How to determine confidence interval using central limit theorem?

Question

How to determine the confidence interval for the unknown theta parameter of the Uniform($[0, \theta]$)-distribution using the central limit theorem, considering that the significance level is given and $\theta > 0$?

When generating the sample, theta should be considered as known and afterwards should be checked whether or not it's in the interval.

I know it can somehow be solved by generating a uniform sample, calculating the value for z and then using a polynomial with coefficients $a, b$ and $c$, then generating a $\delta$ and the roots will be the interval's endpoints, but I don't understand how I can get those coefficients. I have a sample of size $N$ and I calculated the mean and the value for z but I'm stuck at this point. How do I calculate those coefficients for the polynomial?

Answer 1

CLt is a very bad choise to calculate your Confidence interval. I do not know if this is explicitly asked or it is your own choice. Anyway, and in any case, first you have to derive your pivot quantity that is

$$Y=\frac{\text{Max}(X_i)}{\theta}$$

Knowing that, a Confidence interval can be easily calculated basing it on $Y$, amd leading to the following confidence interval:

$$\Bigg[X_{(n)};\frac{X_{(n)}}{\sqrt[n]{\alpha}}\Bigg]$$

Where $\alpha$ is the given significance level and $X_{(n)}$ is the maximum of the $X_i$

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To use CLT I think you can start with estimating $\theta$ with Method of moments and not with ML. Thus you get

$$\hat{\theta}_{MoM}=2\overline{X}_n$$

And now, assuming $n$ great enough, you can use CLT, as $\overline{X}_n$ is asimptotically Normal (but it is a very ugly road...)

Answer 2

For a random sample with large $n$ you could use the CLT to get a less-than-optimal CI for $\theta.$ Here is an outline:

• Notice that $E(\bar X_n) = \theta/2.$ So you might consider the unbiased estimator $\hat \theta = 2\bar X_n.$

• By the CLT, $\hat \theta$ is nearly normal. With what mean and standard deviation?

• Then you can use an approximate 95% CI of the form $\hat\theta \pm 1.96\, \widehat{\mathrm{SE}}(\hat\theta),$ where $\widehat{\mathrm{SE}}(\hat\theta)$ estimates $\mathrm{SD}(2\bar X_n).$

Note: The estimator of @tommik (+1) will tend to be shorter than the intervals above. For $n=20,$ 95% CIs above are about 2 units long, whereas 95% CIs based on the maximum are about 0.612 units long.

Here is a relevant simulation for the case $n = 20, \theta=4.$ (Vertical lines on plots of histograms suggest typical lengths of 95% confidence intervals for $\theta$ made from sample means and sample maximums, respretively.)

set.seed(1128)
m = 10^5;  n = 20;  th = 4
x = runif(m*n, 0, th)
MAT = matrix(x, nrow=m)    # mxn matrix, each row is sample of n
th.est = 2*rowMeans(MAT)
mean(th.est);  sd(th.est)
[1] 4.002066               # aprx E(th.est) = 4
[1] 0.5160839
hist(th.est, prob=T, col="skyblue2")
curve(dnorm(x, mean(th.est), sd(th.est)), add=T, col="red", lwd=2)
pm=c(-1,1)
abline(v = mean(th.est)+pm*1.96*sd(th.est))

mx = apply(MAT, 1, max)  # vector of maximums
mx.unb = 21*mx/20
mean(mx.unb)             # aprx E(unb MLE) = 4
[1] 4.000818     
mean(mx/.05^(1/20) - mx)
[1] 0.6156929            # aprx length of 95% CIs based on max
hist(mx.unb, prob=T, col="skyblue2", xlim=c(2,4.5))
 abline(v = c( mean(mx), mean(mx/.05^{1/20})))