Computation of the probability of dependant uncertain values

Discussion in 'Scientific Statistics Math' started by Santiago Erquicia, Feb 13, 2004.

  1. I have three task to do which the necesary time to perform them
    is uncertain. Each task time is independant of the others. Supose that
    tasks B and C only can start when task A is finished.

    I can calculate the finishing time of tasks B and C by doing the
    combination of tasks A with B and C respectively. I have no problem with
    that.

    Supose that task B and C both have to finish before starting task D. In
    order to know the finishing time of task D, I need to know the starting
    time of it.

    If the ending time of task B and C were independant, the initial time
    of task D would be easy to calculate as cdf_D_initial(t) = cdf_C_final(t) *
    cdf_B_final(t). As both depend on task A, I cannot do that.

    I know that I can combine tasks B and C first and then add task A to get
    the result. My problem is that there are some cases where the solution is
    not so simple.

    Do you know any method to calculate this taking care of the dependency
    between the two pdfs?

    I've been reading something about copulas and I saw
    that they are used to generate dependant random numbers. It looks like
    the inverse I want to get. Is that complicated to use? Consider that I
    am not good at statistics.

    Thanks in advance,
    Santiago
     
    Santiago Erquicia, Feb 13, 2004
    #1
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  2. Yes, you can. time(B) and time(C) are independent -given time(A)-,
    and that's enough.

    Maybe the problem will be clearer if you phrase it in terms of times
    elapsed, i.e., durations, instead of clock or calendar times.
    Then the ending time for all tasks will just be the sum of the durations.

    Now the distribution of the sum of durations may not be neat
    and simple -- but you can always resort to Monte Carlo methods.
    Try taking the analysis as far as possible first.
    Well, then tell us about these more complicated problems. This one
    doesn't seem to present any difficulty.

    For what it's worth,
    Robert Dodier
     
    Robert Dodier, Feb 15, 2004
    #2
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