# Finding a connection between two types of convergence

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.

• 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 $1$... May 19 '20 at 16:17
• 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). May 20 '20 at 9:10
• 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 $= \infty$? May 21 '20 at 19:51
• @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. 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).

• Thank you! It was very helpful! May 20 '20 at 18:29
• Sorry, but can you explain please why we can use $f(x)\equiv x$? Does it meet the conditions for function $f$? May 21 '20 at 9:26
• @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. May 21 '20 at 13:22
• 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. May 21 '20 at 14:39
• @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. May 21 '20 at 19:49