Let N(t) be a Poisson process. Suppose that if N(t) is odd, then at most one event occurs in a very small interval h with probability ₁h.Suppose that if N(t) is even, then at most one event occurs in a very small interval h with probability ₂h.

Under these assumptions, show that:

P₁(t) =P[N(t) = odd] = [₂/( ₁+ ₂)](1-exp{-( ₁+ ₂)t});

P₂(t) =P[N(t) = even] = ₁/( ₁+ ₂) +[₂/( ₁+ ₂)]exp{-( ₁+ ₂)t

Hint: Derive the system of differential equations by obtaining expressions for P₁(t+h) and P₂(t+h) and conditioning on N(t) = odd and N(t) = even, and show that they are satisfied by the proposed solution.