Suppose $g_{n} \to g$ and $X_{n} \to X$. Under what conditions on the above function, random variables, and types of convergence would it be true that $g_{n}(X_{n}) \to g(X)$?


I am aware of the continuous mapping theorem, but doesn't the $g_{n}$ make a difference here? For example, suppose $g_{n}: [0,1] \to [0,1]$ satisfies $g_{n}(x)=x^{n}$. Then each $g_{n}$ is continuous and converges to 0 for $x \in [0,1)$ and 1 for $x=1$.

Now let $U_{n} \stackrel{iid}{\sim} U[0,1]$, and $X_{n}=U_{n}^{\frac{1}{n}}\stackrel{d}{\to} 1$.

Then $g(X)=g(1)=1$ but $g_{n}(X_{n}) = U_{n} \sim U[0,1]$ for all $n$.

  • $\begingroup$ A hint for one of the conditions needed: $g_n$ may be continuous, but the limit function, $g$, is not. $\endgroup$ – Alecos Papadopoulos Oct 18 '16 at 18:17

You basically answered this yourself: one sufficient condition is if $g_n$ converges to $g$ uniformly and also that $X_n,X$ live in a compact set, so that all continuous functions on the range of $X_n,X$ are also uniformly continuous.

Here are the details: it suffices to show that for all bounded continuous functions $f$, $E[f(g_n(X_n))]\rightarrow E[f(g(X))]$.

Notice that $h_n(x):=f(g_n(x))$ is a continuous function for all $n$. As well, for each $\epsilon>0$, by uniform convergence, $|g_n(x)-g(x)|<\epsilon$ for all $n\geq N(\epsilon)$. Similarly by our compact range assumption, $f$ is also uniformly continuous so lets pick an $N$ that makes $g_n$ uniformly close to $g$ and $f$ uniformly continuous:



$$f(g(x))-\epsilon\leq h_n(x)\leq f(g(x))+\epsilon.$$

Now squeeze the limit of $E[f(g_n(X_n)]$ between these two bounds and use the fact that $\epsilon$ is arbitrary, thus proving the claim.

Otherwise, if $X_n,X$ are not compact range, there could be technical problems, specifically with the non-uniformity of $f(x)$. So I think in this case you'd need at least some form of tightness on $g_n(X_n)$.

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I think the most immediate answer is if each function, $g_n$, is continuous. Then, by the continuous mapping theorem, we have that $$X_n\xrightarrow{a.s.}X \Rightarrow g(X_n)\xrightarrow{a.s.}g(X)$$ $$X_n\xrightarrow{P}X \Rightarrow g(X_n)\xrightarrow{P}g(X)$$ $$X_n\xrightarrow{D}X \Rightarrow g(X_n)\xrightarrow{D}g(X)$$

Where the mode of convergence stands for almost sure (a.s.), in probability (P) and in distribution (D).

Also, the Portmanteau theorem might provide some addition insight in regards to convergence of probability measures.

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