0
$\begingroup$

Let $\{X_n\}_n$ be a sequence of i.i.d. RV's, where each $X_n$ is a copy of RV $X: \Omega \rightarrow \mathbb R$. For $n > 0$, I can define $\bar X_n := \frac{1}{n}\sum_{i=1}^n X_i$. Then, central limit theorem states that

$\bar X_n \sim \mathcal N(\mathbb E[X], \sigma^2/n)$.

Now, fix $n > 0$ and let us pick samples of size $n$, and suppose we pick $m$ samples $\{S_1, \cdots, S_m\}$. For each sample $S_k$, I can calculate (mean, variance), denoted by $(\hat \theta_k, \hat \sigma_k^2)$. Then, I construct confidence intervals $\{I_1, \cdots, I_m\}$, where $I_k = (\hat \theta_k - 2\sigma_k, \hat \theta_k + 2\sigma_k)$. Note that for each $k = 1, \cdots, m$, $\sigma_k$ will be different, i.e. each confidence interval would have different length.

Central limit theorem does not give me information about distribution of variances. Without a guarantee that each confidence interval has the same length, how am I guaranteed that 95% of confidence intervals would actually contain the true parameter $\mathbb E[X]$?

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ 1. The confidence intervals having the same length a) will never occur if the random variates are from a continuous distribution and b) is irrelevant. 2. Your definition of the central limit theorem is incorrect: see stats.stackexchange.com/questions/174734/… for some clarification. $\endgroup$
    – jbowman
    Commented Jan 16, 2023 at 22:15
  • $\begingroup$ @James Your notation is off (among other things, you conflate standard deviations with estimates). With CIs write a pivotal quantity (or in this case an asymptotically pivotal quantity) and use not just CLT but also Slutsky's theorem to get a suitable distribution for the pivotal quantity, and hence the CI for the parameter. Different CIs dont need to have the same length, you just need 1-alpha of them to overlap the parameter. $\endgroup$
    – Glen_b
    Commented Jan 16, 2023 at 23:48
  • $\begingroup$ @jbowman Would you elaborate why my definition of CLT is incorrect? en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT $\endgroup$
    – James C
    Commented Jan 17, 2023 at 19:09
  • 1
    $\begingroup$ @JamesC Did you read what Aksakal wrote in the link by Jbowman? "Since your right hand side is changing, it's difficult to prove statements about this convergence relation, and I mean impossible by difficult." If so, what remains unclear? $\endgroup$
    – Dave
    Commented Jan 17, 2023 at 19:11
  • $\begingroup$ @Dave so if I fix my statement as $\sqrt{n} \bar X_n \sim \mathcal N(\mathbb E[X], \sigma)$, then the statement becomes correct. Is this what you mean? $\endgroup$
    – James C
    Commented Jan 17, 2023 at 19:20

0

Reset to default

Your Answer

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