# Example of adapted process that is a martingale w.r.t to one filtration but not another

Can someone give me an intuitive example (best would be using a discrete coin toss generator but Brownian Motion would also do) of a stochastic process that is a martingale with respect to a filtration $F_1$, but not a martingale with respect to a larger filtration $F_2$ (s.t. $F_1 \subset F_2$). Thanks!

-

Let $(B_t)_{t \geq 0}$ be a Brownian motion, then $(B_t)_{t \geq 0}$ is a martingale with respect to the natural filtration $$\mathcal{F}_t := \sigma(B_s; s \leq t).$$ In contrast, for $\mathcal{G}_t := \mathcal{F}_{2t}$ the process $(B_t,\mathcal{G}_t)_{t \geq 0}$ is not a martingale since $$\mathbb{E}(B_t \mid \mathcal{G}_{\frac{t}{2}}) = \mathbb{E}(B_t \mid \mathcal{F}_t)=B_t.$$ Roughly speaking: At time $t$ the $\sigma$-algebra $\mathcal{G}_t$ contains information about the Brownian motion until time $2t$, i.e. it contains information about the future of the process. And that doesn't work ...

-

Here is a typical example.

Take a Brownian Motion $W_t$ within its natural completed filtration $\mathcal{F}_t$ for $t\in [0,1]$. Then let's take $\mathcal{G}_t$ the augmented the filtration of $\mathcal{F}_t$ in the following way :

$$\forall t\in [0,1]~;~ \mathcal{G}_t =\mathcal{F}_t\vee \sigma(W_1)$$

Then it is a classical and not too hard result (try to prove it) about initially augmented filtrations that the process $W_t$ in the filtration $\mathcal{G}_t$ is not anymore a martingale.

Check Protter's book "Stochastic Integration and Differential Equations" Chapter VI for a general presentation of such ideas on augmented filtrations.

Best regards

-