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

$\mathbb E[|X^n_{\tau_n\wedge t}-X^n_{\tau\wedge t}|]^2 = \mathbb E[(\int_{\tau_n \wedge t}^{\tau \wedge t} \sigma_n (s) dW_s]^2)] = \mathbb E[\int_{\tau_n \wedge t}^{\tau \wedge t} \sigma_n (s)^2 ds] <C \mathbb E[|\tau_n \wedge t - \tau \wedge t|],$

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.