# How to show the joint weak convergence?

Given a $$T>0$$, let $$\mathcal{C}[0,T]$$ be the space of continuous functions on $$[0,T]$$. Let $$Y_n(t)$$ be stochastic processes in $$\mathcal{C}[0,T]$$. We define the weak convergence in the sense of uniform metric and denote it as "$$\Rightarrow$$", see Billingsley (1968) Convergence of Probability Measures, Chapter 2.

Suppose we have the joint convergence $$(X_n,Y_n)\Rightarrow (X,c) \quad \text{as} \quad n\to\infty,$$ where $$X_n$$ and $$X$$ are random variables, $$Y_n\in\mathcal{C}[0,T]$$ and $$c$$ is a "constant" process.

Can we prove that $$(X_n,Z_n)\Rightarrow(X,Z) \quad \text{as} \quad n\to\infty,$$ where $$B$$ is a standard Brownian motion being independent of $$X_n$$, $$Z_n(t)=\int_0^t Y_n(s){\rm d}B(s)$$ are stochastic integrals ($$Y_n$$ is adapted to the filtration generated by $$B$$), and $$Z(t)=cB(t)$$.

Notice that, the weak convergence theory of stochastic integrals do show that $$Z_n\Rightarrow Z$$, but I don't find an answer for the above joint weak convergence.

Could you provide a detailed proof ? Or provide a counter-example ?

Thank you !

• well it depends on the type of weak-convergence you have for the field $Y_{n}(t)$. Is it for each fixed $t$? Is it uniformly eg. over Skorokhod topology or over some supremum norm? Even in a deterministic setting you can have pointwise convergence but not L1. Commented Sep 5, 2023 at 3:11
• For convergence theorems for Itô integrals, there are many results in a)Revuz-Yor, b)Stroock-Varadhan and c)the Limit Theorems for Stochastic Processes book by Jacod-Shiryaev Commented Sep 5, 2023 at 3:13
• in your post , can you explicitly write mathematically what you mean in the weak convergence uniform metric? Commented Sep 5, 2023 at 3:25
• @ThomasKojar In their results, they do prove the weak convergence for Ito integrals, i.e., $Z_n\Rightarrow Z$. But I can't find an answer for the joint weak convergence. Commented Sep 5, 2023 at 3:26
• @MikeHWANG : To have the stochastic integrals $Z_n(t)$ defined, you need to assume that the processes $Y_n$ are adapted to the filtration generated by the Brownian motion. Commented Sep 5, 2023 at 13:50

$$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\CC}{\mathcal C}\newcommand{\R}{\mathbb R}$$For any real $$\ep>0$$, let $$\begin{equation*} Y_{n,\ep}(t):=\min(c+\ep,\max(c-\ep,Y_n(t)); \end{equation*}$$ here and else where, $$t\in[0,T]$$. Then the process $$Y_{n,\ep}$$ is adapted to the filtration generated by $$B$$. So, we can introduce $$\begin{equation*} Z_{n,\ep}(t):=\int_0^t Y_{n,\ep}(s)\,dB(s) \end{equation*}$$ and $$\begin{equation*} D_{n,\ep}(t):=Z_{n,\ep}(t)-Z(t)=\int_0^t (Y_{n,\ep}(s)-c)\,dB(s). \end{equation*}$$ Then $$D_{n,\ep}$$ is a martingale and hence $$D_{n,\ep}^2$$ is a nonnegative submartingale. So, for any real $$\de>0$$, by the Doob martingale inequality and the Itô isometry, denoting by $$\|\cdot\|$$ the norm in $$\CC[0,T]$$, we have \begin{equation*} \begin{aligned} P(\|Z_n-Z\|>\de,\|Y_n-c\|\le\ep)&\le P(\|D_{n,\ep}\|>\de) \\ &=P(\max_{t\in[0,T]}D_{n,\ep}(t)^2>\de^2) \\ &\le\frac{ED_{n,\ep}(T)^2}{\de^2} \\ &=\frac1{\de^2}\,\int_0^T E((Y_{n,\ep}(s)-c)^2\,dt \\ &\le\frac{\ep^2 T}{\de^2}\le\de \end{aligned} \end{equation*} if $$\ep^2\le T\de^3$$.
Also, the condition $$(X_n,Y_n)\Rightarrow(X,c)$$ implies $$Y_n\Rightarrow c$$ and hence $$Y_n\to c$$ in probability, so that $$P(\|Y_n-c\|>\ep)\to0$$. So, $$\begin{equation*} \limsup_n P(\|Z_n-Z\|>\de)\le\limsup_n P(\|Z_n-Z\|>\de,\|Y_n-c\|\le\ep)+\lim_n P(\|Y_n-c\|>\ep) \le\de+0 \end{equation*}$$ for any real $$\de>0$$.
So, $$Z_n\to Z$$ in probability and hence $$Z_n\Rightarrow Z$$. So, for any real $$\de>0$$ and some compact $$K_{Z,\de}$$ in $$\CC[0,T]$$ and all $$n$$ we have $$\begin{equation*} P(Z_n\notin K_{Z,\de})+P(Z\notin K_{Z,\de})\le\de. \end{equation*}$$ Similarly, the condition $$(X_n,Y_n)\Rightarrow(X,c)$$ implies $$X_n\Rightarrow X$$ and hence for some compact $$K_{X,\de}$$ in $$\R$$ and all $$n$$ we have $$\begin{equation*} P(X_n\notin K_{X,\de})\le\de. \end{equation*}$$
Now take any bounded (say by $$1$$) continuous function $$f\colon\R\times\CC[0,T]$$. Then $$f$$ is uniformly continuous on the compact set $$K_{X,\de}\times K_{Z,\de}$$. So, for each real $$\ep>0$$ there is some real $$\de>0$$ such that \begin{equation*} \begin{aligned} &E|f(X_n,Z_n)-f(X_n,Z)|\,1(\{(X_n,Z_n),(X_n,Z)\}\subseteq K_{X,\de}\times K_{Z,\de}) \\ &\le\ep+2P(\|Z_n-Z\|>\de)\to\ep. \end{aligned} \end{equation*} So, $$$$\limsup_n E|f(X_n,Z_n)-f(X_n,Z)|\le\ep+2P(Z_n\notin K_{Z,\de})+2P(Z\notin K_{Z,\de}) +2P(X_n\notin K_{X,\de})\le\ep+4\de.$$$$ Also, because $$X_n\Rightarrow X$$ and $$X_n$$ is independent of $$B$$ and hence of $$Z$$, we have $$Ef(X_n,Z)\to Ef(X,Z)$$.
Thus, $$Ef(X_n,Z_n)\to Ef(X,Z)$$. $$\quad\Box$$