Please, help me find connections between two types of convergence:

Let $\{X_n\}_{n\ge1}: (\Omega,F,P) \rightarrow (\mathbb{R},Bor)$ be a sequence of r.v., there are two convergences:

1) $X_n \rightarrow X \hspace{0.2cm}(sLip)$, i.e. $\sum_{n\ge1} E|f(X_n) - f(X)| < \infty \hspace{0.2cm}\forall f \in Lip$ and bounded

2) $X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) = \infty \hspace{0.2cm}\forall \epsilon > 0$

I know several things about other type of "complete convergence" ($X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) < \infty \hspace{0.2cm}\forall \epsilon > 0$) and it's connection with "strong $L^p$" convergence ($X_n \rightarrow X \hspace{0.2cm}(s.-L^p)$, i.e. $\sum_{n\ge1} E(|X_n - X|^p) < \infty$).

Also, I now about the second Borell-Cantelli lemma, but it uses the independence of random variables (which we don't have).

And it is easy to prove that $E|f(\xi_n)-f(\xi)| \le L E|\xi_n - \xi| \le L ||\xi_n - \xi||_{\infty}$ for L-Lipschitz and bounded functions.

But I don't know, how can I apply all these facts to the given situation (or maybe there is another way to solve this problem). If you have any ideas (or some articles to recommend), I will be very pleasant.

  • $\begingroup$ Please make sure to reread your questions carefully before posting. Currently, your second notion of convergence isn't a notion of convergence at all since if I take for $X_n$ a sequence of i.i.d. Bernoullis then it "converges" both to $0$ and to $1$... $\endgroup$ May 19 '20 at 16:17
  • $\begingroup$ Sorry, but this is what really asked at the problem. This definition exists, for example, it's variation is in the second Borel-Cantelli lemma (en.wikipedia.org/wiki/Borel–Cantelli_lemma). May be it's not very good to call it "convergence" but this is what I need to prove: what types of connections are between this two notions (if it is better). $\endgroup$ May 20 '20 at 9:10
  • $\begingroup$ But it's still weird to call it convergence because in the second B-C lemma, (2) is exactly the criterion for the sequence to not converge. Is there any chance you meant (2) to have $< \infty$ instead of $= \infty$? $\endgroup$ May 21 '20 at 19:51
  • $\begingroup$ @NateEldredge : If you replace 2) by not-2), then, in view of my answer, you will get that not-2) implies not-1). So, not-2) will imply a non-convergence. $\endgroup$ May 21 '20 at 19:59

$\newcommand\ep{\epsilon}$We have 1)$\implies$2) but 2)$\kern5pt\not\kern-5pt\implies$1).

Indeed, for each real $a>0$, consider the bounded Lipschitz functions $f_a$ and $g_a$ defined by
\begin{equation*} f_a(x):=a\wedge|x|,\quad g_a(x):=(-a)\vee(a\wedge x) \end{equation*} for real $x$, where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$.

Suppose now that 1) holds. Take any real $a>0$ such that $P(|X|\le a/2)>0$. Note that \begin{multline*} P(|X|\le a/2,|X_n|>a)\le P(f_a(X)\le a/2,f_a(X_n)\ge a) \\ \le P(|f_a(X_n)-f_a(X)|\ge a/2) \le E|f_a(X_n)-f_a(X)|/(a/2), \end{multline*} by Markov's inequality. So, in view of 1), \begin{equation*} \sum_n P(|X|\le a/2,|X_n|>a)<\infty. \end{equation*} Therefore and because of the condition $P(|X|\le a/2)>0$, \begin{multline*} \sum_n P(|X|\le a/2,|X_n|\le a)=\sum_n [P(|X|\le a/2)-P(|X|\le a/2,|X_n|>a)] \\ =\sum_n P(|X|\le a/2)-\sum_n P(|X|\le a/2,|X_n|>a)=\infty. \end{multline*} Hence, \begin{equation*} \sum_n P(|X|\vee|X_n|\le a)=\infty. \tag{*} \end{equation*}

Next, for any real $\ep>0$ \begin{multline*} \sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a) \le\sum_n P(|g_a(X_n)-g_a(X)|>\ep) \\ \le\sum_n E|g_a(X_n)-g_a(X)|/\ep<\infty, \end{multline*} by Markov's inequality and 1).

So, in view of (*), \begin{multline*} \sum_n P(|X_n-X|\le\ep)\ge\sum_n P(|X_n-X|\le\ep,|X|\vee|X_n|\le a) \\ =\sum_n P(|X|\vee|X_n|\le a) -\sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a) =\infty, \end{multline*} so that 2) holds. Thus, 1)$\implies$2).

Now, as suggested in the comment by Martin Hairer, suppose that $X=0$ and $P(X_n=0)=P(X_n=1)=1/2$ for all $n$. Then $P(|X_n-X|\le\ep)\ge1/2$ for all $n$ and hence 2) holds. On the other hand, $E|f_1(X_n)-f_1(X)|=1/2$ and hence 1) does not hold. Thus, 2)$\kern5pt\not\kern-5pt\implies$1).

  • $\begingroup$ Thank you! It was very helpful! $\endgroup$ May 20 '20 at 18:29
  • $\begingroup$ Sorry, but can you explain please why we can use $f(x)\equiv x$? Does it meet the conditions for function $f$? $\endgroup$ May 21 '20 at 9:26
  • $\begingroup$ @IvanPetrov : The condition $f(x)\equiv x$ means $f(x)=x$ for all $x$, that is, $f$ is the identity function. It does meet your conditions on $f$, because the identity function is obviously Lipschitz. $\endgroup$ May 21 '20 at 13:22
  • $\begingroup$ Yes, but the function $f$ should be bounded (from the conditions of strong-Lipschitz convergence. But as I realise, $f(x) = x$ isn't bounded. $\endgroup$ May 21 '20 at 14:39
  • $\begingroup$ @IvanPetrov : Oops! I did not notice that you require the boundedness condition. With it, the result remains the same, but the proof gets quite a bit complicated by a straightforward but nasty truncation argument. $\endgroup$ May 21 '20 at 19:49

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