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Dec 21, 2016 at 13:36 history edited Johan Falkenjack CC BY-SA 3.0
Formatted math formulas latex like
Dec 21, 2016 at 13:27 vote accept Johan Falkenjack
Sep 16, 2015 at 15:57 answer added Johan Falkenjack timeline score: 2
Sep 16, 2015 at 15:52 comment added whuber 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 $X_i$).
Sep 16, 2015 at 15:14 comment added Johan Falkenjack @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?
Sep 16, 2015 at 12:38 comment added Johan Falkenjack @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?
Sep 15, 2015 at 20:33 comment added jlimahaverford 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.
Sep 15, 2015 at 19:54 comment added whuber +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.
Sep 15, 2015 at 19:26 review First posts
Sep 15, 2015 at 19:27
Sep 15, 2015 at 19:21 history asked Johan Falkenjack CC BY-SA 3.0