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Let $(X_t, Y_t)$ be a two-dimensional Markov stochastic process that runs on time interval $[t_0, t_f]$. Its infintesimal generator is described by the functions $\mu_X, \mu_Y, \sigma_X, \sigma_Y$. I have the following information about the process:

(1) $\mu_X(x, y, t)$ and $\sigma_X(x, y, t)$ are known. For both these functions, the parameter $y$ has no effect (so we can write them $\mu_X(x, t)$ and $\sigma_X(x, t)$).

(2) $\mu_Y(x, y, t) = x$

(3) $\sigma_Y(x, y, t) = 0$

(4) $X_{t_0} = x_0$, some known constant. $Y_{t_0} = 0$.

My end goal is: looking forward from $t_0$, find a probability density function that reflects my beliefs about the value of $Y_{t_f}$.

I'm only vaguely familiar with stochastic processes, so I would appreciate a description of how to solve this problem or a reference to a gentle introduction to solving these sorts of problems.

Thank you.

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up vote 2 down vote accepted

In other words, $(X_t)_t$ can be any (autonomous) diffusion solving the stochastic differential equation $\mathrm dX_t=\sigma_X(X_t,t)\mathrm dW_t+\mu_X(X_t,t)\mathrm dt$, and $$ Y_t=\int_0^tX_s\mathrm ds. $$ At this level of generality, I am not sure that much more can be said.

Edit: Recall that the transition probabilities $(p_t)_t$ of $(X_t)_t$ solve the forward Kolmogorov (or Fokker-Planck) equation $$ \partial_tp_t+\partial_x(\mu_Xp_t)-\tfrac12\partial^2_{xx}(\sigma^2_Xp_t)=0. $$ Likewise, the transition probabilities $(P_t)_t$ of $(X_t,Y_t)_t$ such that, for every suitable function $u$, $$ \mathbb E(u(X_t,Y_t))=\iint u(x,y)P_t(x,y)\mathrm dx\mathrm dy, $$ solve the forward Kolmogorov (or Fokker-Planck) equation $$ \partial_tP_t+\partial_x(\mu_XP_t)+x\partial_yP_t-\tfrac12\partial^2_{xx}(\sigma^2_XP_t)=0. $$

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Yeah, time-integration is precisely the motivation behind this post. I got a suggestion over at MO to attempt time-integration by phrasing as a two-dimensional Markov process, but I'm unclear on the procedure for converting between infinitesimal generators and transition functions in multiple dimensions. –  GMB Jan 6 '13 at 7:42
Multiple dimensions do not change a iota to the theory. See Edit. (Is this your question?) –  Did Jan 6 '13 at 11:25
Yes, it's precisely my question. Thanks. –  GMB Jan 7 '13 at 7:49
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