# $f(X_n)$ converges in probability to $f(X)$ implies $X_n$ converges in probability to $X$

Mathematics Asked by Kurt.W.X on January 3, 2022

Show that $$(X_n)_n$$ converges in probability to $$X$$ if and only if for every continuous function $$f$$ with compact support, $$f(X_n)$$ converges in probability to $$f(X).$$

$$implies$$ is very easy, the problem is with the converse. Any suggestions to begin?

We will make use of the following simple observations:

Lemma 1. $$X_n to X$$ in probability if and only if $$mathbb{E}[|X_n-X|wedge1] to 0$$.

Proof. This easily follows from the inequality

$$epsilon mathbb{P}(|X_n-X|>epsilon) leq mathbb{E}[|X_n-X|wedge1] leq epsilon + mathbb{P}(|X_n-X|>epsilon)$$

which holds for any $$epsilon in (0, 1)$$.

Lemma 2. If $$(X_n)$$ is pointwise bounded by an integrable r.v. $$Y$$ and converges in probability to $$X$$, then $$lim_{ntoinfty} mathbb{E}[X_n] = mathbb{E}[X].$$

Proof. For any subsequence of $$(X_n)$$, there exists a further subsequence $$X_{n_k}$$ which converges a.s. to $$X$$. Then by the Dominated Convergence Theorem, $$mathbb{E}[X_{n_k}]tomathbb{E}[X]$$. This implies the desired claim.

Returning to OP's question, fix $$0 and $$chi, varphi in C_c(mathbb{R})$$ such that $$mathbf{1}_{[-a,a]} leq chi leq mathbf{1}_{[-b,b]}$$ on all of $$mathbb{R}$$ and $$varphi(x) = x$$ on $$[-b, b]$$. Then we find that

$$mathbb{P}(|X_n| > b) = 1 - mathbb{E}[mathbf{1}_{[-b,b]}(X_n)] leq 1 - mathbb{E}[chi(X_n)]$$

and

begin{align*} &(|X_n - X|wedge 1)mathbf{1}_{{|X_n|leq b}cap{|X|leq b}} \ &= (|varphi(X_n) - varphi(X)|wedge 1)mathbf{1}_{{|X_n|leq b}cap{|X|leq b}} \ &leq |varphi(X_n) - varphi(X)|. end{align*}

Using this, we may bound $$mathbb{E}[|X_n - X|wedge 1]$$ from above as follows:

begin{align*} mathbb{E}[|X_n - X|wedge 1] &leq mathbb{E}[(|X_n - X|wedge1) mathbf{1}_{{|X_n|leq b}cap{|X|leq b}}] + mathbb{P}(|X_n|>b) + mathbb{P}(|X|>b) \ &leq mathbb{E}[|varphi(X_n) - varphi(X)|] + (1 - mathbb{E}[chi(X_n)]) + mathbb{P}(|X|>b). end{align*}

Taking $$limsup$$ as $$ntoinfty$$ and applying Lemma 1 and 2,

begin{align*} limsup_{ntoinfty} mathbb{E}[|X_n - X|wedge 1] &leq (1 - mathbb{E}[chi(X)]) + mathbb{P}(|X|>b) \ &leq mathbb{P}(|X|>a)+mathbb{P}(|X|>b). end{align*}

Since this limsup is independent of $$a$$ and $$b$$, letting $$btoinfty$$ followed by $$atoinfty$$ shows that the limsup is zero, or equivalently,

$$lim_{ntoinfty} mathbb{E}[|X_n - X|wedge 1] = 0.$$

Therefore the desired conclusion follows by Lemma 1.

Answered by Sangchul Lee on January 3, 2022

Use functions of the form $$f_M(x)=begin{cases}0 & xleq -M \ x+M & -M leq x leq M \ 3M-x & M < x < 3M \ 0 & xgeq 3Mend{cases}$$ for $$M>0$$.

Then begin{align*}mathbb{P}[|X_n-X|>3varepsilon] &leq mathbb{P}[|X_n+M-f_M(X_n)|>varepsilon] + mathbb{P}[|f_M(X_n)-f_M(X)|>varepsilon] + mathbb{P}[|f_M(X)-X-M|>varepsilon] \ &leq mathbb{P}[|X_n|>M] + mathbb{P}[|f_M(X_n)-f_M(X)|>varepsilon] + mathbb{P}[|X|>M]end{align*}

Of the terms in the last line, the last is small for $$M>M_0$$ and the middle is small for $$n>n_0(M,varepsilon)$$. The only term we have to be careful with is the first term.

This is where our choice of class of functions matters and where my previous answer failed. Using the assumption that $$f_M(X_n)xrightarrow{mathbb{P}}f_M(X)$$ for all $$M>0$$ we can control that first term, but I'll leave that to you.

Answered by Brian Moehring on January 3, 2022

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