If $X_i, i = 1,2,3$ are independent exponential random variables with rates $\lambda_i$, $i = 1,2,3,$ find $E[max X_i| X_1 < X_2 < X_3]$

This question from Ross' textbook has been asked many times in mathstackexchange (see comment). After reading all the solutions and discussion in this website, now I understand that why $E[X_3] \neq \frac{1}{\lambda_3}$. I have calculated the integral of three p.m.f's to get the result and verified the answer. However, I still would like to understand the "simpler" way of calculating the expectation.

The solution is given as below:



I don't understand how $E[X_2-X_1|X_1 < X_2 < X_3 ] + E[X_3-X_2|X_1 < X_2 < X_3 ]$ went $E[X_2|X_1< X_2 < X_3 ]+ E[X_3 ]$. I don't understand how it is related to memoryless property of exponential distribution as the solution suggests.

Please someone help....!


1 Answer 1


Let $\lambda_i = \lambda,$ for $i = 1,2,3.$ In this special case, $E(\max_i(X_i)),$ for $X_i \stackrel{iid}{\sim} \mathsf{Exp}(rate = \lambda)$ can be found as follows:

Consider the $X_i$ to be times to failure of three devices. The time to failure of the first device is $\min_i(X_i) = X_{(1)} \sim \mathsf{Exp}(3\lambda),$ with $E(X_{(1)}) = 1/3\lambda.$

Then, by the no-memory property, the additional time to failure of the second device is $D_2 = X_{(2)}-X_{(1)} \sim \mathsf{Exp}(2\lambda),$ with $E(D_2) = 1/2\lambda.$ This is the average minimum time to failure of remaining two devices.

Similarly, the additional time to failure $D_3$ of the (single remaining) third device has $E(D_2) = 1/\lambda.$

Thus the total expected time to failure of the third device is $E(\max(X_i)) = E(X_{(3)}) = 1/3\lambda + 1/2\lambda + 1/\lambda.$

This method cannot be used for the general case in which the rates are unequal because we don't know which devices will fail first and second.

However, with the condition that $X_1 < X_2 < X_3,$ we do know the order of failure, so the conditional expected time to failure can be found as in the solution attached to the question.

Simulation in R for max and min with $\lambda=2:$

set.seed(728);  m=10^6;  lam = 2
x1 = rexp(m,lam);  x2 = rexp(m,lam);  x3 = rexp(m,lam)
v = pmin(x1, x2, x3)
[1] 0.1664693        # aprx E(min) = 1/6
w = pmax(x1, x2, x3)
[1] 0.9167773        # aprx E(max) = 11/12
1/(3*lam) + 1/(2*lam) + 1/lam
[1] 0.9166667        # 11/12
  • 1
    $\begingroup$ Another way to treat this case is to note that the conditional distribution $max| X_1<X_2<X_3$ is equal to the distribution $max| X_{\sigma(1)}<X_{\sigma(2)}<X_{\sigma(3)}$ for any permutation sigma. By the law of total expectation, $Emax=\sum_{\sigma} E(max| X_{\sigma(1)}<X_{\sigma(2)}<X_{\sigma(3)})p(X_{\sigma(1)}<X_{\sigma(2)}<X_{\sigma(3)})$ $=E(max| X_1<X_2<X_3)\sum_{\sigma} p(X_{\sigma(1)}<X_{\sigma(2)}<X_{\sigma(3)})=E(max| X_1<X_2<X_3)$, and the unconditional expectation can be computed by a simple integral. $\endgroup$
    – Mike Hawk
    Jul 29, 2019 at 18:20
  • $\begingroup$ There are only six permutations, would it be feasible to make that into a general answer? $\endgroup$
    – BruceET
    Jul 29, 2019 at 18:28
  • $\begingroup$ Maybe, but the argument is only valid in the case when all rates are equal. $\endgroup$
    – Mike Hawk
    Jul 29, 2019 at 18:31

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