A pair of correlated stochastic processes follow the SDEs \begin{align} dX_t&=a(t,X_t)\,b(t,Y_t)\,dt+c(t,X_t)\,d(t,Y_t)\,dW_t, &&X_0=\bar{x}\\ dY_t&=f(t,Y_t)\,dt+g(t,Y_t)\,dZ_t, &&Y_0=\bar{y} \end{align} where $W_t$ and $Z_t$ are correlated Brownian motions with constant correlation $\rho$ and $a,b,c,d,f,g$ are smooth functions.

We know that the probability density $p(t,x,y)$ of the process reaching the state $X_t=x$, $Y_t=y$, given the initial condition at $t=0$ satisfies the forward Kolmogorov equation (also known as Fokker Planck equation): $$ p_t=-(abp)_x-(fp)_y+\frac{1}{2}(c^2d^2p)_{xx}+\frac{1}{2}(g^2p)_{yy}+\rho(cdgp)_{xy} $$

Let $h(t,x)$ be the marginal probability distribution of the process $X_t$, i.e. $$h(t,x)=\int_y p(t,x,y)dy$$ is it possible to write the partial differential equation in the variables $x$ and $t$ that $h(t,x)$ must satisfy?


1 Answer 1


It is not possible to derive such a PDE as it doesn't exist in general because the process $X$ is not Markovian anymore.

Note that in your case you could derive a PDE for the process $Y$ which is a one dimensional stochastic differential equation and so is Markovian if the solution exists.

Best regards.

  • $\begingroup$ Thanks a lot! It is nice to learn the formal explanation of why it cannot be done. $\endgroup$
    – Fabio
    May 10, 2015 at 2:00

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