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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) $t\ge 0$?

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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.

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  • $\begingroup$ Remark: If I’m not mistaken, the proof shows we even have convergence in probability of $X_{\tau_n \wedge t}^n$ to $X_{\tau \wedge t}$. $\endgroup$
    – Nate River
    Jul 4 at 7:05
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    $\begingroup$ Thanks a lot for the answer. You contribute a lot to my questions indeed :) $\endgroup$
    – GJC20
    Jul 4 at 18:44
  • $\begingroup$ Why $\tau_n\wedge t=\tau\wedge t$ when $|\tau_n-\tau|<c$ and $|\tau-t|>c$? $\endgroup$
    – GJC20
    Jul 4 at 19:05
  • $\begingroup$ Also, in the following line, I do not see why the convergence in probability of $X^n_t$ to $X_t$ yields $\mathbb P(|X^n_{\tau_n\wedge t}-X_{\tau\wedge t}|>\min(\delta,1))<\epsilon/\|f\|_{\infty}$? $\endgroup$
    – GJC20
    Jul 4 at 19:10
  • $\begingroup$ Ah, there is an error there you’re right. This will be true when $\tau > t$, in which case $\tau \wedge t = \tau_n \wedge t = t$, but in the case $\tau < t$ we need to argue differently. I will try to correct it later today, also will add details for the convergence in probability part. $\endgroup$
    – Nate River
    Jul 5 at 0:29

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