Let $X^n$ and $X$ be stochastic processes defined by


$$X^n\_t=1+\int\_0^tb\_n(s)ds+\int\_0^t\sigma\_n(s)dW\_s \quad\mbox{and}\quad X\_t=1+\int\_0^tb(s)ds+\int\_0^t\sigma(s)dW\_s,$$


where $b\_n, \sigma\_n, b, \sigma$ are uniformly bounded measurable functions s.t.


$$\lim\_{n\to\infty}\sup\_{0\le t\le T}|b\_n(t)-b(t)|=0=\lim\_{n\to\infty}\sup\_{0\le t\le T}|\sigma\_n(t)-\sigma(t)|,\quad \mbox{for all } T>0.$$


Consider the stopped processes $(X^n\_{\tau\_n\wedge t})\_{t\ge 0}$ and $(X\_{\tau\wedge t})\_{t\ge 0}$, where $\tau\_n:=\inf\{t\ge 0: X^n\_t\le 0\}$ and $\tau:=\inf\{t\ge 0: X\_t\le 0\}$. Can we prove $X^n\_{\tau\_n\wedge t}$ converges to $X\_{\tau\wedge t}$ in law for all (or almost every) $t\ge 0$?



Convergence in law of stopped stochastic processes
I believe convergence in law holds for all $t \geq 0$. The proof proceeds in three steps.


**Step 1:** Note that by the dominated convergence theorem for stochastic integrals (see, for example [Theorem 7](https://almostsuremath.com/2010/01/11/properties-of-the-stochastic-integral/) here), we have that $X\_n$ converges to $X$ in the ucp topology, that is, $\lim\_{n \to \infty} P(\sup\_{t \in [0, T]} |X\_t^n - X\_t| > \varepsilon) = 0$ for all $\varepsilon > 0$, $T \geq 0$.


**Step 2:** With similar reasoning as Step 3 [here](https://mathoverflow.net/questions/396234/on-the-continuity-of-map-gamma/396269#396269), we can then show that the above convergence implies that $\tau\_n$ converges to $\tau$ in probability. (Can provide details here if needed)


**Step 3:** We claim that the above two convergences combined are enough to imply, for every $t > 0$, convergence in law of $X\_{\tau\_n \wedge t}^n$ to $X\_{\tau \wedge t}$.


The remainder will be dedicated to the proof of Step 3.


To show this, we need to show that $E(f(X\_{\tau\_n \wedge t}^n)) \to E(f(X\_{\tau \wedge t}))$ for all bounded continuous $f$. We argue as follows:


Fix $t \geq 0$, and let $\varepsilon > 0$ be arbitrary. Choose $M > 0$ large enough so that $P(X\_{\tau \wedge t} > M) < \varepsilon ||f||\_{L^{\infty}}$.


By uniform continuity of $f$ on $[0, M+1]$, there exists some $0 < \delta < 1$ such that $|f(x) - f(y)| < \varepsilon$ whenever $x, y \in [0, M+1]$ are such that $|x - y| < \delta$.


By convergence in probability of $\tau\_n$ to $\tau$, we deduce that $\tau\_n \wedge t$ converges in $L^1$ to $\tau \wedge t$.


This implies $|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}^n|$ converges to $0$ in probability.


To see this, note that by the Markov inequality, we have that $\mathbb P(|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|>\epsilon)\le \mathbb E[|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|^2]/\epsilon^2$.


We estimate


$\mathbb E[|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|]^2 = \mathbb E[(\int\_{\tau\_n \wedge t}^{\tau \wedge t} \sigma\_n (s) dW\_s]^2)] = \mathbb E[\int\_{\tau\_n \wedge t}^{\tau \wedge t} \sigma\_n (s)^2 ds] <C \mathbb E[|\tau\_n \wedge t - \tau \wedge t|],$


for some constant $C$ independent of $n$.


Here the last line follows by the uniform boundedness of $\sigma\_n^2$.


Hence we have


$$P(|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|>\epsilon) < C\_0 \mathbb E[|\tau\_n \wedge t - \tau \wedge t|],$$


with the constant $C\_0$ independent of $n$, and so the LHS goes to $0$, as $n \to \infty$, as was to be shown.


Now, by ucp convergence of $X^n$ to $X$, we may find some $N\_0 > 0$ such that $P(\{|X\_{\tau \wedge t}^n - X\_{\tau \wedge t}| > \frac{\delta}{2}\}) < \frac{ \varepsilon}{||f||\_{L^{\infty}}}$.


To see the above, note that by step 1, we may take $N\_0$ to be such that $ P(\sup\_{s \in [0, t]} |X\_s^n - X\_s| > \frac{\delta}{2}) < \frac{ \varepsilon}{||f||\_{L^{\infty}}}$ for all $n > N\_0$. Since $\tau \wedge t \leq t$, and the aforementioned convergence is uniform on $[0, t]$, the statement follows.


By convergence of $|X\_{\tau\_n}^n - X\_{\tau}^n|$ to $0$ in probability, we may choose $N\_1$ such that $P(|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}^n| > \frac{\delta}{2}) < \frac{\varepsilon}{||f||\_{L^\infty}}.$


Thus whenever $n > \max(N\_0, N\_1)$, by the triangle inequality, we have with probability greater than $1 - \frac{3 \varepsilon}{||f||\_{L^{\infty}}}$ that $|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}| < \delta$, and so $|f(X\_{\tau\_n \wedge t}^n) - f(X\_{\tau \wedge t})| < \varepsilon$.


Finally we compute


$E(f(X\_{\tau\_n \wedge t}^n)) - E(f(X\_{\tau \wedge t}))$


$ \leq E(|f(X\_{\tau\_n \wedge t}^n) - f(X\_{\tau \wedge t})|$


$< (1 - 3 \varepsilon||f||\_{L^{\infty}})\varepsilon + 3 \frac{\varepsilon}{ ||f||\_{L^{\infty}}}||f||\_{L^\infty} $


$<\varepsilon + 3 \varepsilon$


$= 4\varepsilon.$


Since $\varepsilon> 0$ was arbitrary, we conclude.


