Poisson counting process & order statistics

[kingwinner]

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 T_i be the time of the i^th event, i=1,2,...n.
Then the (conditional) density function of T_n given that N(t)=n is the exactly the same as the density function of X₍₁₎=min{X₁,X₂,...,X_n}, where X₁,X₂,...,X_n are i.i.d. uniform(0,t).
(similar result holds for the density of the other T_i'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,...,T_N) is equal in distribution to (X₍₁₎,X₍₂₎,...,X_(N)), where the order statistics are coming from X₁,X₂,...,X_N which are i.i.d. uniform(0,T).
=> T₁+T₂+...+T_N is equal in distribution to X₍₁₎+X₍₂₎+...+X_(N) = X₁+X₂+...+X_N
E(Y)=E(total waiting time)
=E[(T-T₁)+(T-T₂)+...+(T-T_N)]
=E(NT) - E(T₁+T₂+...+T_N)
=T E(N) - E(X₁+X₂+...+X_N)
=T E(N) - E[E(X₁+X₂+...+X_N|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+...+X_N]), 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 T_N is equal in distirbution to X_(N)? Why or why not?
In other words, I am questioning the statement "(T₁,T₂,...,T_N) 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!:)

 

[EnumaElish]

Apparently by "N(T)=N" they did not mean N is deterministic.

 

[bpet]

I don't think the theorem above applies...

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.

...Here N itself is a random variable, but in the theorem it is required to be fixed..

To apply the theorem correctly, take the expectation of the conditional expectation.

 

[kingwinner]

If the theorem above does not apply in our problem, is it still true that "(T₁,T₂,...,T_N) 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?