# Continuity and convergence in probability, two sequences case

This problem can be seen as a generalization of the continuous mapping theorem, for two sequences, with arbitrary rate of convergence.

Let $\{X_n\}$ and $\{Y_n\}$ be two possibly non-convergent sequences, and $g$ be a function.

Suppose that

A1) For any $\epsilon > 0$, there exists a set $B$ where $g$ is uniformly continuous satisfying $\limsup\limits_{n\rightarrow \infty} P(X_n, Y_n \notin B ) < \epsilon$.

Then,

\begin{align} \text{1) } X_n - Y_n\xrightarrow{p} 0 \quad\Rightarrow\quad g(X_n) - g(Y_n)\xrightarrow{p} 0 \end{align}

A2) For any $\epsilon > 0$, there exists a set $B$ where $g$ is Lipschitz continuous satisfying $\limsup\limits_{n\rightarrow \infty} P(X_n, Y_n \notin B ) < \epsilon$. then

\begin{alignat}{3} \text{2) } & a_n[X_n - Y_n]\xrightarrow{p} 0 &&\quad\Rightarrow\quad a_n[g(X_n) - g(Y_n)]\xrightarrow{p} 0 \\ \text{3) } & a_n[X_n - Y_n] = Op(1) &&\quad\Rightarrow\quad a_n[g(X_n) - g(Y_n)] = Op(1) \end{alignat}

Note: A1 is automatically satisfied if for all $n$, $Y_n = Y$, where $Y$ is a random variable satisfying $P(Y \in D_g) = 0$, with $D_g$ the set of discontinuity points of $g$.

From (A1), we have a $N_b$ and $B$ such that for $n > N_b$,

$P(X_n, Y_n \notin B) < \epsilon/2$

For (1), by uniform continuity, we have a $\delta > 0$ such that if either $x$ or $y$ are in $B$, $|g(x) - g(y)| < \delta$ when $|x - y | < \eta$.

We know that there exists a $N_{e}$ such that $P(|X_n - Y_n| \ge \eta) < \epsilon/2$ for $n > N_e$.

Then, for $n > max(N_b,N_e)$.

$P(|g(X_n)-g(Y_n)| > \delta) \leq P(X_n, Y_n \notin B) + P(|X_n-Y_n| \ge \eta) = \epsilon$

which proves the result.

For (2) and (3), we have, when $x, y \in B$,

$a_n |g(x) - g(y)| \le a_n K |x - y|$

where $K$ is the Lipschitz constant of $g$.

For (2), we have for any $\delta > 0$ a $N_d$ such that $n > N_k$,

$P(a_n K |X_n - Y_n| > \delta) < \epsilon/2$

For (3), we take a $\delta$ and $N_k$ for which the above holds.

In either case, we have, for $n > max(N_b, N_d)$,

$P(a_n|g(X_n) - g(Y_n)| > \delta) \le P(Y_n, X_n \in B^c) + P(a_n K |X_n - Y_n| > \delta) \le \epsilon$

which proves the result.