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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ian
    Commented Jul 23, 2021 at 20:39
  • $\begingroup$ Also not sure what you're looking for. Anyway, you do have transistion kernels also in the time-continuous setting. $\endgroup$
    – Tobsn
    Commented Jul 23, 2021 at 20:41
  • $\begingroup$ @Tobsn Can you elaborate? I'm looking for a formal definition that gives a hint on how to compute it $\endgroup$
    – Rubi Shnol
    Commented Jul 23, 2021 at 20:54
  • $\begingroup$ $\mathbb{E}[X_{t}]=\int y p_{t}(x,dy)$ assuming $X_{0}=x$. Apart from rare special cases you won't be able to find explicit expressions for the Markov kernel $p$. $\endgroup$
    – Tobsn
    Commented Jul 23, 2021 at 21:55
  • $\begingroup$ @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? $\endgroup$
    – Rubi Shnol
    Commented Jul 23, 2021 at 22:05

2 Answers 2

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  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.

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  • $\begingroup$ 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 $\tau$? $\endgroup$
    – Rubi Shnol
    Commented Jul 23, 2021 at 23:36
  • $\begingroup$ 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/… $\endgroup$
    – WHLin
    Commented Jul 23, 2021 at 23:52
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$X$ 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 $$\mathbb{E}[X_t] = \int_{\Omega} X_t(\omega)d\mathbb{P}(\omega).$$

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  • $\begingroup$ Ok, that's clear, but how is the respective pdf defined? $\endgroup$
    – Rubi Shnol
    Commented Jul 23, 2021 at 22:52
  • $\begingroup$ 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. $\endgroup$ Commented Jul 23, 2021 at 23:03
  • $\begingroup$ It may fail to have a solution in general, as far as I understand. $\endgroup$
    – Rubi Shnol
    Commented Jul 23, 2021 at 23:25

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