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Let $W$ be standard (one-dimensional) Brownian motion and define the process $Y = \left\{ Y_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ by $$ Y_t = (W_t + t)\exp\left\{-W_t - \frac{1}{2}t \right\}. $$ I am trying to show that $Y$ is an $\left\{ \mathcal{F}_t \right\}$-martingale using Ito's Formula.

Here's my approach:

For each $t \ge 0$, put $G_t := W_t + t$, $M_t := -W_t - \frac{1}{2}t$, and $H_t := \exp\{M_t\}$. Then $Y_t = G_t H_t$, and $M$ is an Ito process with $(u_t, v_t) = (-1/2, -1)$. Now, let $f(t,w) = f(w) = e^w$. Then $f(M_t) = f'(M_t) = f''(M_t) = H_t$; and so, by Ito's Formula, $$ d H_t = \left[0 - \frac{1}{2}f'(M_t) d M_t + \frac{1}{2}f''(M_t)\right]dt - f'(M_t) dW_t = 0 - f'(M_t) dW_t = - H_t dW_t. $$ This means that $$ H_t = 1 - \int_0^t H_s \, dW_s, \qquad t \ge 0;$$ and so, the process $H := \left\{H_t, \mathcal{F}_t; 0 \le t < \infty \right\}$ is a martingale.

Similarly, $G$ is an Ito process with $(u_t, v_t) = (1,1)$; i.e., $$ d G_t = d t + d W_t. $$ Now, using the integration-by-parts formula, \begin{align*} d Y_t &= G_t \, dH_t + H_t \, dG_t + dG_t \, dH_t \\ &= -G_t H_t d W_t + H_t\left(d t + d W_t \right) - \left(d t + d W_t \right)\left(H_t d W_t \right) \\ &= 0 + (H_t - G_t H_t)d W_t \\ &= (1 - t - W_t)H_t d W_t. \end{align*} Thus, \begin{align*} Y_t &= \int_{0}^t (1 - s - W_s)H_s d W_s. \end{align*} So, if we put $Q_t := (1 - t - W_t)H_t$, for each $t \ge 0$, then it clear that the process $Q := \left\{Q_t, \mathcal{F}_t ; 0 \le t < \infty \right\}$ is measurable and $\left\{\mathcal{F}_t \right\}$-adapted. Now, we just need to show that \begin{align*} \mathbb{E}\left[\left(\int_{0}^T Q_s^2 \, ds\right)^{1/2} \right] &= \mathbb{E}\left[\left(\int_{0}^T (1 - s - W_s)^2H_s^2 d s \right)^{1/2}\right] \\ &= \mathbb{E}\left[\left(\int_0^T (s^2H_s^2 + 2sW_sH_s^2 - 2sH_s^2 + W_s^2H_s^2 - 2W_sH_s^2 + H_s^2) ds \right)^{1/2}\right] \end{align*} is finite for all $T > 0$. Naturally, this is where I am stuck. Given the ugliness of this integrand, I can only I imagine I've done something wrong here (or that I am overlooking some "slicker" approach)...

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  • $\begingroup$ Isn't it a consequence of the Girsanov theorem? $\endgroup$
    – Chaos
    Dec 1, 2020 at 13:13

2 Answers 2

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Actually you need to show something stronger, that $$\mathbb{E}[\int_0^T Q_s^2ds]< +\infty.$$ This is much easier (and actually implies the thing that you wanted to show, because of Holder). You can do this exchanging the integrals and using the fact that you know the law of $W_s$ for any given $s$.

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  • $\begingroup$ Ah, yes. Interchanging the integrals was, indeed, the key thing I'd overlooked here. Thanks. $\endgroup$ Dec 12, 2020 at 17:58
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It is a consequence of Girsanov's theorem as per what was said earlier. If you define a probability measure $\mathbb{Q}$ such that : $$\mathbb{dQ}_{| F_t} = e^{-W_t - \frac{1}{2}t}\mathbb{dP}_{| F_t}$$

Then it follows from Girsanov's theorem that the process $(B_t)_{t>0}$ defined as : $$B_t = W_t +t$$

is a Brownian motion (hence a martingale) under $\mathbb{Q}$. You can check that this also implies that $e^{-W_t - \frac{1}{2}t}(W_t+t)$ is a martingale under $\mathbb{P}$ using Bayes' conditional expectation formula as follows : $$\mathbb{E}[e^{-W_t - \frac{1}{2}t}(W_t+t)|F_s] = e^{-W_s - \frac{1}{2}s}\mathbb{E_{Q}}[W_t+t|F_s] = e^{-W_s - \frac{1}{2}s}(W_s+s)$$

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