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?


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 '15 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.