5

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.

  • 1
    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 Answers3

7

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.

Did
  • 279,727
  • 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. –  Dec 16 '12 at 16:14
  • 4
    Not happy? Then see Edit. Still confused? – Did Dec 16 '12 at 18:08
  • 1
    @prasenjit You probably don't mean $X_{n}(\omega)=X(\omega)$ in the definition of that set, but rather $X_{n}(\omega)\to X(\omega)$. – T. Eskin 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. –  Dec 17 '12 at 03:58
  • Wrong. What you say holds for discrete probability spaces but not for continuous ones. For example, if $U$ is uniform on $[0,1]$, then the event $[U,\text{irrational}]$ has probability $1$ and is only almost sure since the complementary event $[U,\text{rational}]$ may be non empty. Check this. – Did Dec 17 '12 at 07:05
  • yes, this article says that those events can occur theoretically, but it has probability zero. Hence the difference $|X_{n}(\omega)-X(\omega)|$ can be extremely large with zero probability but not "positive" probability. –  Dec 17 '12 at 07:41
  • You mean, $|X_n-X|$ can be large with zero probability, in my example? NOT. AT. ALL. Each $|X_n-X|$ is large on an event of positive probability. // Since it seems this discussion might be heading slowly to being unpleasant, let me say this: EVERYTHING in my answer is carefully chosen and carefully pondered. So, please read the answer slowly and try to digest it before posting any other comment. – Did Dec 17 '12 at 07:48
  • No I did not mean that. In the wikipedia link you have given in your previous post about almost sure convergence says that in case of almost sure convergence there are some events which can occur theoretically but probability of them to occur is "zero" i.e. $|X_{n}(\omega)-X(\omega)|$ can be extremely large theoretically but probability of this is zero. This is an acceptable statement. But when you say that $|X_{n}(\omega)-X(\omega)|$ is extremely large with "positive" probability in case of a.s. convergence that is creating confusion to me. Hope I have made my point. –  Dec 17 '12 at 08:45
  • 2
    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
  • @Did that last comment deserves to be a separate answer – Alex Jul 05 '20 at 14:55
3

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 $1/n$.

Nate Eldredge
  • 97,710
1

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.