0
$\begingroup$

The following is an exercise in the book Measure Theory and Probability, by Athreya and Lahire.

Let $X_n$ converge to $X$ in probability. If $f$ is continuous, then $f(X_n)$ converges in probability to $f(X)$

This seems like a simple exercise, but I haven’t been able to solve it. This is what I have tried so for.

Assuming that $f$ is uniformly continuous, then for $$\epsilon > 0 \ \exists \delta >0 : \mid X_n(w) - X(w) \mid < \delta \implies \mid f(X_n(w)) - f(X(w)) \mid < \epsilon $$

Hence, we know that $P(\mid X_n - X\mid < \delta ) \leq P(\mid f(X_n) - f(X) \mid < \epsilon )$, and since $X_n \rightarrow_p X$, then $$ 1 = \lim_{n \to \infty } P(\mid X_n - X\mid < \delta ) \leq \lim_{n\to \infty}P(\mid f(X_n) - f(X) \mid < \epsilon ) $$

Now, is the above solution correct? And how does one prove if $f$ is not uniformly continuous?

$\endgroup$
  • 1
    $\begingroup$ You know that $X_n \to X$ in probability iff for every subsequence $(n_k)$ there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely? It is the easiest way of proving your statement $\endgroup$ – Dominik Kutek Aug 4 '20 at 12:57
  • $\begingroup$ I haven’t been introduced to this theorem. Do you have a reference to it’s proof? $\endgroup$ – Davi Barreira Aug 4 '20 at 13:12
  • 1
    $\begingroup$ I've put an answer with the short proof. I thought a little bit about your approach. It is valid for uniformly continuous functions, indeed. However I don't see how to prove it for $f$ only continuous in that way. If you know anything about convergence in distribution and tightness, then you could for any $\varepsilon > 0$ find such $M$ that $\mathbb P(|X_n| > M) < \varepsilon, \mathbb P(|X| > M) < \varepsilon$ for every $n \in \mathbb N$. On $[-M,M]$ continuous function is uniformly continuous, so you can bound it there as you did above and on the rest there is no problem due to tightness $\endgroup$ – Dominik Kutek Aug 4 '20 at 13:40
2
$\begingroup$

Theorem: Let $(X_n)$ be a sequence of random variables. Then $X_n \to X$ in probability, iff for every subsequence $(n_k)$ there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely.

Firstly lemma: If $X_n \to X$ in probability, then there exists subsequence $(n_k)$ such that $X_{n_k} \to X$ almost surely.

Proof of lemma: By convergence in probability, we have a subsequence $(n_k)$ such that $\mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) \le \frac{1}{k^2}$. Hence $\sum_{k=1}^\infty \mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) < \infty$ so by borel cantelli, almost surely we have $|X_{n_k} - X| < \frac{1}{k^2}$ starting from some $k>K$, so $X_{n_k} \to X$ almost surely.

Proof of theorem.

By Lemma we have => direction (since for any subsequence $(n_k)$ we have $X_{n_k} \to X$ in probability, too).

"<=" Assume contrary, that there $X_n \not \to X$ in probability. Hence by definition, there exists $\varepsilon >0, \delta >0$ and subsequence $(n_k)$ such that $\mathbb P(|X_{n_k} - X| > \varepsilon) > \delta$ for every $k \in \mathbb N$. But from that subsequence $(n_k)$ we cannot choose any sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely (because $X_{n_{k_m}} \not \to X$ in probability and this is necessary condition).

Having theorem we can proceed as follows in your question:

Let $f$ be continuous and $X_n \to X$ in probability. We want to show that $Y_n := f(X_n) \to f(X) =: Y$ in probability. Take any subsequence $(n_k)$. We know that there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely. Hence $Y_{n_{k_m}} \to Y$ almost surely (cause continuous functions of pointwise convergent sequences are pointwise convergent). Since $(n_k)$ was arbitrary, by theorem we know that $Y_n \to Y$ in probability.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.