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Let $W$ be a Brownian motion with respect to its natural filtration. How can I show with Itô's formula that $e^{t/2}\cos(W_t)$ is a martingale?

I'm using the following formulation of Itô's formula: Let $X$ be a continuous semimartingale and $f \in C^2(\mathbb{R}^d; \mathbb{R})$. Then $f(X)$ is again a continuous semimartingale, and we $\mathbb{P}$-a.s. simultaneously for all $t \geq 0$

$$f(X_t) = f(X_0) + \sum_{i=1}^d\int_0^t\frac{\partial f}{\partial x^i}(X_s) \ dX^i_s + \frac{1}{2}\sum_{i, j=1}^d \int_0^t\frac{\partial^2 f}{\partial x^i \partial x^j}(X_s) \ d\langle X^i,X^j\rangle_s.$$

And by defining $f(t, x) := e^{t/2}\cos(x)$, I get $$e^{t/2}\cos(W_t) = 1 - \int_0^te^{s/2}\sin(W_s) \ dW_s,$$

and this is where I'm stuck.

I first thought, that I could conclude that $\int_0^te^{s/2}\sin(W_s) \ dW_s$ is a martingale since it's a stochastic integral, but the existence theorem for the stochastic integral that I'm working with, only applies to integrators $M$ that are locally square-integrable and null at zero and integrands $H$ which are predictable and satisfy

$$\mathbb{E}\left[\int_0^{\infty}H^2_s \ d[M]_s \right]<\infty.$$

In my case, I don't believe that the latter condition holds.

I also have a proposition that extends this existence theorem to the set of predictable processes for which there exists a sequence of stopping times $(\tau_n)_{n \in \mathbb{N}}$, s.t. $\tau_n \nearrow \infty$ holds $\mathbb{P}$-a.s. and for each $n$ we have

$$\mathbb{E}\left[\int_0^{\tau_n}H^2_s \ d[M]_s \right]<\infty.$$

Althought this condition is satisfied, since $\mathbb{E}\left[\int_0^{t}e^s \sin^2(W_s) \ d[W]_s \right] = \mathbb{E}\left[\int_0^{t}e^s \sin^2(W_s) \ ds \right]<\infty$ for all $t\geq 0$, this proposition only states that the resulting stochastic integral $H \bullet M$ is a locally square integrable martingale null at zero. How can I now conclude that $\int_0^te^{s/2}\sin(W_s) \ dW_s$ is a true martingale?

Thank you in advance for any clarifications!

Dude3400
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  • @BrianMoehring, are you referring to the fact that "locally square integrable martingale" sometimes means that $M$ itself is a martingale? Sadly, in my notes, we call a process $M$ locally square integrable martingale null at zero if there exists a localising sequence $(\tau_n)_{n \in \mathbb{N}}$ s.t. $M^{\tau_n}$ is a square integrable martingale null at zero. – Dude3400 Aug 06 '25 at 20:58
  • Yes, I was just mentioning the possible ambiguity. So you have a process which is the stochastic integral which coincides with a martingale on $[0,T]$ for each $T>0$? (Since $\mathbb{E}\left[\int_0^{T}H^2_s \ d[M]_s \right]<\infty$). Shouldn't that be enough to conclude? – Brian Moehring Aug 07 '25 at 04:06
  • @BrianMoehring how exactly does the fact that $\mathbb{E}\left[ \int_0^T H^2_s \ d[M]_s\right]$ is finite for all $T >0$ imply that the stochastic integral $H \bullet M$ stopped for each $T>0$ is a true martingale? – Dude3400 Aug 07 '25 at 06:36
  • Because on $t \in [0,T]$, we have $$\int_0^t e^{s/2}\sin(W_s)dW_s = \int_0^t e^{s/2} \sin(W_s)\chi_{[0,T]}(s),dW_s$$ The latter gives a martingale by your first theorem, so the former gives a martingale on $[0,T]$. – Brian Moehring Aug 07 '25 at 07:17
  • @BrianMoehring so your argument is that since $H$ is in $L^2(M^T)$ for all $T>0$, we have by the existence theorem that $H \bullet M^T = (H \bullet M)^T$ is a true martingale for all $T > 0$. Therefore, we conclude that $H \bullet M$ is a true martingale. – Dude3400 Aug 07 '25 at 07:51
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    In short, yes. More generally: Let $(X_t)_{t\in [0,\infty)}$ be a stochastic process with natural filtration $\mathcal{F}$. If $X^T$ is a martingale with respect to $\mathcal{F}$ for every constant $T > 0$, then $X$ is a martingale. – Brian Moehring Aug 07 '25 at 16:15

1 Answers1

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See theorem 4.13 in Le Gall's book. It states that if for all $t\geq 0$, $M_{t}$ is a "true" martingale s.t. $M_{t} \in L^{2}$ if and only if $\Bbb{E}(\langle M,M\rangle_{t})<\infty$ for each $t\geq 0$.

In your case, it follows from Ito's Isometry (and Ito's formula) that for $M_{t}=\int_{0}^{t}e^{s/2}\sin(W_{s})\,dW_{s}$, then $$\langle M,M\rangle_{t}=\int_{0}^{t}e^{s}\sin^{2}(W_{s})\,ds\leq \int_{0}^{t}e^{s}\,ds=e^{t}-1$$

Thus $\Bbb{E}(\langle M,M\rangle_{t})<\infty$ for each $t\geq 0$ and hence is a true Martingale.

Mr. Gandalf Sauron
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