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I have read in some book that in case of almost sure convergence of a sequence of random variable it possible that $|X_{n}(\omega) - X(\omega)|$ can be extremely large for $\omega$ in a small probability set. How is that possible ? Here the sequence {$X_{n}$} of random variables converges to $X$ in almost sure sense.

By almost sure convergence we mean that $P(\omega : X_{n}(\omega)=X(\omega)$ as $n$->infinity)=1 i.e.$P(\omega$ : for any $\epsilon > 0 $there exist an $N$ such that for all $n>=N$ $|X_{n}(\omega)-X(\omega)| < \epsilon)$ = 1 . Then how can $|X_{n}(\omega) - X(\omega)|$ be extremely large for $\omega$ in a small probability set where $|X_{n}(\omega) - X(\omega)| < \epsilon$ with probability 1 ? confused.

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The key is that the rate of convergence can depend on $\omega$. For any $\epsilon$ and any $\omega$ there is an $N$ such that $|X_n(\omega) - X(\omega)| < \epsilon$ for all $n \ge N$, but you might have to choose a different $N$ for different $\omega$. –  Nate Eldredge Dec 16 '12 at 18:21

3 Answers 3

up vote 2 down vote accepted

Let $U$ be a uniform (0,1) random variable, and let $$X_n = \begin{cases} n^{2893}, & U < 1/n \\ 0, & U \ge 1/n \end{cases}.$$ Then $X_n \to 0$ whenever $U > 0$, i.e. almost surely. Indeed, $X_n = 0$ for all $n > 1/U$. But $X_n$ is extremely large on the event $\{U < 1/n\}$ which has positive probability $1/n$.

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If $X_n=x_n\mathbf 1_{A_n}$ for some event $A_n$ with probability $\mathbb P(A_n)=1/n^2$, then $X_n\to0$ almost surely but $[X_n=x_n]$ is an event of positive probability. Pick the rapidly growing sequence $(x_n)$ of your choice.

Edit: To answer the OP's puzzlement as expressed in a comment, note that $\sum\limits_n\mathbb P(A_n)$ converges hence, by Borel-Cantelli lemma, the event $\limsup\limits_nA_n$ has probability zero. This means that the set of $\omega$ such that $X_n(\omega)=0$ for every $n$ large enough has probability $1$. In particular, $X_n(\omega)\to0$ for almost every $\omega$. On the other hand, $X_n(\omega)$ is extremely large, that is, $X_n(\omega)=x_n$, for every $\omega$ in $A_n$ and the probability of $A_n$ is (small but) positive. Thus, the sequence $(X_n)_n$ and its limit $X\equiv0$ seem to fit the conditions asked for in the question.

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Not happy with the answer. By almost sure convergence we mean that $P(\omega : X_{n}(\omega)=X(\omega)$ as $n$->infinity)=1 i.e.$P(\omega$ : for any $\epsilon > 0 $there exist an $N$ such that for all $n>=N$ $|X_{n}(\omega)-X(\omega)| < \epsilon)$ = 1 . Then how can $|X_{n}(\omega) - X(\omega)|$ be extremely large for $\omega$ in a small probability set ? confused. –  sosha Dec 16 '12 at 16:14
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Not happy? Then see Edit. Still confused? –  Did Dec 16 '12 at 18:08
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@prasenjit You probably don't mean $X_{n}(\omega)=X(\omega)$ in the definition of that set, but rather $X_{n}(\omega)\to X(\omega)$. –  Thomas E. Dec 16 '12 at 18:19
    
@did: my confusion is your following statement : "This means that the set of $\omega$ such that $X_n(\omega)=0$ for every $n$ large enough has probability $1$. In particular, $X_n(\omega)\to0$ for almost every $\omega$." If the probability is 1 why are you saying "almost". If it is "almost" then the probability can be very high but not exactly 1. –  sosha Dec 17 '12 at 3:58
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I see. Two different statements are involved here. Either one fixes $n$ and one aks whether $|X_n-X|$ can be large. The answer is yes it can, and, for each $n$, this happens with positive probability. Or, one can ask whether the limit $Y=\lim\limits_{n\to\infty}|X_n-X|$ can be large. The answer is no it cannot, since the event $[Y\ne0]$ has probability zero. Hope this helps. –  Did Dec 17 '12 at 12:39

Consider the probability space $([0,1],{\mathcal B}_{[0,1]},{\bf P})$ where ${\bf P}$ is the uniform probability distribution. Let $$ X_n(s)=\sum_{j=0}^{n}s^j. $$ Then $X_n$ converges in $[0,1)$ with ${\bf P}([0,1))$ =1. But $X_n(1)$ escapes to infinity.

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