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The Problem:

Consider $X_1, ..., X_5$, $5$ iid, exponentially distributed random variables with $\lambda = 1$. Let $Y_1 \lt Y_2 \lt Y_3 \lt Y_4 \lt Y_5$ be their order statistics.

Find the distribution of $ \sum_{i=1}^{4} Y_i $.

SOLUTION:

$$ \sum_{i=1}^{4} Y_i \sim \text{Expo}(5) + \text{Expo}(4) + \text{Expo}(3) + \text{Expo}(2). \qquad (*)$$

How does this work? I understand that $Y_j - Y_{j-1} \sim \text{Expo}((n-j+1)\lambda)$, and therefore, for each $Y_i$, we have that $Y_i = \sum_{j=1}^{i} (Y_j - Y_{j-1})$; but, that means that the desired sum should be a sum of $10$ exponentially distributed random variables (because each $Y_i$ is the sum of $i$ exponential RVs) instead of $4$. What am I missing here?

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    $\begingroup$ You are correct. - need a factor of $i-1$ in front of $\exp(i)$. $\endgroup$ – A.S. Mar 8 '16 at 2:34
  • $\begingroup$ So, are you saying that $(*)$ is, in fact, wrong? $\endgroup$ – thisisourconcerndude Mar 8 '16 at 2:36
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    $\begingroup$ Yes. Write an equivalent of $(*)$ for the sum of all $5$ order statistics and take expected value to triple confirm that. $Y_4\sim (*)$ $\endgroup$ – A.S. Mar 8 '16 at 2:39
  • $\begingroup$ Always the Exponential model. Exponential this. Exponential that. Exponential order stats. Sum of Exponentials. Students might think life is a picnic. They should make these questions more interesting and consider a different model. $\endgroup$ – wolfies Mar 8 '16 at 5:48
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    $\begingroup$ @wolfies Consider life as a sequence of picnics. The duration of each picnic is given by $\text{Expo}(\lambda)$... $\endgroup$ – thisisourconcerndude Mar 8 '16 at 14:56

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