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We have $ X_{1},... , X_{8} $. All independent exponential r.v. with mean $ 1 $. I know how to find the cdf of the smallest among them, but i didn't see how to find the third smallest and its expected value.

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Let $X_{(3)}$ be the third-smallest of $X_1, \ldots, X_8$. Then $X_{(3)} < x$ if and only if at least three of $X_1, \ldots, X_8$ are less than $x$. Now, $P(X_1 < x) = 1 - e^{-x}$ (and similarly for $X_2, \ldots, X_8$). Can you take it from here?

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  • $\begingroup$ Yes I got it, thank you $\endgroup$ – Hal03 Nov 30 '16 at 22:47

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