Convergence in law of stopped stochastic processes 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) $t\ge 0$?
 A: 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 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, 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.
