Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $\omega$.)

share|improve this answer

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 $\mathrm d (B^2)=2B\mathrm dB+\mathrm dt$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.