# When convergence in distribution implies stable convergence

In a previous post I asked help to clarify a property of stable convergence in distribution:

Definition

Let $X_n$ be a sequence of random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with value in $\mathbb{R}^N$. We say that the sequence $X_n$ converges stably in distribution with limit $X$, written $X_n\stackrel{\text{st}}{\longrightarrow} X$, if and only if, for any bounded continuous function $f:\mathbb{R}^N\to\mathbb{R}$ and for any $\mathcal{F}$-measurable bounded random variable $W$, it happens that: $$\lim_{n\rightarrow \infty}\mathbb{E}[f(X_n)\,W]=\mathbb{E}[f(X)\,W].$$

What I need to prove now is the following:

Assume $$(Y_n,Z)\stackrel{\text{d}}{\longrightarrow}(Y,Z),$$

for all measurable random variable $Z$, then

$$(Y_n,Z)\stackrel{\text{st}}{\longrightarrow}(Y,Z)$$ for all measurable random variables $Z$. So I need to prove that, for any bounded continuous function $f$ and for any measurable $Z$ it holds that $$\lim_{n\rightarrow \infty}\mathbb{E}[f(Y_n,Z)\,W]=\mathbb{E}[f(Y,Z)\,W]$$ for all bounded random variables $W$.

I tried unsuccessfully with Portmanteau and Levy continuity theorem…

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In practice I am trying to prove this proposition from the paper by Podolskij and Vetter:

I did this reasoning for (1)=>(3), but I am not so sure of its correctness.

• You will certainly need some assumption on the integrability of $W$; if $W$ is not integrable, then the expectations are not even well-defined. It seems to me that the paper assumes that $W$ is bounded (in the sense that $\|W\|_{L^{\infty}} < \infty$), and this simplifies the proof a lot.
– saz
Apr 13, 2018 at 14:23
• How did you face the problem with the Portmanteau? The idea can be to take as particular $Z$ the $W$, but probably you will need some additional assumptions about $W$ as noticed by @saz.
– Jim
Apr 13, 2018 at 14:43
• Yes, sorry, the $W$ must be bounded. Apr 13, 2018 at 14:46
• In line with the saz comment, let $X$ be any random variable with finite mean but infinite variance, and define $X_n = X/n$. Then $X_n\rightarrow 0$ in distribution but $E[X_nX] = \infty$ for all $n$. Apr 13, 2018 at 14:46
• @AlmostSureUser Does $\mathbb R^N$ mean $N$-dimensional real space or $\mathbb R^{\mathbb N}$?
– Ѕааԁ
Apr 16, 2018 at 7:58

