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I want to show that the Itô-process

$$ Z_t = \text{exp}( \int_0^t \theta_s dB_s - \int_0^t \frac{1}{2} \theta_s^2 ds) $$

is a solution to the SDE

$$ dX_t = \theta(t)X_t dB_t$$

using Itô's formula,

$$ dg(t,B_t) = \frac{\partial g}{\partial x}(t,B_t)dB_t + ( \frac{\partial g}{\partial t} (t,B_t) + \frac{1}{2} \frac{\partial^2 g}{\partial x^2} (t,B_t))dt $$

i.e, I want to find a function $g(t,x)$ s.t $ \frac{\partial g}{\partial x}(t,B_t) = \theta(t) g(t,B_t)$ and $ \frac{\partial g}{\partial t} (t,B_t) + \frac{1}{2} \frac{\partial^2 g}{\partial x^2} (t,B_t) = 0$.

But I'm stuck. How do I proceed?

Kurt G.
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Oskar
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    You need a slightly more general version of Ito's formula: $$dg(t,X_t)=g_t(t,X_t),dt+ g_x(t,X_t),dX_t+\frac{1}{2}g_{xx}(t,X_t),d\langle X\rangle_t,.$$ – Kurt G. Jun 06 '23 at 17:15
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    BTW. Your $\frac{1}{2} \frac{\partial^2 g}{\partial\color{red}t^2} (t,B_t),dt$ should be $\frac{1}{2} \frac{\partial^2 g}{\partial x^2} (t,B_t),dt,.$ – Kurt G. Jun 06 '23 at 18:23
  • Use Itô's lemma with $g(t,x) = \ln x$. – Abezhiko Jun 07 '23 at 06:51

1 Answers1

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Consider the following Itô process given by $$dY_t = - \frac{\theta_t^2}{2} dt + \theta_t dB_t $$ Now, to derive the dynamics of $Z_t$ you can apply Itô's formula to the function $f(x) = e^x$. Indeed, $$dZ_t = df(Y_t) = \Big\{- \frac{1}{2} \theta_t^2e^{Y_t} + \frac{1}{2}\theta_t^2 e^{Y_t} \Big\} dt + \theta_t e^{Y_t} dB_t = \theta_t e^{Y_t} dB_t = \theta_t Z_t dB_t$$ So, indeed $Z_t$ satisfies your SDE.

Oscar
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