Question

Sum of two independent poisson processes is a poisson process proof

Answer 1

First observe that given N(t) = n + m,

N1(t) ∼ Binomial(n + m, p). (why?)

Thus P(N1(t) = n, N2(t) = m)

= P(N1(t) = n, N2(t) = m|N(t) = n + m)P(N(t) = n + m)

=

n + m

n

p

n

(1 − p)

me

−λt

(λt)

n+m

(n + m)!

= e

−λtp

(λpt)

n

n!

e

−λt(1−p)

(λ(1 − p)t)

m

m!

= P(N1(t) = n)P(N2(t) = m).

This proves the independence of N1(t) and N2(t) and that

N1(t) ∼ Poisson(λpt), N2(t) ∼ Poisson(λ(1 − p)t).

Both {N1(t)} and {N2(t)} inherit the stationary and independent

increment properties from {N(t)}, and hence are both Poisson

processes.