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Suppose $\{X_i:i\geq 1\}$ are i.i.d. with mean $\mu$. By the strong law of large numbers, $\bar{X}_n \stackrel{a.s.}{\rightarrow}\mu$. Does this imply the following?

There exists a $\delta>0$, such that for large enough $n$, $$ P(\bar{X}_n\in [\mu-\delta, \mu+\delta]) = 1. $$

In general does this statement hold true if any estimator $\widehat{\theta}_n$ converges almost surely to a parameter $\theta_0$? Any ideas will be appreciated.

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Yes, it does imply that. From the definition of $\lim_{n \rightarrow \infty}X_n = X$, we get :

$1 = \Pr(\lim_{n \rightarrow \infty}\bar{X}_n = \mu) = \Pr( \forall \epsilon > 0 \ \ \exists n_0 \in \mathbb{N}\ s.t.\ \forall n > n_0 \ ; |\bar{X}_n - \mu| < \epsilon)$

Which in turn implies that there exist infinitely many such $\epsilon, n_0$ Such that $\forall n > n_0$

$P(\bar{X}_n\in [\mu-\epsilon, \mu+\epsilon]) = 1$.

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  • $\begingroup$ Then, the $n_0$ should depend on $\omega$. If I understand correctly, $P(\omega: \bar{X}_n(\omega)\rightarrow \mu) = 1$. Hence, for each such $\omega$ and each $\epsilon>0$, there exists a $n_0 = n_0(\epsilon,\omega)\in\mathbb{N}$, such that $|\bar{X}_n(\omega)-\mu|<\epsilon$, for all $n\geq n_0$. If we fix $\epsilon = 1$ (say), then for each $\omega$, we can have different choices of $n_0$. Please clarify if I am getting it wrong somewhere. $\endgroup$
    – cusat15
    Mar 12, 2020 at 16:30
  • $\begingroup$ $\omega$ is an element of the underlying probability space, where $X_i$ are defined. $\endgroup$
    – cusat15
    Mar 14, 2020 at 15:14
  • $\begingroup$ If you write it like that, then it's $P(\{ \omega: \bar{X}_n(\omega)\rightarrow \mu \})$ - you don't have to look at $\omega$s separately. $P(X = k)$ is the probability density of the set of all $\omega$ such that $X(\omega) = k$, not a specific one. Look here $\endgroup$
    – Blueyedaisy
    Mar 14, 2020 at 19:04
  • $\begingroup$ Each sample path $\{\bar{X}_n(\omega): n\geq 1\}$ will eventually lie within a $[\mu\pm \epsilon]$ interval, for large enough $n$. But, for another sample path, $\{\bar{X}_n(\omega_1):n\geq 1\}$, where $\omega_1(\neq \omega)\in \Omega$, the point of 'entering' this $[\mu\pm \epsilon]$ band, can be different. This is what I understand from the explanation given here. $\endgroup$
    – cusat15
    Mar 16, 2020 at 6:25
  • $\begingroup$ Yes, but the almost sure convergence already applies to all possible $\omega$ $\endgroup$
    – Blueyedaisy
    Mar 16, 2020 at 10:58

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