Wikipedia provides the following definition for a confidence interval for a parameter $\theta$:

A confidence interval for the parameter θ, with confidence level or confidence coefficient γ, is an interval with random endpoints (u(X), v(X)), determined by the pair of random variables u(X) and v(X), with the property:

${\Pr}_{\theta,\varphi}(u(X)<\theta<v(X))=\gamma\text{ for all }(\theta,\varphi). $

Here $Pr(θ,φ)$ indicates the probability distribution of X characterised by $(θ, φ)$.

In a specific situation, when x is the outcome of the sample X, the interval $(u(x), v(x))$ is also referred to as a confidence interval for $θ$. Note that it is no longer possible to say that the (observed) interval $(u(x), v(x))$ has probability γ to contain the parameter $θ$. This observed interval is just one realization of all possible intervals for which the probability statement holds.

where, as usual, $X$ here is the random variable representing the sample and $u(x)$ and $v(X)$ refer technically, not just random variables, but to methods for constructing the upper and lower bounds of the confidence interval from one sample.

While I have always known this, I fail to see how exactly one reaches following statement:

... Note that it is no longer possible to say that the (observed) interval $(u(x), v(x))$ has probability $\gamma$ to contain the parameter $θ$.

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    $\begingroup$ There is nothing more here than a [perfect] analog to the conundrum "there's a $1/2$ chance that a fair coin will land heads up, but after you observe that it has landed heads up, it is no longer possible to say that the chance of heads [on this toss] is $1/2$." $\endgroup$
    – whuber
    Jul 9, 2014 at 14:48
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    $\begingroup$ @whuber I sort of follow your example, but technically, as far as I understand it, the definition makes statements about a parameter (the fairness of the coin) that is assumed invariant across tosses, and not about the chance of getting a result on a toss, so when you say "it's no longer possible to say the chance of heads [on this toss] is 1/2", shouldn't that be rephrased as "it's no longer possible to say <something> about the fairness of the coin"? $\endgroup$ Jul 9, 2014 at 15:16
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    $\begingroup$ No--the two statements are entirely different. Statistical estimation concerns the ways to make statements about the fairness of the coin based on observations that have been made: that's the spirit of the rephrased statement. The original phrasing, though, concerns one particular toss of the coin, not the coin itself. $\endgroup$
    – whuber
    Jul 9, 2014 at 15:30

1 Answer 1


As indicated by Wikipedia's reference to Bayesian intervals as an alternative, confidence intervals are really based in a frequentist interpretation of probability.

You are correct that u(X) and v(X) are methods for constructing bounds.

The first definition says that if those methods are applied in very many cases, then in a proportion γ of cases the interval will contain the parameter θ.

The second definition refers to a single application of the methods, the interval either does or does not contain the parameter θ; under the frequentist paradigm it doesn't make sense to talk of the probability of a single completed event.

  • $\begingroup$ Thanks, when you say "it doesn't make sense to talk of the probability of a single completed event." you mean, perhaps, the probability computed from a single completed event, correct? (see my comment on the OP) $\endgroup$ Jul 9, 2014 at 15:26
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    $\begingroup$ "the interval either does or does not contain the parameter θ", but we do not know. The fact the event is completed does not matter, since we do not know the outcome. To me the fact an event is completed make it meanningless to talk of probabilities because the fact it is completed brings a certitude (the probability the event occured was 1 since it occured). But when we do not have the knowledge, we do not have this certitude so to me this make sense to keep talking of probabilities. $\endgroup$
    – Aurelie
    Jul 9, 2014 at 15:26
  • $\begingroup$ @Aurelie: in which case the probability in question is the (unconditional) probability of discovering outcome X. I say unconditional because this only works for models in which there is no material distinction between what state a variable is actually in (for anyone to potentially discover and use) and when we in particular might come to know about it. $\endgroup$ Jul 9, 2014 at 15:32
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    $\begingroup$ Because conversations about confidence intervals can become confusing, it is important to use notation correctly. In this case, $X$ is a random variable; $u$ and $v$ (not $u(X)$ and $v(X)$) are statistical procedures; $u(X)$ is the application of $u$ to the random variable $X$--which is therefore another random variable (as stated in the Wikipedia quotation); $x$ is an outcome or realization of $X$, and $u(x)$ is the application of $u$ to $x$--which is therefore a number. $\theta$, as suggested by your reference to "frequentist paradigm," is simply a number. $\endgroup$
    – whuber
    Jul 9, 2014 at 15:33
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    $\begingroup$ @whuber: I absolutely agree. I didn't see your comment on the original question until after I'd posted my answer. Your comment said the same as my answer tried to but much more succinctly. Aurelie: In relating probability to the state of knowledge I think you're moving to a non-frequentist paradigm. $\endgroup$
    – user20637
    Jul 9, 2014 at 15:40

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