Poisson processes?

Question

Suppose X(t) and Y(t) are two independent poisson processes with the same rate c. Let W_n be the time until the n-th event in process X(t). Find E(W_X(t)) and E(W_Y(t)).

Answer 1

W_X(t) is the waiting time till the last event before time t.
Given that X(t)=n, the conditional distributions of the waiting times till the 1st,2nd,...,nth events are independent and uniform on (0,t), and W_X(t) is the maximum of these.

The cdf of a uniform RV on (0,t) is x/t if 0 Then the cdf of the maximum of these is P(X_1
Thus, the conditional pdf of W_X(t) is the derivative of the above function: n*x^(n-1)/t^n if 0
Then E(W_X(t))=E(E(W_X(t)|X(t))) (the tower property of conditional expectations). Thus, if h(n)=E(W_X(t)|X(t)=n)=
= n/(n+1)*t, then h(X(t)) = t*X(t)/((X(t)+1) =
= t*(1-1/(X(t)+1)), and finally, E(W_X(t)) = E(h(X(t)).

To compute this, apply E(g(X)) = sum of g(x)*p(x) (for discrete RV's), thus E(h(X(t))) = sum (k=0..infinity) of
t*(1-1/(k+1))* exp(-c*t)*(c*t)^k/k! = t-1/c*sum(k=0..infinity) of
exp(-c*t)*(c*t)^(k+1)/(k+1)! = t-1/c*sum(j=1..infinity) of
exp(-c*t)*(c*t)^(j+1)/(j+1)! = t-1/c*(1-exp(-c*t)) =
= 1/c* ((exp(-c*t)+c*t-1).

Similarly, E(W_Y(t))=E(E(W_Y(t)|Y(t))). First we need
h(n) = E(W_Y(t)|Y(t)=n) = E(W_n|Y(t)=n) = E(W_n) (by independence of the two processes) = n/c (because W_n follows a gamma distribution).

Then h(Y(t)) = Y(t)/c, and E(W_Y(t)) = E(h(Y(t)) = E(Y(t)/c) =
= 1/c*E(Y(t)) = 1/c * c*t (since Y(t) follows a Poisson distribution) = t.

Answer 2

According to my 'Larsen and Marx", time between consecutive Poisson event is exponential and the waiting time until the nth event is a Gamma distribution.

lamda has units of 1/time (need to be careful about this parameter, some use the reciprocal). In the Poisson, it should be exp(-lamda*T))

For the Gamma:

E(W) = n/lamda