I will describe an example that seemingly contradicts the following

Theorem For a local martingale $M$, let $[M,M]_t$ be its quadratic variation at $t$. For any $t$, if $E[[M,M]_t]<\infty$, then $E[[M,M]_t]=E[M_t^2]$.

Description: We work on a finite time horizon $[0,T]$ for $T=1$. Let $W_t$ be a Brownian Motion, $J_t$ a single jump process taking the values 0 and 1 with constant hazard rate $h(t)=1$, i.e. $P[J_t=1|\mathcal{F}_t]=\int_0^t h(s) ds=\int_0^t 1 ds$. As $P[J_t=1|\mathcal{F}_t]$ is deterministic, $W$ is also a Brownian motion on the enlarged filtration $\mathcal{G}_t=\mathcal{F}_t\lor\mathcal{H}_t$, where $(\mathcal{H}_t)_t$ is generated by $(J_t)_t$. Define the process $M_t=W_t-\int_0^t W_s dJ_s$. We have that $E[M_t |\mathcal{G}_s]=M_s$, so $M$ is a Martingale with $E[M_t]=0$.

Calculation: As $J$ is quadratic pure jump, we have that $[J,W]_t=0$, and (also using $[W,W]_t=t$ and $[J,J]_t=J_t$ - compare to the Poisson process) we get $$[M,M]_t=t+\int_0^t W_t^2 dJ_t.$$ Taking expectation, we get $$E[[M,M]_t]=t+\int_0^t sh(s) ds=t+t^2/2.$$

I think already here the mistake must have happened, as the (correcting) jump should reduce the variance of the whole process compared to pure Brownian Motion. To back up that intuition, lets calculate $E[M_t^2]$.

$$E[M_t^2]=E[W_t^2-2W_t\int_0^tW_s dJ_s+(\int_0^t W_t dJ_t)^2].$$ As before, $E[W_t^2]=t$ and $$E[(\int_0^t W_s dJ_s)^2]=E[\int_0^t W_s^2 d[J]_s=E[\int_0^t sh(s) ds]=t^2/2.$$ However, $$-2E[W_t\int_0^t W_s dJ_s]=-2\int_0^t E[W_s W_t]h(s)ds=-2\int_0^t s ds=-t^2.$$ Collecting results we get $$E[M_t^2]=t-t^2/2$$.

I checked a couple of time and can't find the mistake in my reasoning and/or calculations. Where does it hide?


1 Answer 1


$M_t$ is not a martingale. In fact, $\mathbb{E}[M_1|\mathcal{G}_s]=0$ on the event $J_s=0$, since at some later time the jump will bring the process back to zero, after which it is just a Brownian motion.


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