Let $P(X_n=n)=1/n^a$ and zero otherwise. Let $Y_n=X_{n+1}\cdot X_n$. Find all $a$ such that $\liminf Y_n=0$ and $\limsup Y_n=\infty$ almost surely.

Question

Remark: My attemps is WRONG as I fasly assumed that the $Y_n$ were independant. I ve corrected this in my own answer to this post. This answer is correct but not enough well wrotten, read the selected answer for a better quality answer.

Question:

Let $a > 0$. For $n \in \mathbb{N}$, let the random variable $X_n$ on $\{0,n\}$ be defined by $$ \begin{cases} P(X_n=0)=1- \frac{1}{n^{a}}\\ P(X_n=n)=\frac{1}{n^{a}}. \end{cases} $$ All $X_n$ are independent. Define the random variables $Y_n=X_{n+1}\cdot X_n$.

Find the set of all $a$ such that $$P(\omega \in \Omega: \liminf_{n \to \infty} Y_n(\omega) = 0 \cap \limsup_{n \to \infty} Y_n(\omega) = \infty) = 1.$$

Attempt:

I have already proved the following statement.

Part 1

Let's denote $$ D = \{ \omega \in \Omega : Y_n(\omega) \neq 0 \text{ for infinity many } n \}.$$

In order for $Y_n(\omega) \neq 0 $ we should have both $ X_n(\omega) \neq 0, X_{n+1} (\omega) \neq 0$, so $$P(Y_n \neq 0 ) = P(X_n\neq 0 \cap X_{n+1} \neq 0 ) = \frac{1}{n^a} \cdot \frac{1}{(n+1)^a}.$$

Hence according to the Borel-Cantelli lemma we have that $$P(D)= 0 \Leftrightarrow \sum_{n \geq 1} P(Y_n \neq 0) = \sum_{n \geq 1} \frac{1}{n^a} \cdot \frac{1}{(n+1)^a} < \infty $$ and this occurs if and only if $a > \frac{1}{2}$.

We also have that $$P(D)= 1 \Leftrightarrow \sum_{n \geq 1} P(Y_n \neq 0) = \sum_{n \geq 1} \frac{1}{n^a} \cdot \frac{1}{(n+1)^a} = \infty $$ and this occurs if and only if $0 < a \leq \frac{1}{2}$.

Part 2

Let's denote $$E = \{ \omega \in \Omega : Y_n(\omega) = 0 \text{ for infinitely many } n\}.$$

In order for $Y_n(\omega) = 0 $ we should have at least one $ X_n(\omega) = 0$ and/or $ X_{n+1} (\omega) = 0 $. Hence we have that $$P(Y_n = 0 ) = 1-P(X_n\neq 0 \cap X_{n+1} \neq 0 ) = 1 - \frac{1}{n^a} \cdot \frac{1}{(n+1)^a}.$$

Hence according to the Borel-Cantelli lemma we have that $$P(E)= 0 \Leftrightarrow \sum_{n \geq 1} P(Y_n \neq 0) = \sum_{n \geq 1} 1 - \frac{1}{n^a} \cdot \frac{1}{(n+1)^a} < \infty $$ and this never occurs for any $a > 0$.

In other words we have that have that for any $a>0$ we have $$P(E)= 1 \Leftrightarrow \sum_{n \geq 1} P(Y_n = 0) = \sum_{n \geq 1} 1- \frac{1}{n^a} \cdot \frac{1}{(n+1)^a} = \infty. $$

Indeed in order for a series $\sum_{n \geq 1 } a_n $ to be convergent we require $\lim_{n \to \infty} a_n = 0$ but here obviously $a_n= 1- \frac{1}{n^a} \cdot \frac{1}{(n+1)^a} \xrightarrow[n \to \infty]{} 1 \neq 0$ for any $a>0$ so the series is necessary divergent.

Part 3

3- We can note that the event $\limsup_{n \to \infty} Y_n(\omega) = \infty$ is equivalent to the event $D$ and that $\liminf_{n \to \infty} Y_n(\omega) = 0$ is equivalent to the event $E$. So we can reformulate the question as follows:

For which $a>0$ do we have that the probability that we get infinitely often $Y_n \neq 0 $ and at the same time that we get $Y_n = 0$ happens too infinitely often at the same time.

Moreover according to Part 2 we have that $D \subseteq E$ as the event $E$ is always realized when we repeat our experience an infinity of times. And it is only the event $D$ that may not be realized if we repeat our experience an infinity of times.

Thus $$P(\omega \in \Omega: \liminf_{n \to \infty} Y_n(\omega) = 0 \cap \limsup_{n \to \infty} Y_n(\omega) = \infty) = P(\omega \in \Omega: \limsup_{n \to \infty} Y_n(\omega) = \infty) = P(D) = 1 $$ for $0<a \leq \frac{1}{2}$ according to Part 1.

