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Math Help - Help: 2 exponential problem

  1. #1
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    Help: 2 exponential problem

    Interesting title, huh?

    I am having trouble solving this problem. Assume X and Y are independent exponential random variables with means 1/x and 1/y, respectively. If Z=min(X,Y): Is Z exponentially distributed as well (if so, how do you know)? What is the expectation of Z? What is the probability that x < y?

    Lastly, with the information from above, show how a M/M/1 queue could be represented as a Markov chain that is continuous-time with transition rates Qn,n+1=L and Qn,n-1=U, n=0,1,2,... M/M/1 queue=Arrival is Poisson/Service is Exponential/1 server (with an infinite buffer size).
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  2. #2
    MHF Contributor matheagle's Avatar
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    Do X and Y have the same mean?
    If so, then the min of these two is another EXP rv.

    Let the density of X be

    {1\over \beta}e^{-x/\beta} where x>0

    and P(X>a)=\int_a^{\infty}{1\over \beta}e^{-x/\beta}=e^{-a/\beta}

    Let Z be the min of X and Y.

    Then P(Z\le a)=1-P(Z>a)=1-P(X>a)P(Y>a)

    And I'll stop here, not because it's 2:12 am,
    but I need to know if you have the same parameter for
    both X and Y or not.
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  3. #3
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    Hey thanks for replying. X and Y don't have the same means. The means are 1/x and 1/y, respectively.
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  4. #4
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    I'm not sure how to compute the Markov chain transition rates either. Is there a base equation you work with and go from there?
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