kingwinner
Oct13-09, 06:12 AM
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).
Solution:
N(T)=N
N(T)~Poisson(λT)
(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[(T-T1)+(T-T2)+...+(T-TN)]
=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!:)
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).
Solution:
N(T)=N
N(T)~Poisson(λT)
(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[(T-T1)+(T-T2)+...+(T-TN)]
=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!:)