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I am absolute newbie to stochastic calculus and have to solve a weighted hazard rates integral, where the hazard rates are stochastic, their logarithm governed by arithmetic Ornstein-Uhlenbeck (OU) process (if it simplifies, asymptotically for large times so that all transients are gone).

I.e. I am looking for the distribution at time $t$ of the stochastic process

$$A(t) = \int_0^t q(s) e^{x_s} ds $$

where $q(s)$ is the weight function of time (let's for simplicity think it polynomial or even constant),

and $x_s = x_0 e^{-as} + (1-e^{-as})\mu + \sigma e^{as} \int_0^s e^{-as} dW_s$ being the OU solution with mean $\mu$ and mean-reversion strength parameter $a$.

After transients, $$ x_s \approx \mu + \sigma e^{as} \int_0^s e^{-as} dW_s = \mu + \beta W_s $$

with $\beta^2 = \frac{\sigma^2}{2a}(1 - e^{- 2 a s}) \approx \frac{\sigma^2}{2a}$


So the simplified version of the initial problem: $$A(t) = \int_0^t q(s) e^{\mu + \beta W_s} ds $$ with constant $\mu$ and $\beta$ and a Wiener process $W_s$ (to be more precise, a random variable with density N(0,1))


Now my problem:

tried to solve $A(t)$ integral by parts and got stuck $\Rightarrow$ need any possible hint top get any further:

With $Q(t) = \int q(s) ds$

$$ A(t) = Q(t)e^{\mu + \beta W_t} - Q(0)e^{\mu} - \beta \int_0^t Q(s) e^{\mu + \beta W_s} dW_s $$

If I am correct so far, how to solve the right integral?

Must be a one-liner well described in the literature but I could not find any information. Maybe solvable by Taylor expansion of $e^W_s$?

Would be very grateful for any hint!!!

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@ user40531 : I think this is the best you can do. But it might depend on what you mean exactly by saying "solving an integral" and what you want to do with it. Best regards –  The Bridge Sep 27 '13 at 12:34

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