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Suppose that $N$ is a homogeneous Poisson process with rate $\lambda$. For $0 \le s \le t < \infty$, how can we find $\mathbb E[N_s\cdot N_t]$?

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48 questions asked and still no clue about the formatting of mathematics on the site? –  Did Apr 13 '12 at 7:51
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If $(N_t)_{t\geq 0}$ is a Poisson process with parameter $\lambda$, then $E[N_t-N_s]=\lambda(t-s)$ for $0\leq s\leq t$. Then if $0\leq s\leq t$ $$ E[N_t N_s]=E[E[N_t N_s\mid \mathcal{F_s}]]=E[N_s E[N_t\mid\mathcal{F_s}]]. $$ Now let's calculate the conditonal expectation using that $N_t-N_s$ is independent of $\mathcal{F}_s$: $$ E[N_t\mid\mathcal{F_s}] = E[N_t-N_s+N_s\mid\mathcal{F_s}]=E[N_t-N_s]+N_s=\lambda(t-s)+N_s. $$ Then $$ E[N_tN_s]=E[N_s(\lambda(t-s)+N_s)]=E[N_s^2+N_s\lambda(t-s)]=E[N_s^2]+\lambda(t-s)E[N_s]\\ =\lambda s+(\lambda s)^2+\lambda(t-s)\lambda s=\lambda^2 t s+\lambda s. $$

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