Suppose that: $$dX_t =\mu(t,X_t)dt+\sigma(t,X_t)dW_t,$$ where $X_t$ is a vector valued stochastic process, $W_t$ is a vector of Brownian motions, $\mu$ is a vector valued function and $\sigma$ is a matrix valued function.

I want to compute the unconditional expectation $\mathbb{E}X_t$ defined by: $$\mathbb{E}X_t := \lim_{s\rightarrow -\infty}{\mathbb{E}[X_t|X_s=0]}.$$ (Assume that this quantity is guaranteed to be well-defined and finite for the particular $\mu$ and $\sigma$ of interest.)

It seems that it is at least true that: $$\frac{d\mathbb{E}X_t}{dt}=\mathbb{E}\mu(t,X_t)?$$ (See: How to find mean of a stochastic differential equation . So, I take it no extra term can enter coming from some kind of instantaneous correlation between $\sigma(t,X_t)$ and $dW_t$.)

Are there any tricks to simplify evaluating $\mathbb{E}\mu(t,X_t)$? Applying Ito's lemma to get the law of motion for $\mu(t,X_t)$ is unlikely to help as that LOM will have a new drift term.

When $\mu$ and $\sigma$ are non-linear, is there any hope of deriving a solution for $\mathbb{E}X_t$ in terms of an ODE?

Does the problem simplify much if $\mu$ and $\sigma$ are not directly functions of $t$?

  • 2
    $\begingroup$ I'm afraid there is no such technique in general, hence a case-by-case treatment. $\endgroup$
    – Abezhiko
    Jan 27 at 18:50


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