Martingales and Integrals

Question

Could someone explain why the following is a Martingale please? \begin{align} M_s = \int_0^s(1+u^2)dW_u \end{align} (where $W_t$ is standard Brownian motion).

I'm used to determining martingales using the expectation operator. But I don't believe that is the approach with this.

Many thanks,

John

Answer 1

$$M_{t+h} - M_t = \int_t^{t+h} (1+u^2)dW_u = \lim \sum_j (1 + t_{j-1}^2) (W_{t_j} - W_{t_{j-1}})$$the limit being in $L^2$, as $\sup [t_j - t_{j-1}] \to 0$.

Now, let $X$ be a random $L^2$ variable, measurable with respect to the $F_t$ filtration.

As $\sum_j (1 + t_{j-1})(W_{t_j} - W_{t_{j-1}})$ depends on the increments of $W$ after $t$, then $$ E \left[X\sum_j (1 + t_{j-1}^2)(W_{t_j} - W_{t_{j-1}})\right] = 0 $$ and now take the $L^2$ limit.

Answer 2

If you discretize the integral, any $M_t-M_s$ is the sum of increments with expectation zero, and so has itself expectation zero. Moreover, the increments are Gaußian, so that the approximate sum is Gaußian too, which is preserved in the limit of discretization to the integral.