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EDIT

Let $X_1,X_2,ldots,X_n$ be a random sample whose distribution is given by $textExp(theta)$, 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 $theta$.

MY ATTEMPT

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 $mathcalN(theta,(nI_F(theta))^-1)$. Another possible approach consists in using the score function, whose distribution is approximately $mathcalN(0,nI_F(theta))$. 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
beginalign*
textbfPleft(c_1 < frac1thetasum_i=1^n X_i < c_2right) = 1 -alpha
endalign*

Can someone help me out? Thanks in advance!

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

$$mathbbP Bigg( c_1 leqslant fracn barXtheta leqslant c_2 Bigg) = 1-alpha
quad quad quad quad quad
int limits_c_1^c_2 textGa(r|n,1) dr = 1 - alpha.$$

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

$$textCI_theta(1-alpha) = Bigg[ fracn barxc_2 , fracn barxc_1 Bigg].$$

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

You don't say how the exponential distribution is
parameterized. Two parameterizations are in common use--mean and rate.

Let $E(X_i) = mu.$ Then one
can show that $$frac 1 mu sum_i=1^n X_i sim
mathsfGamma(textshape = n, textrate=scale = 1).$$

In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsfExp(textrate = lambda = 1/5),$ finds $$Q = frac 1 mu sum_i=1^n X_i =
lambda sum_i=1^n X_i$$ for each sample, and plots the histogram of the one million $Q$'s, The figure
illustrates that $Q sim mathsfGamma(10, 1).$
(Use MGFs for a formal proof.)

set.seed(414) # for reproducibility
q = replicate(10^5, sum(rexp(10, 1/5))/5)
lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
hist(q, prob=T, br=30, col="skyblue2", main=lbl)
 curve(dgamma(x,10,1), col="red", add=T)

Thus, for $n = 10$ the constants $c_1 = 4.975$ and
$c_2 = 17.084$ for
a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsfGamma(10, 1).$

qgamma(c(.025, .975), 10, 1)
[1] 4.795389 17.084803

In particular, for the exponential sample shown below (second row),
a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
has $mu$ in the denominator.

set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
[1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
t = sum(x); t
[1] 37.99
t/qgamma(c(.975, .025), 10, 1)
[1] 2.223614 7.922194

Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.

See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$