2

Let $$dX_t=f(X_t)dt+g(X_t)dB_t,$$ and SDE with $f$ and $g$ nice enough to have existence and unicity of the solution. I'm not sure what mean unicity here. For example, consider the equation $dX_t=dB_t\quad \text{and}\quad X_0=0.\tag{E}$ Does it mean that $X_t$ is a standard brownian motion, or do we really have $\mathbb P(\forall s, X_s=B_s)=1$ ? (or maybe rather $\forall s, \mathbb P(X_s=B_s)=1$). Or does it mean that $(X_t)$ and $(B_t)$ has same finite dimensional distribution ? For example, $(tB_{1/t})_t$ is also a solution of $(E)$ ?

user657324
  • 1,863
  • There are several notions of uniqueness for solutions to SDEs: weak uniqueness (= uniqueness in distribution) and strong uniqueness (=pathwise uniqueness). – saz Apr 23 '19 at 13:25
  • @saz: What do you mean by pathwise uniqueness ? pointwise ? i.e. That $X_t(\omega )=B_t(\omega )$ a.s. for all $t$ ? So in this case, $tB_{1/t}$ won't be a solution as well, right ? – user657324 Apr 23 '19 at 13:46
  • 1
    An SDE has a pathwise unique solution if for any two solutions $(X_t)t$ and $(Y_t)_t$ (with the same initial condition) it holds that $\mathbb{P}(X_t = Y_t , , \text{for all $t$})=1$. The process $t B{1/t}$ is a weak solution but not a strong solution. You might want to take a look at this question – saz Apr 23 '19 at 14:01
  • @saz: thank you for your answer :) (it's crazy to see that you are almost the only one that answer to this sort of question :)) – user657324 Apr 23 '19 at 14:04

0 Answers0