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What's the solution to the following SDE?

$$dX_t = X_t W_t dt + dW_t;\ X_t|_{t=0} = X_0,\ t \in \mathbb{R}_+$$

Joe Shmo
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    If you have $t\in [0,T]$ for any arbitrary $T$ (i.e. you work on any compact) multiply both sides by the integrating factor $e^{-\int_0^t W_s ds}$, and notice that $d(X_t e^{-\int_0^t W_s ds})=dXe^{-\int_0^t W_s ds}-X_tW_tdt$ since $e^{-\int_0^t W_s ds}$ is of bounded variation (it's actually absolutely continuous) – Chaos Apr 01 '21 at 06:27
  • Thank you for your answer, but isn't $(_ ^{−∫_0^t _}) = ^{−∫_0^t _ } − _ _ ^{−∫_0^t _ } $? – Joe Shmo Apr 01 '21 at 15:20
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    yes of course, my bad!, indeed that's what you need! define $Y_t=X_te^{-\int_0^t W_sds}$ and try to solve for $Y$ – Chaos Apr 01 '21 at 15:24
  • You end up with $dY_t = e^{- \int_0^t W_s ds} dW_t$, whose quadratic variation is $[Y, Y]^{(t)} = \int_0^t e^{-2 \int_0^s W_r dr} ds$, but I'm not sure how to evaluate that integral, or the innermost integral $\int_0^s W_r dr$ for that matter.

    I think $\int_0^s W_r dr \sim N(0, \frac{s}{3})$, but I don't know if that helps, and if it does - then what?

     – Joe Shmo Apr 01 '21 at 17:09
  • Some solutions doesn't have a closed form – Chaos Apr 01 '21 at 20:14

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