The continuous mapping theorem says that for $\{X_n\}, X$ random variables, with $g$ a function with set of discontinuity points $D_g$ such that $Pr[X \in D_g] = 0$, we have the three relations
- $X_n \xrightarrow{d}\ X \quad\Rightarrow\quad g(X_n) \xrightarrow{d} g(X)$
- $X_n \xrightarrow{p}\ X \quad\Rightarrow\quad g(X_n) \xrightarrow{p} g(X)$
- $X_n \xrightarrow{a.s}\ X \quad\Rightarrow\quad g(X_n)\xrightarrow{a.s} g(X)$
However, does this generalize to two random sequences $\{X_n\}, \{Y_n\}$? A quick reading of the proof would suggest that, so long that $g$ is continuous and ${X_n}$, ${Y_n}$ are bounded in probability, we have
2'. $X_n - Y_n \xrightarrow{p}\ 0 \quad\Rightarrow\quad g(X_n) - g(Y_n) \xrightarrow{p} 0 $
3'. $X_n - Y_n\xrightarrow{a.s}\ 0 \quad\Rightarrow\quad g(X_n) - g(Y_n)\xrightarrow{a.s} 0$
However, I have not been able to find such fundamental result anywhere. Is it available in a textbook or an article?
This extension of the continuous mapping theorem would be extremely useful to prove that two estimators are asymptotically equivalent, for example.
2' is true. Use uniform continuity of $g$ on bounded sets. Details: let $\epsilon >0$. There exists $T>0$ such that $P\{|X_n|>T\} <\epsilon$ and $P\{|Y_n|>T\} <\epsilon$ for all $n$. There exists $\delta >0$ such that $|g(x)-g(y)| <\epsilon$ if $|x-y| <\delta$ and $|x|\leq T,|y| \leq T$. Now $$P\{|g(X_n)-g(Y_n)| > \epsilon\} \leq P\{|X_n| >T\}+P\{|X_n| >T\}+P\{|X_n-Y_n|>\delta \}$$. The last term $\to 0$ as $n \to \infty$ and the first two terms are each $<\epsilon$. 3' is false. Let $X_n=n$ with probability $\frac 1 n$ and $0$ with probability $1-\frac 1 n$. Let $Y_n=X_n+\frac 1 n$. Then $\{X_n\}$ and $\{Y_n\}$ are bounded in probability and $X_n-Y_N \to 0$ almost surely. Let $g(x)=x^{2}$. Then $g(X_n)-g(Y_n)=n^{2}-(n+\frac 1 n )^{2}=-2+\frac 1 {n^{2}}$ with probability $\frac 1 n$. Assuming independence of $\{X_n\}$ the fact that $\sum P\{g(X_n)-g(Y_n) <-1\} =\infty$ shows that $g(X_n)-g(Y_n)$ does not converge to 0 almost surely.