Let $X_1,\ldots,X_n$ be a random sample from $f(x\mid\theta)=\theta x^{\theta -1}$ for $0<x<1$. Find a confidence interval for $\theta$ using as pivotal quantity a function of the maximum likelihood estimator for $\theta$.

Well, using the logaritmic version of the likelihood function, I got that the MLE of $\theta$ is $T_1=\frac{-n}{\sum\limits_{i=1}^n \log x_i}$. Is this correct? And how could I use a function of it as a pivotal quantity?

I know that a pivotal quantity is a function of the sample and the parameter, whose distribution doesn't depend on $\theta$.

  • $\begingroup$ A relevant question is, what is the pivotal quantity going to help you find? The mean, the variance, or something else? Do you want a confidence interval for $\theta$ itself? $\endgroup$ – OFRBG Oct 26 '15 at 5:52
  • $\begingroup$ I need a confidence interval for $\theta$ itself $\endgroup$ – Fawcett512 Oct 26 '15 at 6:34
  • $\begingroup$ Consider the pivot $-2\theta \sum\limits_{i=1}^n \ln X_i\sim \chi^2_{2n}$, from which it is straightforward to get a confidence interval. $\endgroup$ – StubbornAtom Nov 21 '19 at 19:34

Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that

$$ \hat{\theta}=-\frac{n}{\sum ln\ x_i}. $$

For the expected value we have the integral

$$ E[X] = \int_0^1 \theta x^{(\theta-1)}x\ \text{d}\theta = \frac{\theta}{1+\theta} $$

And for the second moment

$$ \mu_2 = \int_0^1 \theta x^{\theta-1}x^2\ \text{d}\theta = \frac{\theta}{2+\theta} $$

So the variance is

$$ Var(X) = \frac{\theta}{2+\theta} - \left( \frac{\theta}{1+\theta} \right)^2 = \\ = \frac{\theta}{(\theta + 1)^2(\theta+2)}. $$

Now then, we know that the following quotient has a chi-squared distribution, so in this case

$$ C = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \text{ and } C = \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} $$

where $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. Finally we can work with the random variable $C$ distributed as a chi squared. We want a confidence interval of $(1-\alpha)$. Then

$$ P \left( \chi^2_{n-1}(\alpha/2) < C < \chi^2_{n-1}(1-\alpha/2)\right) = 1-\alpha $$

denoting that $\chi^2_{n-1}(\beta)$ is the $\beta$ percentile of the $\chi^2$ which may be found in reference tables.

From here we work with the interval only substituting the percentiles with $\frac{1}{D_l}$ and $\frac{1}{D_s}$, lower and superior.

$$ \frac{1}{D_l} < C < \frac{1}{D_s} \implies \frac{1}{D_l} < \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} < \frac{1}{D_s} \\\ \\\ \\\ D_l > \frac{\frac{\theta}{(\theta + 1)^2(\theta+2)}}{(n-1)S^2} > D_s \\\ \\\ \\\ D_l((n-1)S^2) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > D_s((n-1)S^2) \\\ \\\ \\\ (D_l((n-1)S^2)) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > (D_s((n-1)S^2)) \\\ \\\ \\\ (\hat{\theta} + 1)^2(\hat{\theta}+2)\left( D_l((n-1)S^2) \right) > \theta > (\hat{\theta}+ 1)^2(\hat{\theta}+2)\left( D_s((n-1)S^2) \right) \\\ \\\ \\\ $$

where, taking the before definition and derivation, $\hat{\theta} = -\frac{n}{\sum ln\ x_i}$ and $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!

Hope it helps!

Note: made some edits due to an error :/

  • $\begingroup$ Where does $\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$ come into this? This is a result for normal distribution. $\endgroup$ – StubbornAtom Nov 20 '19 at 20:40

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