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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2016/11/10/variance-of-an-exponential-order-statistics/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong>T</strong>his afternoon, one of my Monte Carlo students at ENSAE came to me with an exercise from <a target="_blank" rel="nofollow" href="http://www.amazon.com/gp/product/1441919392?ie=UTF8&tag=chrprobboo-20&linkCode=as2&camp=1789&creative=390957&creativeASIN=1441919392">Monte Carlo Statistical Methods</a> that I did not remember having written. And I thus “charged” <a target="_blank" rel="nofollow" href="http://george%20casella/">George Casella</a> with authorship for that exercise!</p>
<p><a target="_blank" rel="nofollow" href="https://i0.wp.com/xianblog.files.wordpress.com/2016/11/exerc33-e1478288127856.jpg?ssl=1"><img sizes="(max-width: 450px) 100vw, 450px" alt="" class="wp-image-31931" src="https://i0.wp.com/xianblog.files.wordpress.com/2016/11/exerc33-e1478288127856.jpg?resize=450%2C234&ssl=1" data-recalc-dims="1"></a>Exercise 3.3 starts with the usual question (a) about the (Binomial) precision of a tail probability estimator, which is easy to answer by iterating simulation batches. Expressed via the empirical cdf, it is concerned with the <em>vertical</em> variability of this empirical cdf. The second part (b) is more unusual in that the first part is again an evaluation of a tail probability, but then it switches to find the .995 quantile by simulation and produce a precise enough [to three digits] estimate. Which amounts to assess the <em>horizontal</em> variability of this empirical cdf.</p>
<p>As we discussed about this question, my first suggestion was to aim at a value of N, number of Monte Carlo simulations, such that the .995 x N-th spacing had a length of less than one thousandth of the .995 x N-th order statistic. In the case of the Exponential distribution suggested in the exercise, generating order statistics is straightforward, since, as suggested by <a target="_blank" rel="nofollow" href="https://xianblog.wordpress.com/2016/10/31/ratio-of-uniforms-2/">Devroye</a>, see Section V.3.3, the i-th spacing is an Exponential variate with rate (N-i+1). This is so fast that <a target="_blank" rel="nofollow" href="https://xianblog.wordpress.com/2016/10/31/ratio-of-uniforms-2/">Devroye</a> suggests simulating Uniform order statistics by inverting Exponential order statistics (p.220)!</p>
<p>However, while still discussing the problem with my student, I came to a better expression of the question, which was to figure out the variance of the .995 x N-th order statistic in the Exponential case. Working with the density of this order statistic however led nowhere useful. A bit later, after Google-ing the problem, I came upon <a target="_blank" rel="nofollow" href="http://math.stackexchange.com/a/80485/14893">this Stack Exchange solution</a> that made use of the spacing result mentioned above, namely that the expectation and variance of the k-th order statistic are</p>
<p><img title="\mathbb{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}" alt="\mathbb{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}" class="latex" src="https://s0.wp.com/latex.php?latex=%5Cmathbb%7BE%7D%5BX_%7B%28k%29%7D%5D%3D%5Csum%5Climits_%7Bi%3DN-k%2B1%7D%5EN%5Cfrac1i%2C%5Cqquad+%5Cmbox%7BVar%7D%28X_%7B%28k%29%7D%29%3D%5Csum%5Climits_%7Bi%3DN-k%2B1%7D%5EN%5Cfrac1%7Bi%5E2%7D&bg=000000&%23038;fg=B0B0B0&%23038;s=0"></p>
<p>which leads to the proper condition on N when imposing the variability constraint.</p>
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