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One of my finance professors claims that the following is a meaningful SDE.

$$dX_t = \delta_t\mu X_tdt + \delta_t\sigma X_tdW_t$$

Here $W$ is BM and $\mu$ and $\sigma$ are positive real constants. $(\delta_t)$ is a stochastic process such that $\delta_t \sim U[0,1]$ for each $t$. Furthermore, it is independent of $W$ and $\delta_t$ and $\delta_s$ are independent whenever $t\neq s$. There are no path properties imposed on $(\delta_t)$.

My claim is that there is no solution to this SDE. But I don't know how to show this. All I know is that $(\delta_t)$ is not exactly well-defined. The professor thinks you can just discretize the SDE and pass to the limit from that. I know that passing to the limit is just a routine operation for these people and they think things always work out when you interchange limit with anything else. That is why I find his argument hard to believe. Can someone show who is right here?

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  • $\begingroup$ What do you mean $\delta_t$ is not exactly well enough? It's well defined - not worse than $W_t$ - in fact much simpler. $\endgroup$
    – A.S.
    Mar 14, 2016 at 12:12
  • $\begingroup$ @A.S. what i meant is that nothing is said about its paths. $\endgroup$
    – Calculon
    Mar 14, 2016 at 12:13
  • $\begingroup$ These are everywhere discontinuous path. You already listed all the definition needed. $\endgroup$
    – A.S.
    Mar 14, 2016 at 12:14
  • $\begingroup$ @A.S. Exactly. So then the integrals are not well-defined. It was a mistake on my part to say that $(\delta_t)$ itself is not well defined. $\endgroup$
    – Calculon
    Mar 14, 2016 at 12:16
  • $\begingroup$ Indeed I don't see a way to interpret $X_t=\int_0^t\delta_s X_s\,ds$ - there is too much freedom in $\delta_t$'s for it to be defined. $\endgroup$
    – A.S.
    Mar 14, 2016 at 12:36

1 Answer 1

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In defining $\int_0^t \delta_s X_s\,ds$ or $\int_0^t X_s\,dW_s$, one requires, at a minimum, that the process $\delta$ be progressively measurable. This condition is incompatible with the assumed independence of $\delta_s$ and $\delta_t$ for all $s\not=t$. See, for example, my answer in Prove that the stochastic process can not have continuous paths.

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