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Assume a Wiener process W and a bounded F-adjusted stochastic process a. Show that the following process is a martingale on F

$$X(t)=(\int_{0}^{t}a(s)dW(s))^{2}-\int_{0}^{t}a^{2}(s)ds,\ t\geq0$$

Can someone help me on the above exercise? I tried to apply Ito's lemma but I got stuck

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$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$ \langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s} $$ by definition of the stochastic integral with respect to the Wiener process.

Using Itō's lemma, $$ d \left[\left[Y \left(t\right)\right]^2\right] = 2 Y \left(t\right) d Y\left(t\right) + d \langle Y \left(\cdot\right)\rangle_t $$ Subtracting the differential of the time integral, i.e. $a^2 \left(t\right) \, dt$, removes the drift term due to $d \langle Y \left(\cdot\right)\rangle_t$ and you are done.

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  • $\begingroup$ Thank you! Your answer was really helpful $\endgroup$ Apr 16 at 22:07
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Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively):

$$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right] $$

$$ = X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \int_s^t a_u^2du \big|{\cal F}_s \right] =X_s$$

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  • $\begingroup$ Great! Thank you! $\endgroup$ Apr 17 at 14:15

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