5.1 The pivotal quantity method

Definition 5.1 (Confidence interval) Let X be a rv with induced probability P(;θ), θΘ, where ΘR. Let (X1,,Xn) be a srs of X. Let T1=T1(X1,,Xn) and T2=T2(X1,,Xn) be two unidimensional statistics such that

P(T1θT2;θ)1α,θΘ.

Then, the interval CI1α(θ):=[T1(x1,,xn),T2(x1,,xn)] obtained for any sample realization (x1,,xn) is referred to as a confidence interval for θ at the confidence level 1α.

The value α is denoted as the significance level. T1 and T2 are known as the inferior and the superior limits of the confidence interval for θ, respectively. Sometimes the interest lies in only one of these limits.

Definition 5.2 (Pivot) A pivot Z(θ)Z(θ;X1,,Xn) is a function of the sample (X1,,Xn) and the unknown parameter θ that is bijective in θ and has a completely known distribution.

The pivotal quantity method for obtaining a confidence interval consists in, once fixed the significance level α desired to satisfy (5.1), find a pivot Z(θ) and, using the pivot’s distribution, select two constants c1 and c2 such that

PZ(c1Z(θ)c2)1α.

Then, solving60 for θ in the inequalities we obtain an equivalent probability to (5.2). If θZ(θ) is increasing, then (5.2) equals

PZ(Z1(c1)θZ1(c2))1α,

so T1=Z1(c1) and T2=Z1(c2). If Z is decreasing, then T1=Z1(c2) and T2=Z1(c1). In any case, [T1,T2] is a confidence interval for θ at confidence level 1α.

Usually, the pivot Z(θ) can be constructed from an estimator ˆθ of θ. Assume that making a transformation of the estimator ˆθ that involves θ we obtain ˆθ. If the distribution of ˆθ does not depend on θ, then we have that ˆθ is a pivot for θ. For this process to work, it is key that ˆθ has a known distribution61 that depends on θ: otherwise the constants c1 and c2 in (5.2) are not computable in practice.

The interpretation of confidence intervals has to be done with a certain care. Notice that in (5.1) the probability operator refers to the randomness of the interval [T1,T2]. This random confidence interval is said to contain the unknown parameter θ “with a probability of 1α. Yet, in reality, either θ belongs or does not belong to the interval, which seems contradictory. The previous quoted statement has to be understood in the frequentist sense of probability:62 when the confidence intervals are computed independently over an increasing number of samples,63 the relative frequency of the event “θCI1α(θ)” converges to 1α. For example, suppose you have 100 samples generated according to a certain distribution model depending on θ. If you compute CI1α(θ) for each of the samples, then in approximately 100(1α) of the samples the true parameter θ would be actually inside the random confidence interval. This is illustrated in Figure 5.2.

Illustration of the randomness of the confidence interval for \(\theta\) at the \(1-\alpha\) confidence. The plot shows 100 random confidence intervals for \(\theta,\) computed from 100 random samples generated by the same distribution model (depending on \(\theta\)).

Figure 5.2: Illustration of the randomness of the confidence interval for θ at the 1α confidence. The plot shows 100 random confidence intervals for θ, computed from 100 random samples generated by the same distribution model (depending on θ).

Example 5.1 Assume that we have a single observation X of a Exp(1/θ) rv. Employ X to construct a confidence interval for θ with a confidence level 0.90.

We have a srs of size one and we need to find a pivot for θ, that is, a function of X and θ whose distribution is completely known. The pdf and the mgf of X are given by

fX(x)=1θex/θ,x>0,mX(s)=(1θs)1.

Then, taking Z=X/θ, the mgf of Z is

mZ(s)=mX/θ(s)=mX(s/θ)=(1θsθ)1=(1s)1.

Therefore, mZ does not depend on θ and, in addition, is the mgf of a rv Exp(1) with pdf

fZ(z)=ez,z>0.

Then, we need to find two constants c1 and c2 such that

P(c1Zc2)0.90.

We know that

P(Zc1)=c10ezdz=1ec1,P(Zc2)=c2ezdz=ec2.

Splitting the probability 0.10 evenly in two,64 then

1ec1=0.05,ec2=0.05.

Solving for the c1 and c2, we obtain

c1=log(0.95)0.051,c2=log(0.05)2.996.

Therefore, it is verified

P(c1X/θc2)=0.9.

Solving θ from the inequalities, we have

P(X/2.996θX/0.051)0.9,

so the confidence interval for θ at significance level 0.10 is

CI0.90(θ)[X/2.996,X/0.051].


  1. Therefore, it is key that Z is bijective in θ to be invertible.↩︎

  2. If the distribution of ˆθ is only known asymptotically, then one can build an asymptotic confidence interval through the pivot method; see Section 5.4.↩︎

  3. See Definition 1.7.↩︎

  4. For fixed sample size! The n does not change. What is repeated is the extraction of new samples.↩︎

  5. Other splittings are possible.↩︎