Solved – How do we build a confidence interval for the parameter of the exponential distribution


Let $X_{1},X_{2},ldots,X_{n}$ be a random sample whose distribution is given by $text{Exp}(theta)$, where $theta$ is not known. Precisely, $f(x|theta) = (1/theta)exp(-x/theta)$ Describe a method to build a confidence interval with confidence coefficient $1 – alpha$ for $theta$.


Since the distribution in discussion is not normal and I do not know the size of the sample, I think we cannot apply the central limit theorem. One possible approach is to consider the maximum likelihood estimator of $theta$, whose distribution is approximately $mathcal{N}(theta,(nI_{F}(theta))^{-1})$. Another possible approach consists in using the score function, whose distribution is approximately $mathcal{N}(0,nI_{F}(theta))$. However, in both cases, it is assumed the CLT is applicable.

The exercise also provides the following hint: find $c_{1}$ and $c_{2}$ such that
textbf{P}left(c_{1} < frac{1}{theta}sum_{i=1}^{n} X_{i} < c_{2}right) = 1 -alpha

Can someone help me out? Thanks in advance!

Taking $theta$ as the scale parameter, it can be shown that ${n bar{X}}/{theta} sim text{Ga}(n,1)$. To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:

$$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha quad quad quad quad quad int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 – alpha.$$

Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:

$$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$

This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.

Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.

#Set the objective function for minimisation OBJECTIVE <- function(c1, n, alpha) {     pp <- pgamma(c1, n, 1, lower.tail = TRUE);     c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);     1/c1 - 1/c2; }  #Find the minimum-length confidence interval CONF_INT <- function(n, alpha, xbar) {     START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);     MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);     c1 <- MINIMISE$estimate;     pp <- pgamma(c1, n, 1, lower.tail = TRUE);     c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);     c(n*xbar/c2, n*xbar/c1); }  #Generate simulation data set.seed(921730198); n     <- 300; scale <- 25.4; DATA  <- rexp(n, rate = 1/scale);  #Application of confidence interval to simulated data n     <- length(DATA); xbar  <- mean(DATA); alpha <- 0.05;  CONF_INT(n, alpha, xbar);  [1]  23.32040 29.24858 

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