If $X$ is a random variable from the uniform distribution on the interval $(\theta, 2\theta)$, how would one construct a 90% confidence interval for the method of moments estimator of $\theta$?

The answer should be in the form: $$ \frac{\bar{x}}{a±\frac{b}{\sqrt{c*n}}} $$

Where $a,b,c$ are constants and $n$ is the number of samples of X (assume n is large).

I've determined the MOM estimator to be $\hat{\theta} = \frac{2}{3}\bar{x}$, but can't figure out how to get a confidence interval in that above form.

  • 1
    $\begingroup$ Use the central limit theorem. Under the CLT, the sample mean is approximately normal with mean equal to 1.5 * theta and variance equal to theta^2/(12n). Therefore, your MOM estimator is approximately normally distributed with mean equal to 2/3 * 3/2 * theta = theta, and variance equal to 4/9 * theta^2/(12n) = theta^2/(27n). Now you can construct your approximate 90% CI using the nornal distribution, and rearrange it into the format you need for a, b, and c. $\endgroup$
    – aranglol
    Dec 4, 2022 at 20:42
  • $\begingroup$ @aranglol Okay so I'd end up with something along the lines of theta ± 1.645 * theta/(sqrt(27n)). How does the n end up in the denominator of a fraction in the denominator? $\endgroup$
    – Mike K
    Dec 4, 2022 at 21:05

1 Answer 1


Confidence intervals are typically formed by finding a pivotal quantity that involves the parameter of interest, writing a probability statement for that pivotal quantity falling within an interval, and then "inverting" that probability statement to get it in terms of an interval for the parameter. As a first step, you should be trying to come up with a quantity $f(\hat{\theta}, \theta)$ that has a fixed distribution that does not depend on the parameter:

$$f(\hat{\theta}, \theta) \overset{\text{Approx}}{\sim} \text{Fixed Dist}.$$

Have a think about how the central limit theorem might be applied here to get an approximate distribution for an appropriately standardised version of $\bar{x}$. This ought to lead you to a pivotal quantity, which you can then use to write a probability statement for your confidence interval. (You can find some examples of the derivation of confidence intervals in this related answer).

Updated based on your proposed solution: Using the CLT, for large $n$ we have:

$$\hat{\theta} = \frac{2}{3} \bar{X}_n \overset{\text{Approx}}{\sim} \text{N} \bigg( \theta, \frac{12 \theta^2}{n} \bigg).$$

You can use this to form a quasi-pivotal quantity with a standard normal distribution and then form a confidence interval using that quantity. That should give you a confidence interval of the stipulated form, but it's not the best way to do this. Instead, I would recommend building the confidence interval with the MOM estimator using the quasi-pivotal quantity:

$$n \cdot \frac{(\hat{\theta} - \theta)^2}{12 \theta} \overset{\text{Approx}}{\sim} \text{ChiSq}(\text{df} = 1).$$

You can now derive the confidence interval in a manner analogous to the Wilson score interval. To facilitate this analysis, let $\chi_{1,\alpha}^2$ denote the critical point of the chi-squared-one distribution with upper tail area $\alpha$. For any confidence level $1-\alpha$ we can write the probability interval:

$$\begin{align} 1-\alpha &\approx \mathbb{P} \bigg( n \cdot \frac{(\hat{\theta} - \theta)^2}{12 \theta} \leqslant \chi_{1, \alpha}^2 \bigg) \\[6pt] &= \mathbb{P} \bigg( n (\hat{\theta} - \theta)^2 \leqslant 12 \chi_{1, \alpha}^2 \theta \bigg) \\[6pt] &= \mathbb{P} \bigg( n \Big[ \theta^2 - 2 \Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big) \theta + \hat{\theta}^2 \Big] \leqslant 0 \bigg) \\[6pt] &= \mathbb{P} \bigg( \Big( \theta - \hat{\theta} - 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 \leqslant \Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n} \bigg) \\[6pt] &= \mathbb{P} \bigg( \Bigg| \theta - \hat{\theta} - 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Bigg| \leqslant \sqrt{\Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n}} \bigg) \\[6pt] &= \mathbb{P} \bigg( \theta \in \bigg[ \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \pm \sqrt{\Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n}} \bigg] \bigg). \\[6pt] \end{align}$$

Upon substitution of the actual data into the estimator, you get the confidence interval:

$$\text{CI}_\theta(1-\alpha) \equiv \bigg[ \hat{\theta}_n + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \pm \sqrt{\Big( \hat{\theta}_n + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}_n^2}{n}} \bigg].$$

This is probably about as good as you're going to get based on the MOM estimator. However, it's worth noting that this is a pretty crappy confidence interval estimator in this case. In cases where the parameter of interest determines the bounds of the support, it is almost always better to use an estimator based on the extreme order statistics rather than the mean.

  • $\begingroup$ Thanks. So if my intuition is correct, assuming theta hat is unbiased, it should follow N(theta, theta/sqrt(27n)), standardizing this, it follows that (theta_hat - theta)/(theta/sqrt(27n)) ~ N(0,1) right? Assuming n is large, the z-value is 1.645, I got a confidence interval of: (2/3)xbar ± 1.645(xbar*2/3)/sqrt(27n). How does this go in the format mentioned in the question? Doesn't n have to be in the numerator (in order to go in the denominator of the denominator)? Have I done something wrong? $\endgroup$
    – Mike K
    Dec 5, 2022 at 2:55
  • $\begingroup$ Rather than trying to give a solution in comments, please update your question to include your proposed solution (and then ping me to update my answer). $\endgroup$
    – Ben
    Dec 5, 2022 at 4:41

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