# Moment of Ito diffusion computationally

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.

• I don't really understand the question. Do you mean how do you evaluate it numerically? Draw a bunch of trajectories up to time $t$, record the value at time $t$, average the results.
– Ian
Jul 23, 2021 at 20:39
• Also not sure what you're looking for. Anyway, you do have transistion kernels also in the time-continuous setting. Jul 23, 2021 at 20:41
• @Tobsn Can you elaborate? I'm looking for a formal definition that gives a hint on how to compute it Jul 23, 2021 at 20:54
• $\mathbb{E}[X_{t}]=\int y p_{t}(x,dy)$ assuming $X_{0}=x$. Apart from rare special cases you won't be able to find explicit expressions for the Markov kernel $p$. Jul 23, 2021 at 21:55
• @Tobsn I am not talking about explicit methods. How is $p_t$ defined then? And why do we condition on $x$ there? What if $X_t$ "drifted" far from the initial state? Jul 23, 2021 at 22:05

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.

• Let's say we fixed a small time interval $[0, \tau]$. How close are $\mathbb E[\hat X_t]$ and $\mathbb E[X_t]$, where $\hat X_t$ is the Euler-Maruyama approximate solution with the discretization step $\tau$? Jul 23, 2021 at 23:36
• I think it will depend on how $f$ and $\sigma$ behave on $[0, \tau]$ (e.g. Lipchitz condition etc.) There are some discussion here: math.stackexchange.com/questions/4058518/… Jul 23, 2021 at 23:52

$$X$$ 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 $$\mathbb{E}[X_t] = \int_{\Omega} X_t(\omega)d\mathbb{P}(\omega).$$

• Ok, that's clear, but how is the respective pdf defined? Jul 23, 2021 at 22:52
• The pdf is defined the same way as for any other random variable, as the derivative of the CDF: $p_t(z) = \frac{d}{dz} \mathbb{P}(X_t \le z)$. In general, this is not something that can be computed explicitly. As @WHLin mentions in their answer, you can derive a PDE for it, but in general that will still not have an explicit solution. Jul 23, 2021 at 23:03
• It may fail to have a solution in general, as far as I understand. Jul 23, 2021 at 23:25