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Suppose $u(t,\omega), t \geq 0$ is a continuous time stochastic process with smooth paths so $\frac{d}{dt} u(t,\omega)$ exists for all $\omega$. Suppose you know the distribution of $\frac{d}{dt} u(t,\omega)$ and the initial distribution of $u$. Can you use this information to determine the distribution of $u(t,\omega)$ for all $t \geq 0$? That is, you know

  1. $P(\frac{d}{dt}u(t,\omega) \leq A)$ for all $A \in R, t\geq 0$
  2. $P(u(0,\omega) \leq A)$ for all $A \in R$.

Can you use 1 and 2 to determine $P(u(t,\omega) \leq B)$ for all $B \in R$?

Can you go the other direction? That is, if you know $P(u(t,\omega) \leq A$ for all $A \in R$, what is $P(\frac{d}{dt} u(t,\omega) \leq B)$?

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I would think the answer is "no" to both questions.

Let me abbreviate $v(t)=du/dt$, so $u(t)=u(0)+\int_0^t v(t')dt'$. Now ask for the expectation of $u^2(t)$. You'll need to know how $v(t')$ and $v(t'')$ are correlated for any $t',t''$ in the interval $(0,t)$. That information is not given.

Similarly, in the other direction, to find the distribution of the velocity $v(t)$ you need to know how the position $u(t)$ at nearby times is correlated, and that information is not given.

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  • $\begingroup$ Makes sense. A simple counterexample to the first question would be $v_1(t) = X$, and $v_2(t) = Y$ for $t < 1, v_2(t) = Z$ for $t >= 1$ with $X, Y, Z$ all being independent standard normals. $v_1$ and $v_2$ are equal in distribution but their integrals are not. Thank you very much for your help! $\endgroup$
    – bryan
    Commented May 2, 2013 at 2:42

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