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Let $(B_t)_{t\geq 0}$ be a Brownian motion and $f(t)$ a square integrable deterministic function. Then: $$ \mathbb{E}\left[e^{\int_0^tf(s) \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^t f^2(s) \, ds}\right] $$ Now assume $(X_t)_{t\geq 0}$ is such that $\left(\int_0^tX_sdB_s\right)_{t\geq 0}$ is well defined. Does $$ \mathbb{E}\left[e^{\int_0^tX_s \, dB_s}\right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^tX_s^2 \, ds}\right] $$ still hold?

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No, in general this is not correct.

Applying Itô's formula, one can actually show that $$(t,w) \mapsto M(t,w) := \exp \left( \int_0^t X_s \, dB_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$ is a local martingale. There are several sufficient conditions (on the process $(X_t)_t$) implying that $(M_t)_{t \geq 0}$ is indeed a martingale. In this case, one obtains

$$1=\mathbb{E}M_0 = \mathbb{E}M_t = \mathbb{E} \exp \left( \int_0^t X_s \, dB_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$

Note that this is not equivalent to $$\mathbb{E} \exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right) = \mathbb{E}\exp \left( \int_0^t X_s \, dB_s \right)$$

(It works fine for deterministic functions, since in this case the left-hand side of the last equation does not depend on $\omega$.)

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If $X$ and $B$ are independent, yes (use the first result to compute the expectation conditional on $X$, then take the expectation).

Otherwise, no. For a counterexample, consider $X=B$ and use Itô's formula $\mathrm d (B^2)=2B\mathrm dB+\mathrm dt$.

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