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Let $X(t)$ denote standard Brownian motion $dX(t) = a X dt + X dW(t)$ with solution $X(t) = e^{a t + W(t)}$. I want to consider the time-integrated process \begin{equation} Y(t) := \int_0^t d\tau~ X(\tau) \end{equation} which itself is again a stochastic process. My question is: How can I write a stochastic differential equation for $Y$ that is of the form \begin{equation} dY(t) = \ldots dt + \ldots dW(t) \end{equation} Whenever I use Ito's Lemma, I trivially obtain $dY(t) = X(t) dt$. Closely related to this problem: what would be the corresponding Fokker-Planck equation?

user56643
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    Indeed, $$dY=Xdt=e^{at+W}dt.$$ – Did May 27 '15 at 09:32
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    but then how would you obtain the Fokker-Planck equation? Because generally, for a stochastic process $dY = f(Y,t) dt + g(Y,t) dW(t)$ with $f$ the drift and $g$ the volatility, the FPeq is given by $\partial_t p + \partial_x (f p) = \frac{1}{2} \partial_x^2 g$ where $p$ denotes the probability density $p(y,t)$. Setting here $g = 0$ yields a strange first-order PDE that I would not know how to handle... How would you calculate $\partial_x f$? – user56643 May 27 '15 at 11:22
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    The approach you recall works for diffusion processes, here $(Y_t)$ is not a diffusion, it is not even Markov. Note that $dY_t=e^{at+W_t}dt$ is not of the form $dY_t=\sigma(Y_t,t)dB_t+b(t,Y_t)dt$ because $e^{at+W_t}$ is not a function of $(t,Y_t)$, due to the factor $e^{W_t}$. The same reamarks would apply to the simpler setting of the process $(Z_t)$ solving $dZ_t=W_tdt$. – Did May 27 '15 at 11:52
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    However, $(X_t,Y_t)$ is Markov, and there is a Fokker-Planck equation for that. – S.Surace Dec 17 '16 at 21:26

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