Let $X$ be a continuous 1-d diffusion:

$$ dX_t = a(X_t)dt + b(X_t)dW_t, X_0 = x. $$ W is a standard Brownian Motion and $a(\cdot)$ and $b(\cdot)$ can have nice regularity properties.

Let $Z^n_1,Z^n_2,\ldots$ be i.i.d. standard normal random variables. The Euler discretization $X^n$ with step size $\frac{1}{n}$ is the process given by

$$ X^n_\frac{k}{n} = X^n_\frac{k-1}{n} + \frac{1}{n}\cdot a(X^n_\frac{k-1}{n}) + \frac{1}{\sqrt{n}}b(X^n_\frac{k-1}{n}) \cdot Z^n_k, X^n_0 = x. $$

A (at least to me) reasonable way to embed $X$ and $X^1,X^2,\ldots$ on the same probability space is by taking the $Z_i$ generated by the increments of the Brownian Motion $W$: $Z^n_i = \sqrt{n} \cdot (W_\frac{i}{n} - W_\frac{i-1}{n})$.

The reason why I want all process defined on the same space is that I want convergence rates, and I have heard that weak convergence results aren't appropriate for this, although I don't really understand why.

It's believable that in some sense $X^n \rightarrow X$. There are several different types of convergence available (expectation over product space, expectation of time supremum). I'm interested in what kind of convergence there is, and specifically in the convergence rates.

Motivation: the problem I'm studying is an optimal stopping problem on a diffusion when there is perfect information. When information comes at discrete times, it becomes a weird discretization, which can be approximated by the one above. I'd like to see how fast there is convergence to the perfect information case.