I know that $M_t=N_t-\lambda t$ is a martingale for $N_t$ a rate $\lambda$ poisson process and that for a brownian motion, $B_t^2-t$ is a martingale. I'm wondering, is there something similar for $M_t^2$? Like some function of t, say $a_t$, such that $M_t^2-a_t$ is a martingale. If so how would you go about finding one?

Any help is appreciated!

  • $\begingroup$ Why "pink" hammer? $\endgroup$
    – Did
    Feb 24, 2014 at 15:18

1 Answer 1


Note that $(M_t)_{t \geq 0}$ has independent stationary increments and its mean equals

$$\mathbb{E}M_t = \mathbb{E}N_t - \lambda t = \lambda t - \lambda t = 0$$

since $(N_t)_{t \geq 0}$ is a Poisson process. Hence,

$$\begin{align*} \mathbb{E}(M_t^2 \mid \mathcal{F}_s) &= \mathbb{E}((M_t-M_s)^2 \mid \mathcal{F}_s) + 2 M_s \mathbb{E}(M_t-M_s \mid \mathcal{F}_s) + M_s^2 \\ &= \mathbb{E}(M_{t-s}^2) + 0 + M_s^2 \tag{1}\end{align*}$$


$$\begin{align*} \mathbb{E}(M_{t-s}^2) &= \text{var}(N_{t-s}) = \lambda(t-s) \end{align*}$$

as $N_{t} \sim \text{Poi}(\lambda t)$. Therefore, $(1)$ shows that $(M_t^2-\lambda t)_{t \geq 0}$ is a martingale.


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