Almost sure convergence

Point-wise convergence

Consider a random sequence X:\Omega\to\R^\N defined on a probability space (\Omega, \sF, P), then each X_n \triangleq \pi_n\circ X: \Omega \to \R is a random variable. There are many possible definitions for convergence of a sequence of random variables. One idea is to consider X(\omega)\in \R^\N as a real valued sequence for each outcome \omega, and consider the \lim_nX_n(\omega) for each outcome \omega.

Definition 1. A random sequence X:\Omega\to\R^\N defined on a probability space (\Omega, \sF, P) converges point-wise to a random variable X_\infty:\Omega \to \R, if for all outcomes \omega \in \Omega, we have

Remark 1. This is a very strong convergence. Intuitively, what happens on an event of probability zero is not important. We will strive for a weaker notion of convergence, where the sequence of random variable converge point-wise on a set of outcomes with probability one.

Almost sure statements

Definition 2. A statement holds almost surely (a.s.) if there exists an event called the exception set N \in \sF with P(N) = 0 such that the statement holds for all \omega \notin N.

Example 3 (Almost sure equality). Two random variables X, Y defined on the probability space (\Omega, \sF, P) are said to be equal a.s. if the following exception set has probability measure P(N) = 0. Then Y is called a version of X, and we can define an equivalence class of a.s. equal random variables.

Example 4 (Almost sure monotonicity). Two random variables X, Y defined on the probability space (\Omega, \sF, P) are said to be X \le Y a.s. if the exception set N \triangleq \set{\omega \in \Omega: X(\omega) > Y(\omega)} \in \sF has probability measure P(N) = 0.

Almost sure convergence

Definition 5 (Almost sure convergence). A random sequence X:\Omega\to\R^\N defined on the probability space (\Omega, \sF, P) converges almost surely, if the following exception set has zero probability. Let X_\infty be the point-wise limit of the sequence of random variables X:\Omega\to\R^\N on the set N^c, then we say that the sequence X converges almost surely to X_\infty, and denote it as

Example 6 (Convergence almost surely but not everywhere). Consider the probability space ([0,1], \sB([0,1]), \lambda) such that \lambda([a,b]) = b-a for all 0 \le a \le b \le 1. For each n \in \N, we define the scaled indicator random variable X_n: \Omega \to \set{0,1} such that Let N = \set{0}, then for any \omega \notin N, there exists m = \lceil\frac{1}{\omega}\rceil \in \N, such that for all n > m, we have X_n(\omega) = 0. That is, \lim_nX_n = 0 a.s. since \lambda(N) = 0. However, X_n(0) = n for all n \in \N.

Convergence in probability

Definition 7 (convergence in probability). A random sequence X:\Omega\to\R^\N defined on the probability space (\Omega, \sF, P) converges in probability to a random variable X_\infty:\Omega\to\R, if \lim_nP(A_n(\epsilon)) = 0 for any \epsilon > 0, where

Remark 2. \lim_nX_n = X_\infty a.s. means that for almost all outcomes \omega, the difference X_n(\omega)-X_\infty(\omega) gets small and stays small.

Remark 3. \lim_nX_n = X_\infty i.p. is a weaker convergence than a.s. convergence, and merely requires that the probability of the difference X_n(\omega)-X_\infty(\omega) being non-trivial becomes small.

Example 8 (Convergence in probability but not almost surely). Consider the probability space ([0,1], \sB([0,1]), \lambda) such that \lambda([a,b]) = b-a for all 0 \le a \le b \le 1. For each k \in \N, we consider the sequence S_k = \sum_{i=1}^ki, and define integer intervals I_k \triangleq \set{S_{k-1}+1, \dots, S_k}. Clearly, the intervals (I_k:k \in \N) partition the natural numbers where \abs{I_k} = k. It follows that each n \in \N lies in some I_{k_n}, such that n = S_{k_n-1}+i_n for i_n \in [k_n]. Therefore, for each n \in \N, we define indicator random variable X_n: \Omega \to \set{0,1} such that For any \omega \in [0,1], we have X_n(\omega) = 1 for infinitely many values since there exist infinitely many (i,k) pairs such that \frac{(i-1)}{k} \le \omega \le \frac{i}{k}, and hence \lim\sup_nX_n(\omega) = 1 and hence \lim_nX_n(\omega) \neq 0. However, \lim_nX_n(\omega) = 0 in probability, since

Infinitely often and all but finitely many

Lemma 9 (infinitely often and all but finitely many). Let A \in \sF^\N be a sequence of events.

1. For some subsequence (k_n: n \in \N) depending on \omega, we have

2. For a finite n_0(\omega) \in \N depending on \omega, we have

Proof. Proof. Let A \in \sF^\N be a sequence of events.

1. Let \omega \in \lim\sup_nA_n = \cap_{n \in \N}\cup_{k \ge n}A_k, then \omega \in \cup_{k \ge n}A_k for all n \in \N. Therefore, for each n \in \N, there exists k_n \in \N such that \omega \in A_{k_n}, and hence Conversely, if \sum_{j \in \N}\Ind{A_j}(\omega) = \infty, then for each n \in \N there exists a k_n \in \N such that \omega \in A_{k_n} and hence \omega \in \cup_{k \ge n}A_k for all n \in \N.

2. Let \omega\in \lim\inf_nA_n = \cup_{n\in\N}\cap_{k\ge n}A_k, then there exists n_0(\omega) such that \omega \in A_n for all n \ge n_0(\omega). Conversely, if \sum_{j \in \N}\Ind{A_j^c}(\omega) < \infty, then there exists n_0(\omega) such that \omega \in A_n for all n \ge n_0(\omega).

