Is the solution of $dX_t=f(X_t)dt+dW_t$ continuous?

Question

Let $f:\mathbb R\to \mathbb R$ measurable s.t. $|f(x)|\leq K(1+|x|)$ for some $K>0$. Then, $$dX_t=f(X_t)dt+dW_t,$$ has a weak solution, i.e. there is $(X_t,W_t,\mathcal F_t)$ s.t. $$X_t=X_0+\int_0^t f(X_s)ds+W_t.$$

Does $(X_t)$ will necessarily be a continuous process ? Or it may happen some situation where there are no continuous solution $(X_t)$ ?

score 1 · Accepted Answer · answered Dec 10 '19 at 13:46

1

For any Ito-integrable process $\xi$, the integral $\int_0^t \xi_s dW_s$ has continuous version. So yes, $X$ is continuous (has continuous version).

answered Dec 10 '19 at 13:46

zhoraster

26,086

What if $f$ is not continuous? – UBM Dec 10 '19 at 14:35
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$X_t$ is continuous whatever the regularity of $f$ because $\int_0^t{f(X_s) ds}$ is the integral of an (almost surely) locally bounded term. – Quentin Dec 10 '19 at 14:48

Is the solution of $dX_t=f(X_t)dt+dW_t$ continuous?

1 Answers1