Moment of Ito diffusion computationally

Question

Say, we have an SDE

$$ \mathrm d X_t = f(X_t) \mathrm d t + \sigma(X_t) \mathrm d W_t $$

where $W_t$ is a Wiener process.

Assuming a strong solution exists globally (so the 1st and 2nd moments should be bounded), what is exactly

$$\mathbb E [X_t]$$

from the computation standpoint?

In discrete-time processes, if we have transition pdfs, it's quite clear, but in time-continuous case it seems difficult.

I tried to look up a pdf of $X_t$ knowing that of the driving noise, but couldn't find anything.

Answer 1

$X$ can be thought of as just a collection of random variables $(X_t)_{t \in [0,\infty)},$ and their expected value can be defined the same way as any other random variable's.

More formally, let $(\Omega, \mathcal F, \mathbb P)$ be a probability space with a Brownian motion $(W_t)_{t \in [0,\infty)} = (W_t(\omega))_{t \in [0,\infty)}$. Since we assume $f$ and $\sigma$ satisfy sufficient regularity conditions to guarantee a strong solution exists, and strong existence of a solution implies pathwise uniqueness, we have a unique process $(X_t)_{t \in [0,\infty)} = (X_t(\omega))_{t \in [0,\infty)}$ satisfying $$dX_t = f(X_t)dt + \sigma(X_t)dW_t.$$ This process $X$ is the solution to the SDE, and the expected value of $X_t$ is defined by $$\mathbb{E}[X_t] = \int_{\Omega} X_t(\omega)d\mathbb{P}(\omega).$$

Answer 2

1. One way is to numerically integrate the SDE similar to ODE. The expectation could be obtained by averaging many trajectories. See: https://en.wikipedia.org/wiki/Euler%E2%80%93Maruyama_method

2. Another way is to calculate the corresponded Fokker-Planck equation and solve this PDE numerically. Then, the expectation at given time can be obtained by taking the moment of the PDE solution at that time.