Gebelein's Inequality and convergence of distribution

We know that for a bivariate standard normal vector $Z=(Z_1,Z_2)$ it holds that \begin{align*} \operatorname{Cov}(1\{Z_1\leq u),1\{Z_2\leq u))\leq \operatorname{Cov}(Z_1,Z_2). \end{align*} This result is known as Gebelein's Inequality. Now we consider two sequences $X_i$ and $Y_i$ $(i\in\mathbb{N})$ of real valued random variables with finite variance and it holds that $X_i$ converges to $Z_1$ in distribution and $Y_i$ to $Z_2$ as $i\rightarrow\infty$. Can we follow that there exists an $l\in\mathbb{N}$ that for all $i>l$ \begin{align*} \operatorname{Cov}(1\{X_i\leq u),1\{Y_i\leq u))\leq\operatorname{Cov}(X_i,Y_i) \end{align*}or \begin{align*} \operatorname{Cov}(1\{X_i\leq u),1\{Y_i\leq u))\leq \operatorname{Cov}(Z_1,Z_2). \end{align*}

If you have the stronger hypothesis that the sequence of pairs $(X_n, Y_n)$ converges in law to $(Z_1, Z_2)$ then you'll have that by continuous mapping property** we have that $(1_{X_n \leq u}, 1_{Y_n \leq u})$ converges in law to $(1_{Z_1 \leq u}, 1_{Z_2 \leq u})$. So any expectation-like operator of it will converge also, and thus for any $\varepsilon$ there exist $l$ such that for $n > l$ $$\operatorname{Cov}(1_{X_n \leq u}, 1_{Y_n \leq u}) \leq \operatorname{Cov}(1_{Z_1 \leq u}, 1_{Z_2 \leq u}) + \varepsilon$$
So for $\varepsilon$ smaller than $\operatorname{Cov}(Z_1, Z_2) - \operatorname{Cov}(1_{Z_1 \leq u}, 1_{Z_2 \leq u}) > 0$ you have your bound.
** : $1_{\cdot \leq u}$ is indeed a discontinuous mapping. But there exists a slightly general version of the continuous mapping theorem that allow measurable functions with discontinuities, such that this set has zero measure under the limit law. See Billingsley's "Convergence of Probability Measures" Th.2.7
• Thank you very much! Would it be enough to assume that $\operatorname{Cov}(X_n,Y_n)\xrightarrow{n\rightarrow\infty}\operatorname{Cov}(Z_1,Z_2)$? Would than $(X_n,Y_n)$ converge in distribution to $(Z1,Z2)$? – stroem Jul 10 '13 at 11:52
• An other question concerning your answer. Why will "any expectationlike operator converge also"? I understand that the expectation of each component will converge because it is uniformly integrable. Is it possible to use the continuous mapping theorem again to see that $1_{X_n \leq u}\cdot 1_{Y_n \leq u}$ converges in law to $1_{Z_1 \leq u}\cdot 1_{Z_2 \leq u}$? – stroem Jul 11 '13 at 8:53
• I guess is a bit sloppy :). With expectation-like I meant that weak convergence of random elements $X_n$ to $X$ (all living in a metric space $S$) is equivalent to convergence of $E(f(X_n))$ to $E(f(X))$ for any $f$ continuous and bounded is $S$. For the case of uniformly bounded random vectors we can consider a compact subset of $\mathbb{R}^d$ as space $S$ and we will have that any continuous function defined there will be bounded. The latter is your case since the vectors $(1_{X_n \leq u}, 1_{Y_n \leq u})$ are uniformly bounded. – Bunder Jul 11 '13 at 13:32