgroup$
    – A.S.
    Oct 6, 2015 at 2:31
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    $\begingroup$ @A.S. I see no obvious way to. $\endgroup$
    – Did
    Oct 6, 2015 at 7:33
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    $\begingroup$ Could you please explain why $X_{(2)}$ is distributed like the sum of $X_{(1)}$ and an independent exponential random variable with parameter $n-1$? $\endgroup$
    – PSPACEhard
    Apr 13, 2016 at 10:26
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    $\begingroup$ @NP-hard Devroye (1986, p.219) invokes the memoryless property of the exponential distribution. $\endgroup$ Nov 4, 2016 at 20:52
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The probability density function for the $k$th order statistic of a sample of size $n$ from a distribution with pdf $f(x)$ and distribution function $F(x)$ is

$$f_k(x) = n {n-1\choose k-1} F(x)^{k-1} (1-F(x))^{n-k} f(x)$$

and, since the exponential distribution with mean 1 has pdf $f(x)=e^{-x}$ and distribution function $F(x)=1-e^{-x}$, we can compute

$$f_k(x) = n {n-1\choose k-1} (1-e^{-x})^{k-1} e^{-(n-k+1)x}$$

Computing the expectation and variance of the $k$th order statistic can now be done via the usual method.

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    $\begingroup$ but how would you calculate the expectation of this? It involves a lot of calculus $\endgroup$ Nov 9, 2011 at 10:22
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    $\begingroup$ It's quite likely that there isn't a closed form. Since it involves terms like $y^a(1-y)^b$ the integrals can probably be expressed using hypergeometric functions, although maybe you could give it a crack by constructing a recurrence for $n$ and $k$. I'll have a look at that later. $\endgroup$ Nov 9, 2011 at 10:25
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    $\begingroup$ "It's quite likely that there isn't a closed form" See other answer. $\endgroup$
    – Did
    Sep 3, 2016 at 7:47

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