# 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) $$t\ge 0$$?

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]

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.

• Remark: If I’m not mistaken, the proof shows we even have convergence in probability of $X_{\tau_n \wedge t}^n$ to $X_{\tau \wedge t}$. Jul 4 at 7:05
• Thanks a lot for the answer. You contribute a lot to my questions indeed :) Jul 4 at 18:44
• Why $\tau_n\wedge t=\tau\wedge t$ when $|\tau_n-\tau|<c$ and $|\tau-t|>c$? Jul 4 at 19:05
• Also, in the following line, I do not see why the convergence in probability of $X^n_t$ to $X_t$ yields $\mathbb P(|X^n_{\tau_n\wedge t}-X_{\tau\wedge t}|>\min(\delta,1))<\epsilon/\|f\|_{\infty}$? Jul 4 at 19:10
• 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. Jul 5 at 0:29