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Let $(B_t)$ be a standard Brownian motion on some probability space and let $X_t$ be the process defined by the SDE $dX_t = \mu_t dt + dB_t$, where $\mu_t$ is adapted, deterministic, and only takes the values $0$ or $1$. (For example, $\mu_t = 0$ on $(2n, 2n+1]$ for all $n \geq 0$ and $1$ everywhere else.)

Does it follow that $X_t$ has continuous sample paths? (Under the same probability measure that makes $(B_t)$ a standard BM?)

I think it does -- use the Girsanov theorem to get a new probability measure so that with $X_t$ is now a standard BM. Because the new probability measure is mutually absolutely continuous with the old one, it should follow that $X_t$ is also continuous almost everywhere.

Am I right with this line of thought, or am I missing something?

Victor
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1 Answers1

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Applying Girsanov's theorem is overkill. Note that, by definition,

$$X_t = X_0 + \int_0^t \mu_s \, ds+ B_t, \qquad t \geq 0.$$

We know that $t \mapsto B_t$ is continuous (almost surely); moreover, it is well-known that mappings of the form

$$t \mapsto I(t) := \int_0^t \mu_s \, ds$$

are continuous whenever the integral is well-defined. (Just consider e.g. $\mu(s) = 1_{[1,2]}(s)$; draw a picture to see that $t \mapsto I(t)$ is continuous, it doesn't have any jumps.)

This means that $t \mapsto X_t$ is continuous almost surely since it is the sum of the continuous functions.

saz
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