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Let $X_t$ be such that $X_t$ is bounded continuous martingale adapted to the filtration $\mathcal{F}_t$ such that

$$\Bbb{E} \bigg[\int_0^t e^{X_s} \, d\langle X\rangle_s\bigg] = 0$$

and $X_0=0$. Does it follow that $X_t$ is identically zero?

remark: This question came from the following $$\forall \lambda >0 ; e^{\lambda X_t} \text{ is a martingale} \Rightarrow X_t \equiv 0$$

I see two ways to solve this:

0) $\Bbb{E}[e^{\lambda (X_t - X_s)}\vert \mathcal {F}_s] = 1$ therefore

$$\Bbb{E}[e^{\lambda (X_t - X_s)} - 1\vert \mathcal {F}_s] = 0\\ \Bbb{E}\bigg[\frac{e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0\\ \lim_{\lambda\to 0}\Bbb{E}\bigg[\frac{e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0\\ \Bbb{E}\bigg[\frac{\lim_{\lambda\to 0}e^{\lambda (X_t - X_s)}-1}{\lambda}\vert \mathcal {F}_s\bigg] = 0 \\ \Bbb{E}[X_t - X_s\vert \mathcal {F}_s] = 0$$

therefore $X_t$ is a martingale.

A) Since the exponential transform (extend it to complex you get characteristic functions, or just consider the laplace transform) allows one to recover the distribution, we get that $$\Bbb{E}[e^{\lambda X_t}] = 1 = \Bbb{E}[e^{\lambda 0}] $$

B) by Itô

$$e^{X_t} = 1 + \int_0^t e^{X_s} \, dX_s + \int_0^t e^{X_s}\, d\langle X\rangle_s $$ take expectations

$$\Bbb{E}[e^{X_t}] = 1 + \int_0^t \Bbb{E}[e^{X_s}]\, d\langle X\rangle_s $$

So basically I am asking how to proceed along the lines of $B)$

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  • $\begingroup$ What about $X_t := 1$? (Or have you forgotten mentioning that $X_0=0$?) $\endgroup$ – saz Jul 5 '15 at 19:51
  • $\begingroup$ Yes $X_0 = 0$, thank you $\endgroup$ – Conrado Costa Jul 5 '15 at 19:59
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Recall that for any function $f \geq 0$ and any measure $\mu$ we have

$$\int f \, d\mu = 0 \implies f=0 \quad \mu\text{-a.s.} $$

Therefore, as $(\langle X \rangle_t)_{t \geq 0}$ is an increasing process,

$$\mathbb{E} \left( \int_0^t e^{X_s} \, d\langle X \rangle_s \right)=0$$

implies

$$\int_0^t e^{X_s} \, d \langle X \rangle_s =0$$

$\mathbb{P}$-almost surely. Since $e^x$ is strictly positive and $(\langle X \rangle_t)_{t \geq 0}$ an increasing continuous process, this already implies

$$\langle X \rangle_t = \text{const} = \langle X \rangle_0 = 0$$

almost surely. Indeed: Suppose that it is not constant. Then we can pick $t_0 >0$ such that $\langle X \rangle_{t_0}(\omega) > \langle X \rangle_{0}(\omega)$. By the very definition of the Riemann-Stieltjes integral and the strict positivity of $e^x$, this implies

$$\int_{0}^{t_0} e^{X_s(\omega)} \, d\langle X \rangle_s(\omega)>0$$

in contradiction to our assumption.

Consequently,

$$\mathbb{E}(X_t^2) = \mathbb{E}(\langle X \rangle_t) = 0$$

which implies $X_t = 0$ almost surely.

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