\def\dto{\xrightarrow{\mathrm{d}}}\def\stto{\xrightarrow{\mathrm{st}}}\def\mto{\xrightarrow{\mathrm{m}}}(3) \Rightarrow (2): Trivial. (2) \Rightarrow (1): For any g \in C_b(\mathbb{R}^N) and bounded \mathscr{F}-measurable W, suppose |W| \leqslant M. Take\begin{align*} f: \mathbb{R}^N × \mathbb{R} &\longrightarrow \mathbb{R},\\ (y, z) &\longmapsto g(y) · \frac{1}{2} (|z + M| - |z - M|). \end{align*} Because (Y_n, W) \dto (Y, W) and f \in C_b(\mathbb{R}^{N + 1}), then E(g(Y_n) W) = E(f(Y_n, W)) \to E(f(Y, W)) = E(g(Y) W). \quad n \to \infty $$Therefore, Y_n \stto Y. (1) \Rightarrow (3): Suppose Z and W are \mathscr{F}-measurable and W is bounded. First, for any A \in \mathscr{B}(\mathbb{R}^N) and B \in \mathscr{B}(\mathbb{R}), there exists \{g_k\} \subseteq C_b(\mathbb{R}^N) such that g_k \mto I_A, i.e.$$ m(\{ x \in \mathbb{R}^N \mid g_k(x) \neq I_A(x)\}) \to 0. \quad k \to \infty $$For any k \geqslant 1, because Y_n \stto Y and I_B(Z) W is \mathscr{F}-measurable and bounded, then$$ E(g_k(Y_n) I_B(Z) W) \to E(g_k(Y) I_B(Z) W). \quad n \to \infty $$Note that g_k \mto I_A and I_B(Z) W is bounded, thus$$ E(I_A(Y_n) I_B(Z) W) \to E(I_A(Y) I_B(Z) W). \quad n \to \infty \tag{1} Now, for any C \in \mathscr{B}(\mathbb{R}^{N + 1}), there exists \{A_{k, j}\} \subseteq \mathscr{B}(\mathbb{R}^N) and \{B_{k, j}\} \subseteq \mathscr{B}(\mathbb{R}) such that \{h_k\} defined by\begin{align*} h_k : \mathbb{R}^N × \mathbb{R} &\longrightarrow \mathbb{R},\\ (y, z) &\longmapsto \sum_{j = 1}^{s_k} I_{A_{k, j}}(y) I_{B_{j, k}}(z) \end{align*} satisfies h_k \mto I_C. For any k \geqslant 1, from (1) there is E(h_k(Y_n, Z) W) \to E(h_k(Y, Z) W). \quad n \to \infty $$Because h_k \mto I_C and W is bounded, then$$ E(I_C(Y_n, Z) W) \to E(I_C(Y, Z) W). \quad n \to \infty \tag{2} $$Now, for any f \in C_b(\mathbb{R}^{N + 1}), there exists a sequence of simple functions \{f_k\} such that f_k \rightrightarrows f. For any k \geqslant 1, from (2) there is$$ E(f_k(Y_n, Z) W) \to E(f_k(Y, Z) W). \quad n \to \infty $$Because f_k \rightrightarrows f and W is bounded, then$$ E(f(Y_n, Z) W) \to E(f(Y, Z) W). \quad n \to \infty $$Therefore, (Y_n, Z) \stto (Y, Z). • The proof of (1)\Rightarrow (3) I had in mid was much simpler: Assume$$Y_n\stackrel{st}{\rightarrow} Y.$$Then, by definition,$$ E[g(Y_n)\,W]\to E[g(Y)\,W] $$for any bounded continuous function g(y) and for any bounded random variable W. Now consider any bounded continuous function f(y,z), an arbitrary \mathcal{F}-measurable variable Z and note that$$E[f(Y_n,Z)\,W]=E[E[f(Y_n,c)\,W | Z=c]]\rightarrow E[E[f(Y,c)\,W | Z=c]]= E[f(Y,Z)\,W].Apr 16, 2018 at 14:29 • @AlmostSureUser This works only when Z is a continuous or discrete random variable. For Z in general, the conditinal expectation is hard to rigorously characterize. – Ѕааԁ Apr 16, 2018 at 14:41 • I do not understand the meaning of the double arrow. Apr 23, 2018 at 7:25 • @AlmostSureUser It means uniform convergence. – Ѕааԁ Apr 23, 2018 at 7:40 • @AlmostSureUser Suppose M_k=\|f_k-f\| and |W|\leqslant M. Because\begin{align*}|E(f(Y_n, Z) W)-E(f(Y, Z) W)|&\leqslant|E(f(Y_n, Z) W)-E(f_k(Y_n, Z) W)|\\&\quad+|E(f_k(Y_n, Z) W)-E(f_k(Y, Z) W)|\\&\quad+|E(f_k(Y, Z) W)-E(f(Y, Z) W)|\\&\leqslant M_kM+|E(f_k(Y_n, Z) W)-E(f_k(Y, Z) W)|+M_kM,\end{align*}and M_k→0 (k→∞), so\varlimsup_{n→∞}|E(f(Y_n, Z) W)-E(f(Y, Z) W)|\leqslant2M_kM,$$and make k→∞ to get E(f(Y_n, Z) W)→E(f(Y, Z) W). – Ѕааԁ Apr 23, 2018 at 8:16 What I suggested in the comment was the following idea: by Portmanteau$$ (Y_n,Z)\stackrel{\text{d}}{\longrightarrow}(Y,Z), $$IFF$$ \lim_{n\rightarrow \infty}\mathbb{E}[f(X_n, \, Z)]=\mathbb{E}[f(X,\,Z)] $$for any measurable bounded continuous function f:\mathbb{R}^{N+1}\to\mathbb{R}. Then take as particular Z := W. Then you should have that$$ (Y_n)\stackrel{\text{st}}{\longrightarrow}(Y). $$since you can seen f(Y_n,W)\,W for any f C.B. as a particular f_1(Y_n,W) C.B. under the assumption on W. Moreover$$ (Y_n,Z)\stackrel{\text{st}}{\longrightarrow}(Y,Z). $$should follow from$$ (Y_n)\stackrel{\text{st}}{\longrightarrow}(Y).$applying bounded convergence theorem. Is it right? • why bounded convergence? It requires point wise convergence: math.stackexchange.com/questions/235511/… Apr 15, 2018 at 20:58 • @AlmostSureUser You are right, I was too sloppy: bounded convergence is not immediate. Your argument seems right to me. Anyway, you can also prove it, as in proposition 2.5 (i) of Podolskij and Vetter paper, taking the sequence$V_n$that converges in probability to$V$constant and equal to$Z\$.