24K (Applied Probability)

A rate-

λ

Poisson process is a counting process starting at zero, with right-continuous nondecreasing integer-valued paths, independent stationary increments, and

N_{t} - N_{s} \sim Pois (λ (t - s))

for

0 \leq s < t

. Its jumps have size one almost surely.

For independent Poisson processes

N, M

, increments of

N + M

on disjoint intervals are independent, because all the corresponding increments of the two processes are jointly independent. On an interval of length

t

, multiplying their probability generating functions gives

exp (λ t (z - 1)) exp (μ t (z - 1)) = exp ((λ + μ) t (z - 1)) .

(1)

Thus the increment is Poisson with mean

(λ + μ) t

. The path requirements hold, and independent processes almost surely have no simultaneous jumps. Therefore $N + M$ is a Poisson process of rate $λ + μ$ .

ii

 0  0

Solution

 0  0

The random clock is understood to be independent of the discrete-time Markov chain; without this hypothesis its law need not be determined by

K

alone. Conditional on

N_{t} = n

, the transition matrix is

K^{n}

, so

P_{t} = e^{- λ t} \sum_{n \geq 0} \frac{( λ t ) ^{n}}{n !} K^{n} = e^{λ t (K - I)}, Q = λ (K - I) .

(1)

Hence

q_{x y} = λ K_{x y}

for

x \neq = y

and

q_{xx} = - λ (1 - K_{xx})

. If

π K = π

, then

π P_{t} = π

by the series, proving that the invariant distribution is unchanged.

Let

ξ_{i}

be the independent exponential interarrival times of the Poisson process. The absorption time is

T_{a} = \sum_{i = 1}^{τ_{a}} ξ_{i}

. By independence of the chain and clock and Tonelli theorem,

E_{x} T_{a} = \sum_{i \geq 1} E [ξ_{i}] P_{x} (τ_{a} \geq i) = \frac{1}{λ} E_{x} τ_{a} .

(2)

This includes infinite expectations and absorption at time zero. The same argument proves the hinted random-sum identity for an independent nonnegative integer-valued count