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Theorem 10 (Convergence a.s. implies in probability). If a sequence of random variables X: \Omega \to \R^\N defined on a probability space (\Omega, \sF, P) converges a.s. to a random variable X_\infty:\Omega\to\R, then it converges in probability to the same random variable.

Proof. Proof. Let \lim_nX_n = X_\infty a.s. and \epsilon > 0. We define events A_n \triangleq \set{\omega \in \Omega: \abs{X_n(\omega)-X_\infty(\omega)} > \epsilon} for each n \in \N. We will show that \lim_nP(A_n) = 0. To this end, let N be the exception set such that For \omega \notin N, there exists an n_0(\omega) such that \abs{X_n-X_\infty} \le \epsilon for all n \ge n_0. That is, \omega \in A_n^c for all n \ge n_0(\omega) and hence N^c \subseteq \lim\inf_nA_n^c. It follows that 1 = P(\lim\inf_nA_n^c). Since \lim\inf_nA_n^c = (\lim\sup_nA_n)^c, we get 0 = P(\lim\sup_nA_n) = \lim_nP(\cup_{k \ge n}A_k) \ge \lim_nP(A_n) \ge 0. ◻

Borel-Cantelli Lemma

Proposition 11 (Borel-Cantelli Lemma). Let A \in \sF^\N be a sequence of events such that \sum_{n \in \N}P(A_n) < \infty, then P\set{A_n \text{ i.o.}} = 0.

Proof. Proof. We can write the probability of infinitely often occurrence of A_n, by the continuity and sub-additivity of probability as The last equality follows from the fact that \sum_{n \in \N}P(A_n) < \infty. ◻

Proposition 12 (Borel zero-one law). Let A \in \sF^\N be a sequence of independent events, then

Proof. Proof. Let A \in \sF^\N be a sequence of independent events.

1. From Borel-Cantelli Lemma, if \sum_nP(A_n) < \infty then P\set{A_n\text{ i.o.}} = 0.

2. Conversely, suppose \sum_nP(A_n) = \infty, then \sum_{k \ge n}P(A_k) = \infty for all n \in \N. From the definition of \lim\sup and \lim\inf, continuity of probability, and independence of sequence of events A \in \sF^\N, we get Since 1-x \le e^{-x} for all x \in \R, from the above equation, the continuity of exponential function, and the hypothesis, we get

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Example 13 (Convergence in probability can imply almost sure convergence). Consider a random Bernoulli sequence X: \Omega \to \set{0,1}^\N defined on the probability space (\Omega, \sF, P) such that P\set{X_n = 1} = p_n for all n \in \N. Note that the sequence of random variables is not assumed to be independent, and definitely not identical. If \lim_np_n = 0, then we see that \lim_nX_n = 0 in probability.

In addition, if \sum_{n\in \N}p_n <\infty, then \lim_nX_n = 0 a.s. To see this, we define event A_n \triangleq \set{X_n = 1} \in \sF for each n \in \N. Then, applying the Borel-Cantelli Lemma to sequence of events A \in \sF^\N, we get That is, \lim_nX_n = 0 for \omega \in \lim\inf_nA_n^c, implying almost sure convergence.

Limits of sequences

For any real valued sequence a \in \R^\N, we can define

2 &_n a_n _n_k na_k,& &_n a_n _n_k na_k.

We define e_n \triangleq \sup_{k \ge n}a_k and f_n \triangleq \inf_{k \ge n}a_k, and observe that f_n \le a_k for all k \ge n. That is, f_1, \dots, f_{n-1} \le a_n and f_k \le a_k for all k \ge n. It follows that \sup_nf_n \le \sup_{k \ge n}a_k = e_n for all n \in \N, and hence \lim\inf_n a_n = \sup_nf_n \le \inf_n e_n = \lim\sup_na_n.

A sequence a \in \R^\N is said to converge if \lim\sup_n a_n = \lim\inf_n a_n and the limit is defined as a_n \triangleq \lim_n a_n = \lim\sup_n a_n = \lim\inf_n a_n.

Theorem 14. A sequence a \in \R^\N converges to a_\infty \in \R if for all \epsilon > 0 there exists an integer N \in \N such that for all n \ge N, we have \abs{a_n-a_\infty} < \epsilon.

Proof. Proof. Let \epsilon > 0 and find the integer N_\epsilon \in \N such that a_n \in (a_\infty-\epsilon, a_\infty+\epsilon) for all n \ge N_\epsilon. It follows that a_\infty-\epsilon \le f_n \le e_n \le a_\infty+\epsilon for all n \ge N_\epsilon, and hence a_\infty-\epsilon \le \lim\inf_n a_n \le \lim\sup_n a_n \le a_\infty+\epsilon. Since \epsilon was arbitrary, it follows that \lim_na_n = a_\infty. ◻

For any sequence a \in \R_+^\N, the following statements are true.

If \sum_{n\in\N}a_n < \infty then \lim_{n \to \infty}\sum_{k \ge n}a_k = 0.

If \sum_{n\in\N}a_n = \infty then \sum_{k \ge n}a_k = \infty for all k \in \N.

Proof. Proof. We observe that (\sum_{k < n}a_k: n \in\N) is a non-decreasing sequence, and hence \lim_{n \to \infty}\sum_{k < n}a_k = \sup_n\sum_{k< n}a_k = \sum_{n \in \N}a_n.

It follows that \sum_{k \ge n}a_k = \sum_{n \in \N}a_n - \sum_{k < n}a_k is a non-increasing sequence with limit 0.

We can write \sum_{n \in \N}a_n = \sum_{k < n}a_k + \sum_{k \ge n}a_k. Since the first term is finite for all n \in \N, it follows that the second term must be infinite for all n \in \N.

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