Given the standart stochastic differential equation setting (i.e. a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\in > [0,\infty]},P)$ with complete filtration and ($\mathcal{F}_t$)-Brownian motion $B$), we consider $$X_t=X_0+\int_0^tb(s,X_s)ds+\int_0^t \sigma(s,X_s)dB_s$$ It is well known, that in the Lipschitz case (i.e. $\sigma(t,x),\ b(t,x)$ are continuous on $\mathbb{R}_+ \times \mathbb{R}^d$ and Lipschitz continuous in $x$ uniformly for $t\ge 0$) the solution exists uniquely for $\mathcal{F}_0$-measurable $X_0$.

My Question: Many sources use solutions to this SDE for initial conditions $X_0$, distributed according to some distribution (e.g. uniform or normal). How to overcome the following problem:

Can we even define a random variable $X_0$ distributed to a given distribution, which is $\mathcal{F}_0$-measurable?

  1. If one just augments the filtration by $\tilde{\mathcal{F}}_t=\sigma({\mathcal{F}_t,X_0})$, we would lose, that $B$ is a $(\mathcal{F}_t)$-Brownian motion.
  2. If $X_0$ is not $\mathcal{F}_0$-measurable, then the solution cannot be adapted.
  • 2
    $\begingroup$ If $X_0$ is independent from $(B_t)_{t \geq 0}$, then $\tilde{\mathcal{F}}_t := \sigma(\mathcal{F}_t,X_0)$ is an admissible filtration for $(B_t)_{t \geq 0}$... $\endgroup$ – saz Dec 15 '17 at 13:41
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    $\begingroup$ If (and only if) $\mathcal F_0$ is atomless, then it contains a random variable with arbitrary distribution. $\endgroup$ – zhoraster Dec 15 '17 at 14:51
  • $\begingroup$ Thank you very much. @saz that means my solution will then be $\tilde{\mathcal{F}}_t$-adapted. Would the notion of a "strong solution" in this case mean, that $X_t$ is also $\sigma(X_0,(B_s)_{s\le t})$ adapted? And could one just apply the usual product measure trick to construct $B$ independent of $X_0$? $\endgroup$ – nehemoro Dec 15 '17 at 15:39
  • $\begingroup$ Yes, that's correct. $\endgroup$ – saz Dec 15 '17 at 15:45
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    $\begingroup$ It may be written in some good textbooks. Shortly, if $\mathcal F_0$ has an atom, then it cannot support a continuous distribution. Otherwise, it is possible construct a random variable uniformly distributed on $[0,1]$ (using an argument similar to the proof of Urysohn's lemma), whence, with the help of quantile transformation, it is possible to get arbitrary distribution. $\endgroup$ – zhoraster Dec 16 '17 at 9:59

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