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Exponential Distribution Problem

  1. Sep 18, 2009 #1
    I am having trouble solving this problem. I'm not sure how to solve 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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  3. Sep 18, 2009 #2

    EnumaElish

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    "Is Z exponentially distributed as well (if so, how do you know)?"

    Prob {Z < z} = Prob {min(X,Y) < z} = 1 - Prob {min(X,Y) > z} = 1 - Prob {X > z and Y > z}. Using this logic, you can derive the distribution of Z and decide whether it looks exponential.

    "What is the probability that x < y?"

    You can use either route:

    1. Prob {X < Y} = Prob {X/Y < 1}

    2. Prob {X < Y} = Prob {X - Y < 0}

    In either case, you need to derive the distribution of X/Y or X - Y.
     
  4. Sep 18, 2009 #3
    Thanks, I'm working on that and it looks likes since X and Y are independent, I would be able to add them together in the denominator for the new mean (1/(x+y)).

    Does anyone know about the Markov part? That sort of came out of nowhere.
     
  5. Sep 18, 2009 #4

    EnumaElish

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    A Markov chain describes a system that is in one state (out of two or more states) at each period (e.g., at the end of each day); the probability of going from state s today to state s' tomorrow is independent of yesterday's state. For MM1, what would those states be, and how would you characterize these transition probabilities?
     
  6. Sep 19, 2009 #5
    For M/M/1, given that arrival rates is Poisson with mean L and service (exit) rate is Exponential with mean U, I would think that the probability of getting an additional person in the next state is L and losing a person in the next state is U, which is just from logic. I'm not sure how to compute that though...?
     
  7. Sep 20, 2009 #6

    EnumaElish

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    It appears as if you need to compute the probability of acquiring N new arrivals between the end of this period and the end of the next period, conditional on having acquired A arrivals by the end of this period. This will give you the transition probability between today's state (A) and tomorrow's (A+N).
     
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