3
$\begingroup$

If we have a sequence $X_n$ that converges almost surely, it makes sense to say that $\mathbb{P}(\lim_{n \to \infty} {X_n} = \mu) = 1$. However, does this imply that the number of failures, i.e., $|{X_n}-\mu| > \epsilon$ is necessarily finite for any variation of the sequence?

Additionally, I'm also curious about the way that probabilities tie into a.s. convergence with an infinite number of possible sequences. For example, I understand that the probability of the events where $X_n$ does not converge to $\mu$ is 0. However, the number of sequences where $X_n$ does not converge also seems infinite. Is there any insight to be gained from thinking about the infinities here, or is it better to just rely on the fact that as defined by the probability measure, the probability of convergence is 1, hence the sequences where we don't converge is ignored?

$\endgroup$
3
  • $\begingroup$ Could you explain what you mean by "variation of the sequence"? $\endgroup$
    – whuber
    Commented Jun 14, 2022 at 22:51
  • $\begingroup$ @whuber I just meant every possible sequence $X_n$. $\endgroup$ Commented Jun 21, 2022 at 22:56
  • $\begingroup$ Then trivially the answer is "no," because (assuming this is a sequence of independent non-constant random variables) there can be "variations" where the "number of failures" is infinite. They just won't have very high probability ;-). $\endgroup$
    – whuber
    Commented Jun 22, 2022 at 11:57

1 Answer 1

1
$\begingroup$

Almost sure convergence is a limiting statement that occurs with probability one. Occurrence with probability one is a bit weaker than saying something "necesarily" happens in the logical sense, but it is close. I will sow you how you might frame the first part of your question concerning the number of "failures" at a given deviation level $\varepsilon$.

To get something like the result you're looking for, first remember that the definition of the limiting statement $\lim_{n \rightarrow \infty} X_n = \mu$ is that there exists a function $N: \mathbb{R}_{+} \rightarrow \mathbb{N}$ such that:

$$n \geqslant N(\varepsilon) \quad \quad \implies \quad \quad |X_n-\mu| \leqslant \varepsilon \quad \quad \quad \text{for all } \varepsilon > 0.$$

Consequently, if this limiting statement holds then for any given $\varepsilon>0$ we can write the number of "failures" (in the sense you use the term) as:

$$\begin{align} \text{Failures}(\varepsilon) &\equiv \sum_{n \in \mathbb{N}} \mathbb{I}( |X_n-\mu| > \varepsilon) \\[6pt] &= \sum_{n=1}^{N(\varepsilon)-1} \mathbb{I}( |X_n-\mu| > \varepsilon) + \sum_{n=N(\varepsilon)}^{\infty} \mathbb{I}( |X_n-\mu| > \varepsilon) \\[6pt] &= \sum_{n=1}^{N(\varepsilon)-1} \mathbb{I}( |X_n-\mu| > \varepsilon) \\[12pt] &< N(\varepsilon). \\[6pt] \end{align}$$

That is, if $\lim_{n \rightarrow \infty} X_n = \mu$ then the number of "failures" at any level $\varepsilon$ is always finite (strictly less than $N(\varepsilon)$). Under almost sure convergence this limit holds with probability one, so we can say that the number of "failures" is finite with probability one, for any $\varepsilon >0 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.