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Poisson counting process & order statistics

  1. Oct 13, 2009 #1
    Theorem: Let {N(t): t≥0} be a Poisson process of rate λ. Suppose we are given that for a fixed t, N(t)=n. Let Ti be the time of the ith event, i=1,2,...n.
    Then the (conditional) density function of Tn given that N(t)=n is the exactly the same as the density function of X(1)=min{X1,X2,...,Xn}, where X1,X2,...,Xn are i.i.d. uniform(0,t).
    (similar result holds for the density of the other Ti's)

    Example: Let {N(t): t≥0} be a Poisson process of rate λ. The points are to be thought of as being the arrival times of customers to a store which opens at time t=T. The customers arriving between t=0 and t=T have to wait until the store opens. Let Y be the total times that these customers have to wait. Calculate E(Y).

    (T1,T2,...,TN) is equal in distribution to (X(1),X(2),...,X(N)), where the order statistics are coming from X1,X2,...,XN which are i.i.d. uniform(0,T).
    => T1+T2+...+TN is equal in distribution to X(1)+X(2)+...+X(N) = X1+X2+...+XN
    E(Y)=E(total waiting time)
    =E(NT) - E(T1+T2+...+TN)
    =T E(N) - E(X1+X2+...+XN)
    =T E(N) - E[E(X1+X2+...+XN|N)]
    =T E(N) - E[N E(X1)]
    =T E(N) - E[N T/2]
    =T E(N) - (T/2) E[N]
    =(1/2) T E(N)
    =(1/2) T λT
    In the theorem, we require N(t)=n where n is a FIXED number. But throughout the solution (for example when they calculate E[X1+X2+...+XN]), N(t)=N is being treated as a RANDOM VARIABLE rather than a fixed number. Why? If N is a random variable, I don't think the theorem above applies...will we still have that TN is equal in distirbution to X(N)? Why or why not?
    In other words, I am questioning the statement "(T1,T2,...,TN) is equal in distribution to (X(1),X(2),...,X(N))" in the solution of the example. Here N itself is a random variable, but in the theorem it is required to be fixed (in the theorem we're GIVEN the value of N(t))...

    Can someone please explain? Any help is greatly appreciated!:)
  2. jcsd
  3. Oct 16, 2009 #2


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    Apparently by "N(T)=N" they did not mean N is deterministic.
  4. Oct 17, 2009 #3
    That's right, you'll need the theorem that the joint distribution of the arrival times conditional on N is the same as the order stats of the uniform distribution.

    To apply the theorem correctly, take the expectation of the conditional expectation.
  5. Oct 18, 2009 #4
    If the theorem above does not apply in our problem, is it still true that "(T1,T2,...,TN) is equal in distribution to (X(1),X(2),...,X(N))" ?(here N=N(T) is a random variable, I believe) Why or why not?
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