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My question applies to regression estimates. The formulae for confidence interval: $$ \hat y \pm t_{\alpha/2, n-2} \sqrt{MSE} \sqrt{1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2}} $$

and prediction interval: $$ \hat y \pm t_{\alpha/2, n-2} \sqrt{MSE} \sqrt{1 + 1/n + \frac{(x-\bar x)^2}{\sum (x_i - \bar x)^2}} $$

both show that they decrease with increasing n and they both don't seem to tend to 0. But I remember vaguely someone telling me that CI shrinks to a point.

It would be great if someone could clarify this.

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    $\begingroup$ A prediction interval better not shrink to a point! Otherwise there's no random behavior involved. $\endgroup$ – whuber Oct 12 '15 at 16:33
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    $\begingroup$ Re the edit (which introduced the equations): it is evident that under mild conditions the confidence interval formula does tend to zero as $n$ grows, because (a) the $t$ term tends to the $1-\alpha/2$ quantile of the standard Normal distribution, which is finite, (b) $1/n$ tends to $0$, and (c) the other fraction in the square root tends to zero provided the $x_i$ don't all tend rapidly towards $\bar x$. $\endgroup$ – whuber Oct 12 '15 at 19:31
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Recall that consistency means that the estimator converges in probability to the parameter. This means that all the distribution of the estimator is arbitrarily concentrated around the parameter.

If you construct a confidence interval based on a consistent estimator, is will thus shrink infinitesimally as the sample size grows.

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Formally, repeating a measurement N times decreases the random part of the confidence interval by SQRT(N) so with huge*huge N you get tiny random errors. That part of the error on its own would become vanishingly small eventually. This is useful until the nonrandom part of the errors gets comparable. Because for almost any measurement there is a least a little bit of nonrandom error, you won't get to tiny enough to call it a point.

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