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Given $A_n\rightarrow \infty$ almost surely (a.s). Show that $\forall\ N > 0$, $P\left\{A_n<N\ \text{infinitely often}\right\} = 0$.

My thought: By the sake of contradiction, assume there exists an $\infty> N >0$ such that $P\left\{A_n<N\ \text{infinitely often}\right\} > 0$. This is equivalent to: $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} (A_k-N) < 0\right\} > 0$. But this means there exists a $k_1\geq n$ such that $A_{k_1} < N$ (1)

On the other hand, since $A_n\rightarrow \infty$ a.s, for every $\epsilon > 0$, $\lim_{n\rightarrow \infty} P\left\{\cup_{k\geq n} |A_k- \infty| \geq \epsilon\right\} = 0$. This means $|A_k-\infty|<\epsilon$ for every $k\geq n$. Since $k_1\geq n$, $\infty - A_{k_1} < |A_k- \infty| < \epsilon$ for any $\epsilon > 0$. Let $\epsilon = N$, we get $\infty < N+A_{k_1} < 2N$ (due to (1)). This is a contradiction, since $\infty > N$.

My question: I don't find my mathematical notation above legit, because I am doing subtraction with $\infty$. But I would like to know if my solution above is correct, so could someone please help verify? Also, I think there should be a shorter/slicker way to solve it, as my solution above is quite complicated and use more of real analysis. Anyone wants to give it a try? Any thoughts about my solution or this problem would be really appreciated anyway.

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    $\begingroup$ @BCLC: That makes A LOT OF sense:) Thank you very much for your help. The rest of my proof is correct? You have any better solution? $\endgroup$ – user177196 Oct 4 '16 at 4:50
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    $\begingroup$ For a proof, note that the problem is not measure related (in particular, Borel-Cantelli lemmas are offtopic) but simply a consequence of some inclusions between events, namely, $$[A_n<N\ \text{infinitely often}]\subseteq[\liminf A_n\leqslant N]\subseteq\Omega\setminus[A_n\to\infty]$$ $\endgroup$ – Did Oct 4 '16 at 7:08
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    $\begingroup$ The first inclusion is pure logic: if $(x_n)$ is a real valued sequence such that $x_n<N$ for infinitely many $n$ then $\inf\limits_{k\geqslant n}x_k<N$ for every $n$ hence $\liminf x_n=\lim\limits_{n\to\infty}\inf\limits_{k\geqslant n}x_k\leqslant N$. The second inclusion is pure logic as well: if $\liminf x_n\leqslant N$ then $x_n<N+1$ for infinitely many indexes $n$ hence $x_n\to\infty$ is impossible. Re the tentative proof in your question, sorry but I have no idea what the chain of characters $$\cup_{k\ge n}|A_k|\ge M$$ can mean. Please explain. $\endgroup$ – Did Oct 4 '16 at 11:35
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    $\begingroup$ @user177196 I think it's supposed to be $$\bigcup_{k \ge n} (|A_k| \ge M)$$ instead of $$(\bigcup_{k \ge n} |A_k|) \ge M$$ $\endgroup$ – BCLC Oct 4 '16 at 21:57
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    $\begingroup$ @BCLC: that's what I meant. $\endgroup$ – user177196 Oct 5 '16 at 2:14
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Important inequalities (Probability w/ Martingales):

1, 2

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  1. If $$\liminf x_n > z$$ then $$(x_n > z)$$ eventually

  2. If $$\liminf x_n < z$$ then $$(x_n < z)$$ infinitely often


$$P([\lim X_n] = \infty) = 1$$

$$\to P([\lim X_n] > N) = 1$$

$$\to P([\liminf X_n] > N) = 1$$

$$\to P(\liminf [X_n > N]) = 1$$

$$\to P(\limsup[X_n \le N]) = 0$$

QED

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$$X_n \to \infty$$ means $\forall M > 0, \exists L_M > 0$ s.t. $X_n > M$ wh $n > L_M$

We want TS $\forall N > 0, P(\limsup(X_n < N)) = 0$.

Pf: Given $N > 0$, $\exists L_N > 0$ s.t. $X_n > N$ wh $n > L_N$

Now consider $$\sum_{n=1}^{\infty} P(X_n < N) \tag{*}$$

The summand is $0$ for $n > L_N$. Thus,

$$(*) = \sum_{n=1}^{L_N} P(X_n < N) \le L_N < \infty.$$ By BCL1, $P(\limsup(X_n < N)) = 0.$ QED

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