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I am currently learning about SDEs and the different notions of solutions to them.

I get the general idea that weak solutions are a tuple $(X_t, B_t)$ of a process together with a Brownian motion, whereas a strong solution is defined only for a specific choice of Brownian motion. I also get that for weak solutions it does not make sense to talk about initial values but only about initial distributions (as illustrated in this answer).

However, none of the sources I looked at discusses strong solutions to SDEs for a given initial distribution. I wonder why that is? Is it because for some reason the concept of a strong solutions for a given initial distribution does not make sense? Or is it simply because if we know the strong solution for an arbitrary initial value, we immediately get a strong solution for the SDe with initial condition? It seems that one could first sample a starting value and then solve the SDE conditioned on this value.

I am aware that a similar question was already asked and answered on this site. However, it seems that the answer just repeats some general stuff about weak and strong solutions but does not answer the question. Probably thats why the answer is not accepted.

Peter
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A weak solution even consists of more than just $(X_t,B_t)$. If you have a weak solution, you are free to choose the whole probability space, the starting random variable, that is distributet like your given distribution and your brownian motion and the process itself. For a strong solution you are given the whole probability space, the brownian motion, a random variable $X_0$ and the equation and then need to find a solution for that particular case. So you already have the given random variable $X_0$ fixed. Now if you want to have strong existence, then you talk about a starting distribution. Because then you need to find for every probability space, starting random variable $X_0$ that has the given starting distribution, given brownian motion and your equation a strong solution. is it clear whan I meant?

JakobGFF
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  • Thanks. Sorry, but it is not clear ;). You say that even for a strong solution, we are given a random variable $X_0$. However, in what I have seen, $X_0=x$ is always deterministic; it is a fixed point. Do strong solutions still exist if $X_0\sim P$ is non-deterministic? – Peter Jun 22 '23 at 21:04
  • Yes they are. I learned it in the most general way and there you also consider starting values that are non deterministic. But this is really general. Many books are only using deterministic starting values. – JakobGFF Jun 23 '23 at 11:27
  • Thanks. Would you recommend a book which considers the general case? – Peter Jun 24 '23 at 09:46