Is this correct? I have mostly doubt on my Part 3.
Thank for your help.

Answer 1

You can be much more compact and precise in writing your answer. Let $i.o.$ stand for infinitely often.

Let $D=\{\omega:Y_{n}(\omega)=0\,\,i.o.\}$ and let $E=\{\omega:Y_{n}\neq 0\,\,i.o.\}$

The obvious case of $a=0$ is ommitted.

If $a>\frac{1}{2}$, then $\displaystyle\sum_{n=1}^{\infty}P(|Y_{n}|>\epsilon)=\sum_{n=1}^{\infty}\dfrac{1}{n^{a}(n+1)^{a}}<\infty$ for all $\epsilon>0$. Hence by Borel-Cantelli Lemma 1, you have that $Y_{n}$ converges to $0$ almost surely and hence $\lim\sup Y_{n}=\lim\inf Y_{n}=0$ almost surely.

So consider $0<a\leq \frac{1}{2}$.

Note that the sequence $(Y_{2n})_{n\geq 3}$ is a sequence of independent random variables. Now, you have $$\sum_{n=3}^{\infty}P(Y_{2n}=2n(2n+2))=\sum_{n=3}^{\infty}\dfrac{1}{(2n(2n+1))^{a}}=\infty$$

and hence by Borel-Cantelli Lemma 2, you have that $P(\{Y_{n}=2n(2n+2)\,\,i.o.\})=1$. As $\{Y_{n}=2n(2n+2)\,\,i.o.\}\subset E$, you have that $P(E)=1$.

And $\displaystyle\sum_{n=1}^{\infty}P(Y_{n}=0)=\infty\implies P(D)=1$ again due to Borel Cantelli Lemma 2.

Hence $P(D\cap E)=1$ which means that almost surely $\lim\sup Y_{n}=\infty$ and $\lim\inf Y_{n}=0$. That is, for each $\omega\in D\cap E$, you have $\lim\sup Y_{n}(\omega)=\infty$ and $\lim\inf Y_{n}(\omega)=0$.

Answer 2

I thought again on this question and on my answer and I have found that it is incorrect for different reasons (one of them is that I ve used incorrectly the BC lemma ). So below you will be able to read a new (and I think and hope correct) answer of mine.

Answer:

1- According to the answer I ve posted here we have that $P(D) = 1$ with $D = \{ Y_n(\omega) \neq 0 $ for an infinity of indexes $n \in \{1,2,3,.... \} \} $ for $0 < a \leq \frac{1}{2} $.
Moreover it is easy to notice that $ P(limsup Y_n = \infty) $ correspond to the probability of $P(D)$ as the event $ D $ and $\{ \omega \in \Omega : limsup_ {n \to \infty }Y_n = \infty \} $ are a.s. equals.

2- Now let's focus on the event $ E= \{ ω∈Ω:Y_n(ω)=0 $ for infinitely many $ n \}. $ that correspond to the event $ \{\omega \in \Omega : \liminf _{n \to \infty } Y_n = 0 \}$
We want to prove that $P(E)=1 , \forall a>0$ that will mean that $ E = \Omega$ a.s.
But again here we can not use directly BC lemma as the sequence of events $ \{ A_k \} = \{Y_k = 0 \}$ are not independent,but we can note that $P(A_k) = 1 - P(\{Y_k \neq 0 \}) = 1 - (\frac{1}{n(n+1)})^a$
So let focus on the following set $ E_{even}= \{ ω∈Ω:Y_{2n}(ω)=0 $ for infinitely many $ n \} $ and on the following sequence of events $ \{ A_{2k} \} = \{Y_{2k} = 0 \} $. That are independant (trivial) thus we can use the corresponding BC lemma and we get that $P(E_{even})=1$ because $ \sum_{n>0} P(A_{2n}) = \sum_{n>0} 1 - (\frac{1}{2n(2n+1)})^a = \infty $ as the necessary condition of convergence for serie is not even respected, indeed $1 - (\frac{1}{2n(2n+1)})^a \xrightarrow[n \to \infty]{} 1 \neq 0$
Obviously $ E_{even} \subseteq E \Rightarrow P(E_{even}) = 1$ thus for any $a>0$ we necessarly have $P(E)=1$ and $\Omega = E $ a.s.

3- Combining "1-" and "2-" we have that $ D \cap E = D$ a.s. and so that $P(D \cap E ) = P(D) $ thus $ P(D \cap E )=1 $ for $ 0< a \leq \frac{1}{2}$