5.4 Asymptotic confidence intervals

We assume now that the rv X that represents the population follows a distribution that belongs to a parametric family of distributions {F(;θ):θΘ}, and that we want to obtain a confidence interval for θ. This family does not have to be normal and may be even unknown. In this situation, we may have an asymptotic pivot of the form

Z(θ)=ˆθθˆσ(ˆθ)dN(0,1).

Then, an asymptotic confidence interval for θ is given by

ACI1α(θ):=[ˆθzα/2ˆσ(ˆθ)].

This confidence interval has a confidence level approximate to 1α, that is

P(θACI1α(θ))1α.

This approximation becomes more accurate as the sample size n grows due to (5.6).

5.4.1 Asymptotic confidence interval for the mean

Let X be a rv with E[X]=μ and Var[X]=σ2, both being unknown parameters, and let (X1,,Xn) be a srs of X. In Example 3.17 we have seen that

Z(μ)=ˉXμS/ndN(0,1).

Therefore, an asymptotic confidence interval for μ at the confidence level 1α is

ACI1α(μ)=[ˉXzα/2Sn].

Example 5.8 The shopping times of n=64 random customers at a local supermarket are measured. The sample mean and the quasivariance were 33 and 256 minutes, respectively. Estimate the average shopping time, μ, of a customer with a confidence level 0.90.

The asymptotic confidence interval for μ at significance level α=0.10 is

ACI0.90(μ)=[ˉXz0.05Sn][331.64525664][29.71,36.29].

Example 5.9 Let (X1,,Xn) be a srs of a rv with distribution Pois(λ). Let us compute an asymptotic confidence interval at significance level α for λ.

In Exercise 3.11 we have seen that

ˉXλˉX/ndN(0,1).

Therefore, the asymptotic confidence interval for λ is

ACI1α(λ)=[ˉXzα/2ˉXn].

5.4.2 Asymptotic confidence interval for the proportion

Let X1,,Xn be iid rv’s with Ber(p) distribution. We are going to obtain an asymptotic confidence interval at confidence level 1α for p.

In Example 3.18 we have seen that

Z(p)=ˆppˆp(1ˆp)/ndN(0,1).

Therefore, an asymptotic confidence interval for p is

ACI1α(p)=[ˆpzα/2ˆp(1ˆp)n].

This asymptotic confidence interval can be easily extended to the two proportions case.

Example 5.10 Two brands of refrigerators, A and B, have a warranty of one year. In a random sample of nA=50 refrigerators of A, 12 failed before the end of the warranty period. From the random sample of nB=60 refrigerators of B, 12 failed before the expiration of the warranty. Estimate the difference of the failure proportions during the warranty period at the confidence level 0.98.

Let pA and pB be the proportion of failures for A and B, respectively. By the CLT we know that, for a large sample size, the sample proportions ˆpA and ˆpB verify

ˆpAN(pA,pA(1pA)nA),ˆpBN(pB,pB(1pB)nB).

Then, for large nA and nB it is verified

ˆpAˆpBN(pApB,pA(1pA)nA+pB(1pB)nB).

Applying Corollary 3.3 and Theorem 3.4, a pivot is

Z(pApB)=(ˆpAˆpB)(pApB)ˆpA(1ˆpA)nA+ˆpB(1ˆpB)nBdN(0,1)

so an asymptotic confidence interval for pApB is

ACI0.98(pApB)=[(ˆpAˆpB)z0.01ˆpA(1ˆpA)nA+ˆpB(1ˆpB)nB].

The sample proportions for the refrigerators are ˆpA=12/50=0.24 and ˆpB=12/60=0.20, so the above asymptotic confidence interval is

ACI0.98(pApB)[(0.240.20)2.330.24×0.7650+0.20×0.8060][0.145,0.225].

This confidence interval contains zero. This indicates that pA=pB with a confidence of 0.98, that is, that the probabilities of failure of the refrigerators of the two brands are not different.

5.4.3 Asymptotic maximum likelihood confidence intervals

Theorem 4.1 and Corollary 4.2 provide a readily usable asymptotic pivot for θ based on the very general maximum likelihood estimator:

Z(θ)=ˆθMLEθ1/nI(ˆθMLE)=nI(ˆθMLE)(ˆθMLEθ)dN(0,1).

The result can be precised as follows.

Theorem 5.1 (Asymptotic maximum likelihood confidence intervals) Let Xf(;θ). Under the conditions of Theorem 4.1 and Corollary 4.2, the asymptotic maximum likelihood confidence interval

ACI1α(θ)=[ˆθMLEzα/2nI(ˆθMLE)]

is such that P(θACI1α(θ))1α as n.

Remark. An analogous result can be established for a discrete rv XF(;θ).

Remark. Due to Corollary 4.3, I(ˆθMLE) can be replaced with ˆI(ˆθMLE) in (5.7) without altering the asymptotic coverage probability. This is especially useful when I(θ) is difficult to compute, since ˆI(θ)=1nni=1(logf(Xi;θ)/θ)2 is straightforward to compute from the srs (X1,,Xn) from Xf(;θ).

Example 5.11 Let us compute the asymptotic maximum likelihood confidence interval for λ in a Exp(λ) distribution.

We know from Example 4.12 that ˆλMLE=1/ˉX and I(λ)=1/λ2. Therefore, the asymptotic confidence interval is

ACI1α(λ)=[ˆλMLEzα/2ˆλMLEn].

To illustrate this confidence interval, let us generate a simulated sample of size n=100 for λ=2 and compute the asymptotic 95%-confidence interval:

# Sample from Exp(2)
set.seed(123456)
n <- 100
x <- rexp(n = n, rate = 2)

# MLE
lambda_mle <- 1 / mean(x)

# Asymptotic confidence interval
alpha <- 0.05
z <- qnorm(alpha / 2, lower.tail = FALSE)
lambda_mle + c(-1, 1) * z * lambda_mle / sqrt(n)
## [1] 1.374268 2.044294

In this case, λ=2 belongs to the confidence interval.