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I am trying to learn some stochastic calculus, and I have gotten stuck trying to answer what I thought would be a simple question. I think I am misunderstanding something fundamental about stochastic calculus, and I hope that answering this question will help me figure out what that is!

Let $Y:\Omega \times [0,\infty)$ be a diffusion process with $$dY_t = \mu(Y_t)dt + \sigma(Y_t)dW_t$$ where $W_t$ is a Wiener process. Suppose $Y_t$ represents a stochastic "flow" value, such as the amount of water flowing in or out of a tub. Then, for any $\omega \in \Omega$, the "stock" resulting from this flow (e.g. the random amount of water in the tub) is $$X(t,\omega) \equiv \int_0^t Y(s,\omega) ds$$

Is $X$ also a diffusion process? I keep trying to apply Ito's lemma, but $X(t,\omega)$ is a function of $(Y(s,\omega))_{s\leq t}$ rather than of $Y(t,\omega)$. Am I doing something stupid?

Frank
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    In this situation $dX_t$ is just $Y_t dt$, that's it. – Ian Mar 09 '23 at 18:59
  • Ah, of course! This comment (along with your comment on Snoop's answer about combining X and Y into a higher dimensional diffusion) completely cleared up my confusion – Frank Mar 09 '23 at 19:46

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Ito diffusions $dY_t=b(Y_t)dt+\sigma(Y_t)dW_t$ have SDE coefficients which satisfy the Lipschitz condition $$|b(x)-b(y)|+|\sigma(x)-\sigma(y)|\leq D|x-y|$$ for all $x,y\in \mathbb{R}$ and are all Markov processes (ref: Oksendal, Th.7.1.2. p. 111). The simple example $dY_t=dW_t$ yields $X_t=\int_0^tY_sds=\int_0^tW_sds$ which is not Markov, so it can't be a Ito diffusion.

Snoop
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    True, it's not a diffusion because the drift is not a function only of the current state. You can still write down the DE, though, and you can assemble a higher dimensional diffusion by adjoining $Y$ itself to the state. – Ian Mar 09 '23 at 19:37
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    @Ian yes, you are correct :) – Snoop Mar 09 '23 at 19:39
  • Thank you! This answer, along with Ian's comment here and on my original question, completely cleared up my confusion. – Frank Mar 09 '23 at 19:46