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I am wondering if the following expression can be processed a bit analytically, $$ E \left[ e^{aX} \int_0^X e^{bu}dW(u)\right], $$ where $W_u$ is the normal Brownian motion (1D Wiener process), and $X$ is a random variable.

I know that $E \left[ \int_0^T e^{bu}dW(u)\right]$ is zero, so $E \left[ \int_0^X e^{bu}dW(u)\right]$ should also be zero. However $e^{aX}$ and $\int_0^X e^{bu}dW(u)$ are clearly correlated, so the expectation of their product is not trivial.

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    $\begingroup$ What do the angle brackets inside the expectation denote? Usually angle brackets would mean "quadratic variation" but that doesn't seem to make sense here. Are you just using them as parentheses? Anyway, without knowing more about the random variable $X$ (such as how it relates to the Brownian motion) I don't see how one is going to be able to say anything. $\endgroup$
    – Nate Eldredge
    Commented Feb 20, 2015 at 15:19
  • $\begingroup$ @Nate Yes, the angle brackets are used as parenthesis for the expectation. I'll change them to square brackets. I see now how $X$ needs to be specified a more. I am looking into hitting times for SDEs, in particular the Ornstein-Uhlenbeck $dY(x) = (ax+b)dx + \sigma dW(x)$ because it has an explicit solution. The random variable X is defined as the values x for which Y(x)=y. First exit times can be studied via the Backward Kolmogorov with specified region boundaries, but I was wondering if some information about hitting times (not "first" anymore) can be obtained directly from the SDE solution. $\endgroup$
    – lkdo
    Commented Feb 20, 2015 at 15:39
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    $\begingroup$ By the way, it's not true that $\mathbb{E} \int_0^X e^{bs} dB_s = 0$ for general X. Let, for example, $X = \inf \{ t > 0 : B_t = c \}$. Then with $b = 0$, we have $\mathbb{E} \int_0^X dB_s = \mathbb{E} B_X = c \neq 0$. Even if $X$ is bounded, this may easily be false, let $X = \min \{ t \in [0, T] \colon B_t = \max_{s \in [0, T]} B_s \}$. Then $\mathbb{E} \int_0^X dB_s = \mathbb{E} B_X = \mathbb{E} |B_T| \neq 0$. $\endgroup$
    – tsnao
    Commented Apr 1 at 3:14

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Based on the comment of being interested in the hitting time for OU, it turns out that finding its density is still open as mentioned here: On the First Hitting Time Density of an Ornstein-Uhlenbeck Process.

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