Convergence of expectations of bivariate functions

Question

Suppose I have a continuous bivariate function $g:\mathbb{R}^2\rightarrow [0,\infty)$ that is increasing in both arguments and uniformly bounded by some constant.

I also have three sequences of random variables $X_n, Y_n, Z_n$ such that $X_n \rightarrow X, Y_n \rightarrow X, Z_n \rightarrow Z$ when $n\rightarrow \infty$, with $X_n\leq_{st}X_{n+1}$, $Y_n\leq_{st}Y_{n+1}$, and $Z_n\leq_{st}Z_{n+1}$, where $\leq_{st}$ means less than or equal to in the usual stochastic order.

My question is now whether the following holds:

$\lim_{n\rightarrow \infty}\left( \mathbb{E}[g(X_n,Z_n)] - \mathbb{E}[g(Y_n,Z_n)] \right) = 0$

I believe this must be true because of some argument involving the monotone convergence theorem ($g$ is monotone, continuous and bounded), but I have no idea how to start.. Does anybody know how to prove this?

Thank you!

Answer 1

Since $g$ is bounded, you can directly apply DCT.

Use continuity to conclude that $g(X_{n},Z_{n})-g(Y_{n},Z_{n})\xrightarrow{n\to\infty} 0$ almost surely and that $|g(X_{n},Z_{n})-g(Y_{n},Z_{n})|\leq 2C$ where $|g(x,y)|\leq C$.

Since you have not mentioned what type of convergence holds for $X_{n}$,$Y_{n}$ and $Z_{n}$, I have just assumed almost sure convergence. But, just convergence in Probability suffices (as DCT only requires convergence in Probability).

If in an even weaker tone, you just have joint convergence in distribution of $(X_{n},Z_{n})\to(X,Z)$ and $(Y_{n},Z_{n})\to (Y,Z)$, then too the result directly follows from the very definition of joint convergence(or if you want to prove this without appealing to the functional analytic definition of weak convergence, then Skorokhod's Representation Theorem will let you conclude the result).