Let $\{M_{i}(t), t \geq 0\}$, $i=1,2,3$ be independent Poisson processes with respect rates $\lambda_{i}$, $i=1,2,3$ and set $$N_{1}(t)=M_{1}(t)+M_{2}(t), \quad N_{2}(t)=M_{2}(t)+M_{3}(t)$$ The stochastic process $\{(N_{1}(t), N_{2}(t)),t \geq 0\}$ is called a bivariate Poisson process.
Find $\text{Cov}(N_{1}(t),N_{2}(t))$.
Hello, I am trying to solve this problem and I did the following \begin{align*} \text{Cov}[N_{1}(t), N_{2}(t)] &= \text{Cov}[M_{1}(t)+M_{2}(t), M_{2}(t)+M_{3}(t)]\\ &= \text{Cov}[M_{2}(t), M_{2}(t)] \quad (\text{By independence})\\ &= \text{Var}(M_{2}(t)). \end{align*} But I don't know how to proceed.
We can compute this directly using the identity $$ \mathrm{Cov}(X,Y) = \mathbb E[XY] - \mathbb E[X]\mathbb E[Y] $$ and the fact that for independent random variables with finite expectation, $$ \mathbb E[XY] = \mathbb E[X]\mathbb E[Y]: $$ \begin{align} \mathrm{Cov}(N_1(t),N_2(t)) &= \mathbb E[(M_1(t)+M_2(t))(M_2(t)+M_3(t))] - \mathbb E[M_1(t)+M_2(t)]\mathbb E[M_1(t)+M_2(t)]\\ &= \mathbb E[M_1(t)M_2(t) + M_2(t)^2+M_1(t)M_3(t)+M_2(t)M_3(t)] - (\lambda_1+\lambda_2)(\lambda_2+\lambda_3)\\ &= \lambda_1\lambda_2+\lambda_2^2+\lambda_1\lambda_3 + \lambda_2\lambda_3 - (\lambda_1\lambda_2+\lambda_2^2+\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)\\ &= \lambda_2. \end{align}