为下面题目写出详细证明推到过程Suppose that Xt and Yt are independent Poisson processes with parameter 1 and 2 respectively measuring the number of calls arriving at two different service center. Let Zt = Xt + Yt (1) Find the distribution of Zt (2) Derive the distribution of Zt+s − Zs (3) Is Zt+s − Zs independent of Zs? Why.

(1) The distribution of Zt can be found by considering that it is the sum of two independent Poisson processes with parameters 1 and 2 respectively. The probability mass function of a Poisson process with parameter λ is given by:

P(X=k) = (λ^k / k!) * e^(-λ)

Therefore, the probability mass function of Zt is:

P(Zt=k) = P(Xt+Yt=k) = Σi=0^k P(Xt=i) * P(Yt=k-i)
= Σi=0^k (e^(-1) * 1^i / i!) * (e^(-2) * 2^(k-i) / (k-i)!)
= e^(-3) * Σi=0^k (1^i * 2^(k-i) / (i! * (k-i)!))
= e^(-3) * Σi=0^k (k! / (i! * (k-i)!) * 1^i * 2^(k-i))
= e^(-3) * (1+2)^k / k!
= e^(-3) * 3^k / k!

Therefore, Zt follows a Poisson distribution with parameter 3.

(2) To derive the distribution of Zt+s − Zs, we need to consider the behavior of the Poisson process over the interval [s, t+s]. We know that Zt = Xt + Yt, and that Xt and Yt are independent Poisson processes with parameters 1 and 2 respectively. Therefore:

Zt+s = X(t+s) + Y(t+s)
Zs = Xs + Ys

Zt+s − Zs = (X(t+s) + Y(t+s)) − (Xs + Ys)
= (X(t+s) − Xs) + (Y(t+s) − Ys)

Since Poisson processes are memoryless, the increments X(t+s) − Xs and Y(t+s) − Ys are independent of Xs and Ys respectively. Therefore, we can treat them as independent Poisson processes with parameters 1 and 2 respectively. The difference between two independent Poisson processes with parameters λ1 and λ2 is itself a Poisson process with parameter λ1 − λ2. Therefore:

Zt+s − Zs = (X(t+s) − Xs) + (Y(t+s) − Ys)
= X(t+s) − Xs + Y(t+s) − Ys
~ Poisson(1+2) = Poisson(3)

Therefore, Zt+s − Zs follows a Poisson distribution with parameter 3.

(3) Zt+s − Zs is independent of Zs because Poisson processes are memoryless. The value of Zs tells us nothing about the future behavior of Zt+s − Zs. Therefore, Zt+s − Zs is independent of Zs.