1
$\begingroup$

EDIT

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 $\theta$.

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 $\mathcal{N}(\theta,(nI_{F}(\theta))^{-1})$. 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 \begin{align*} \textbf{P}\left(c_{1} < \frac{1}{\theta}\sum_{i=1}^{n} X_{i} < c_{2}\right) = 1 -\alpha \end{align*}

Can someone help me out? Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess. $\endgroup$ – Glen_b Apr 15 at 2:03
  • $\begingroup$ Thanks for the comment and sorry for the inconvenience. I edited the question. $\endgroup$ – user1337 Apr 15 at 2:34
  • 1
    $\begingroup$ Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong. $\endgroup$ – Glen_b Apr 15 at 2:40
  • $\begingroup$ For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$ $\endgroup$ – BruceET Apr 15 at 2:49
  • $\begingroup$ There are so many options here because there are different choices of pivots. A C.I. could also be found using $\min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $\sum X_i$. $\endgroup$ – StubbornAtom Apr 15 at 6:18
2
$\begingroup$

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
$\endgroup$
  • $\begingroup$ In the first place, thanks for the answer. Could you please provide the demonstration that the given pivotal quantity has gamma distribution? $\endgroup$ – user1337 Apr 25 at 22:38
1
$\begingroup$

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 $$\frac 1 \mu \sum_{i=1}^n X_i \sim \mathsf{Gamma}(\text{shape} = n, \text{rate=scale} = 1).$$

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 $\mathsf{Exp}(\text{rate} = \lambda = 1/5),$ 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 $Q \sim \mathsf{Gamma}(10, 1).$ (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)

enter image description here

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 $Q \sim \mathsf{Gamma}(10, 1).$

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 $\mu.)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.