Question

Show that if \(\left\\{N_{i}(t), t \geqslant 0\right\\}\) are independent Poisson processes with rate \(\lambda_{i}, i=1,2\), then \([N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\) where \(N(t)=N_{1}(t)+\) \(N_{2}(t)\)

Short answer

In summary, we've shown that if \(N_1(t)\) and \(N_2(t)\) are independent Poisson processes with rates \(\lambda_{1}\) and \(\lambda_{2}\), then their sum \(N(t) = N_{1}(t) + N_{2}(t)\) is also a Poisson process with rate \(\lambda_{1}+\lambda_{2}\), by verifying that \(N(t)\) satisfies the independent and stationary increments property and its counts follow a Poisson distribution.

Step-by-step solution

Verify Independent and Stationary Increments

For independent and stationary increments, we want to show that for any choice of time intervals \((t_1, t_2], (t_3, t_4], \dots, (t_{2n}, t_{2n+1}]\), where \(0 \leq t_1 < t_2 < t_3 < \dots < t_{2n+1}\), the increment counts \(N(t_2) - N(t_1), N(t_4) - N(t_3), \dots, N(t_{2n+1}) - N(t_{2n})\) are independent. Each of these increments can be written as the sum of the corresponding increments in \(N_1(t)\) and \(N_2(t)\): \[ N(t_{2i})-N(t_{2i-1}) = (N_1(t_{2i})-N_1(t_{2i-1})) + (N_2(t_{2i})-N_2(t_{2i-1})) \\ \] By independence of \(N_1(t)\) and \(N_2(t)\), \((N_1(t_{2i})-N_1(t_{2i-1}))\) and \((N_2(t_{2i})-N_2(t_{2i-1}))\) are independent. Since each increment in \(N_1(t)\) and \(N_2(t)\) are independent, their sum \(N(t)\) also has independent increments.

Verify Poisson Distribution

To prove that \(N(t)\) follows a Poisson distribution, we can use the fact that the sum of independent Poisson random variables is Poisson. Let \(X_1 \sim\) Poisson\((\lambda_1t)\) and \(X_2 \sim\) Poisson\((\lambda_2t)\) be the random variables representing the counts in the processes \(N_1(t)\) and \(N_2(t)\). Then, \(N(t) = X_1 + X_2 \sim\) Poisson\((\lambda t)\), where \(\lambda = \lambda_1+\lambda_2\). Now, we've verified that \(N(t)\) has both independent and stationary increments, and the counts follow a Poisson distribution with rate \(\lambda_{1}+\lambda_{2}\). Therefore, we have proven that \(N(t) = N_{1}(t) + N_{2}(t)\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\).