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I'm sorry if this question is trivial but I'm new at this and I'm learning this stuff on my own from a book + youtube tutorials so I haven't really got anyone to ask. I have the following exercise from a book of basic probability theory:

"A process is divided into 3 sub-processes which are carried out in parallel. Each sub-process takes an Exponential amount of time, 5 minutes on the average, independent of the other sub-processes. The process is finished when all sub-processes are finished. What is the expected time for the full process to complete?"

If I understand this correctly we have three random variables $X_1, X_2, X_3 \sim Exponential(1/5)$. My naive guess would be that the expected time is 5 minutes as the sub-processes are independent and parallel but I'm probably missing something. I guess I could compute the joint distribution $f(Y) = f(X_1, X_2, X_3)$ and calculate $E(Y)$ from this. However, this entails calculating a triple integral and I'm hoping something in the simple setup (independence and identical distributions) means there is a simpler solution. Can anyone give me any advice?

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    $\begingroup$ +1 Begin by writing down a formula for the time to completion based on the actual times needed by the three subprocesses. For instance, when those times are $1$, $8$, and $9$ minutes, your formula should give $9$ minutes. This formula, when applied to $X_1,X_2,X_3$, gives a univariate random variable. Find its distribution function (it doesn't require a triple integral), then compute the expectation. $\endgroup$
    – whuber ♦
    Commented Sep 15, 2015 at 19:54
  • $\begingroup$ As Whuber suggested, "the process is finished when all sub-processes are finished" leads to a mathematical description of one variable of interest, depending on these three random variables. $\endgroup$
    – jlimahaverford
    Commented Sep 15, 2015 at 20:33
  • $\begingroup$ @whuber I'm sorry but I'm not sure I understand. Are you saying I should construct a formula f(X1,X2,X3) = max(X1,X2,X3) and calculate it's density? $\endgroup$
    – Johan Falkenjack
    Commented Sep 16, 2015 at 12:38
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    $\begingroup$ @whuber Okay, I think I got it. P(Y<t) = P(X1<t, X2<t, X3<t)=/by independence/=P(X1<t)P(X2<t)P(X3<t). The CDF of Y would then be (1-e^(-t/5))^3. I derive this to get PDF, f(t), and calculate E(Y) the usual way by integrating t*f(t) to infinity. This gave me ~9.17 which doesn't seem unreasonable. Is my reasoning correct? $\endgroup$
    – Johan Falkenjack
    Commented Sep 16, 2015 at 15:14
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    $\begingroup$ Sounds right to me. A good way to get more feedback is to post your solution as an answer, then wait for votes and comments. Incidentally, there's an easier way to obtain the expectation: integrate $1-F(x)^3$ from $0$ to $\infty$ (where $F$ is the common CDF of the $X_i$). $\endgroup$
    – whuber ♦
    Commented Sep 16, 2015 at 15:52

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Okay, I think I got it. $P(Y<t) = P(X_1<t, X_2<t, X_3<t) = /\text{by independence}/ = P(X_1<t)P(X_2<t)P(X_3<t)$

The CDF of $Y$ would then be $(1-e^{-t/5})^3$. I derive this to get PDF, $f(t)$, and calculate $E(Y)$ the usual way by integrating $t \times f(t)$ to infinity. This gave me ~9.17 which doesn't seem unreasonable. Is my reasoning correct?

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