# stochastic process: multiply two poisson process

Consider the process $\{X(t),t\ge0\}$ defined by $$X(t)=N(t)-\lambda t$$ where $N$ is a Poisson process with rate $\lambda \gt 0$ I have multiple questions about this one:

1. Compute, for $t_1, t_2, n_1, n_2 \gt 0$, the second-order probability mass of $N, G(t_1,t_2;n_1,n_2)$.
2. Compute $\mathbb E[X(t_1)X(t_2)]$, for $0 \lt t_1 \le t_2$.

For the problem 2, I don't know if there is correlation between $N(t_1)\cdot N(t_2)$ to compute its expectation.

• Your question is phrased in language suitable for assigning homework. That tends to be frowned on here. – Michael Hardy Oct 6 '17 at 3:18

The Poisson process has a kind of Markov property: the distribution of $N(s_2) - N(s_1)$, and hence the distribution of $X(s_2) - X(s_1)$, depends only on the length of the interval and is in particular independent of $N(s_1)$ or $X(s_1)$. (This property comes from the definition of the Poisson process.)

To exploit that here, use the identity $X(t_1) X(t_2) = X(t_1) \left[( X(t_2) - X(t_1)) + X(t_1)\right].$

• Thank you! So actually $\mathbb E[X(t_1)X(t_2)]=\mathbb E[X(t_1)]\mathbb E[X(t_2)-X(t_1)]+\mathbb E[(X(t_1)^2]$ and since $\mathbb E[X(t_1)]=0$, it equals to $\mathbb E[(X(t_1)^2]=\lambda t_1$ Is this correct? – Qing Oct 6 '17 at 4:17
• Indeed, and you can go further; one of the terms you've written is 0. – Aaron Montgomery Oct 6 '17 at 4:18
• Thank you so much! Is the $Var[X(t_1)]=Var[N(t_1)]=\lambda t_1$? – Qing Oct 6 '17 at 4:27

No way I'm doing parts (1)-(5) of a problem, but you did ask one actual question about the problem so I'll give a hint for that. We can write $$N(t_1)N(t_2) = N(t_1)^2 + N(t_1)(N(t_2)-N(t_1)) = N(t_1)^2 + (N(t_1)-N(0))(N(t_2)-N(t_1))$$ and you know (from the definition of a Poisson process) a great deal about the joint distribution of the factors in the second term.

• Thank you so much! but actually I have get the answer to be $\lambda t_1$ It's a bit strange to me. – Qing Oct 6 '17 at 4:39
• can you also give a hint of how to prove it is a martingale? – Qing Oct 6 '17 at 4:56
• @Qing Why is that strange? (It's right). Use (again) the fact that by independent increments $N(t_2)-N(t_1)$ is independent of $N(t_1) = N(t_1)-N(0).$ (You'll be using this fact a lot). – spaceisdarkgreen Oct 6 '17 at 5:11
• Thank you so much, below is my approach to prove the martingale, but I don't know if it is right: A process is said to be a martingale if only satisfies three conditions, the first two are easy to prove, I'm more focused on the third one: $\mathbb E[X_t|\mathcal F_s]=X_s$for all $s \lt t$ – Qing Oct 6 '17 at 5:18
• $\mathbb E[X_t|\mathcal F_s]=X_s$for all $s \lt t$ $$\mathbb E[X_t|\mathcal F_s]=\mathbb E[N(t)-\lambda t|\mathcal F_s]=\mathbb E[(N(t)-\lambda t)-(N(s)-\lambda s)+(N(s)-\lambda s)||\mathcal F_s]=\mathbb E[(N(t)-N(S)-\lambda (t-s)|\mathcal F_s]+\mathbb E[X_s|\mathcal F_s]=X(s)$$ – Qing Oct 6 '17 at 5:25