Solution: Compensator of the Poisson process

Here we need to use that \(M^2(t) = N^2(t)−2\lambda t N(t)+(\lambda t)^2\) and the properties of the Poisson process \(N(t)\). Specifically we need to use that \begin{align*}E[N(t)| \mathcal{F}_s] &= E[N(t)-N(s) + N(s)| \mathcal{F}_s] \\ &= E[N(t)-N(s)| \mathcal{F}_s] + E[N(s)| \mathcal{F}_s] \\ &= E[N(t-s)] + N(s) \\ &= \lambda (t-s) + N(s) \\ &= (N(s) - \lambda s) + \lambda t,\end{align*} and that \begin{align*}E[N^2(t)| \mathcal{F}_s] &= Var[N(t)| \mathcal{F}_s] + E[N(t)| \mathcal{F}_s]^2,\end{align*} where \begin{align*} Var[N(t)| \mathcal{F}_s] &= Var[N(t)-N(s) + N(s)| \mathcal{F}_s] \\ &= Var[N(t)-N(s)| \mathcal{F}_s] + Var[N(s)| \mathcal{F}_s] \\ &= Var[N(t-s)] + 0 \\ &= \lambda (t-s).\end{align*} Now we can calculate \begin{align*} E[M^2 (t)−\lambda t |\mathcal{F}_s] &= E[N^2(t)−2\lambda t N(t)+(\lambda t)^2 -\lambda t|\mathcal{F}_s] \\ &= E[N^2(t)|\mathcal{F}_s] −2\lambda t E[N(t)|\mathcal{F}_s] +(\lambda t)^2 -\lambda t \\ &= Var[N(t)| \mathcal{F}_s] + E[N(t)| \mathcal{F}_s]^2−2\lambda t \left((N(s) - \lambda s) + \lambda t\right) +(\lambda t)^2 -\lambda t\\ &= \lambda (t-s) + ((N(s) - \lambda s) + \lambda t)^2 −2\lambda t (N(s) - \lambda s) - 2(\lambda t)^2 + (\lambda t)^2 -\lambda t\\ &= \lambda (t-s) + (N(s) - \lambda s)^2 + 2\lambda t (N(s) - \lambda s) + (\lambda t)^2 −2\lambda t (N(s) - \lambda s) - 2(\lambda t)^2 + (\lambda t)^2 -\lambda t \\ &= (N(s) - \lambda s)^2-\lambda s \\ &= M^2(s) - \lambda s.\end{align*} Thus we have shown that \(M^2 (t)−\lambda t\) is a martingale.