We are confronted with an exercise that asks us to find a function $h(t)$ such that: $$M_t = [B_t + h(t)] \cdot e^{-B_t - \frac{t}{2}}$$ is a martingale wrt to its natural filtration. After applying Ito's lemma using the function $$f(t, x) = [x + h(t)] \cdot e^{-x - \frac{t}{2}}$$ we noted that if $h(t) = t$ we obtain a nice result: $$[B_t + t] \cdot e^{-B_t - \frac{t}{2}} = \int_0^t(1 - B_s - s) \cdot e^{-B_s - \frac{s}{2}} dB_s$$ So our aim is to show that the stochastic integral is a Martingale. To do that, we observed ($\epsilon_t = e^{-B_t - \frac{t}{2}}$):
- $E\left[ \int_0^t \epsilon_s^2 ds \right] < \infty$
- $E\left[ \int_0^t s^2 \cdot \epsilon_s^2 ds \right] < \infty$
- We are then stuck in showing $E\left[ \int_0^t B_s^2 \cdot \epsilon_s^2 ds \right] < \infty$
If we show 3. we get that the stochastic integral is sum of martingales, which will prove $M_t$ to be a martingale under $h(t) = t$. Are we on the right track? Any ideas to show 3.??
To show (3) Exchange expectation and integration using Tonelli's Theorem. All that is left is to denote $s=\sigma^2$ and bound
$$ E[B_s^2 \cdot \epsilon_s^2] \le \int_{-\infty}^{\infty} (\sigma x)^2 e^{-2\sigma x-\sigma^2} e^{-x^2/2} \, dx \, \, , $$ then integrate the result over $s$.