Pivotal Method

A very useful method for finding confidence intervals uses a pivotal quantity that has 2 characteristics.

1. It is a function of the sample measurements and the unknown parameter $\theta$ (where $\theta$ is the only unknown).
2. It has a probability distribution which does not depend on the parameter $\theta$.

Suppose that T$=$t(X) is a reasonable point estimate of $\theta$, then we will denote this pivotal quantity by $p(T,\,\theta)$, and we will use the known form of the probability distribution of $p(T,\,\theta)$ to make the following statement.

For a specified constant $\gamma$, ( $0 < \gamma < 1$), and constants $a$ and $b$, ($a < b$),

$$P(a < p(T, \theta) < b)=\gamma.$$ (8.16)

So, given $T$, the inequality (8.16) is solved for $\theta$ to obtain a region of $\theta$-values which is a confidence region (usually an interval) for $\theta$ corresponding to the observed T-value. This rearrangement, of course, results in an equation of the form (8.15).

Example 8..9
For random variable $X \sim U(0, \theta)$, construct a 90% confidence interval for $\theta$.

Now we know that $Y_n$, the largest order statistic from a sample of size $n$ from this distribution, is sufficient for $\theta$ and has pdf

$$f_{Y_n}(y)=n\,y^{n-1}/\theta^n, \, 0 \leq y \leq \theta.$$

Let $Z=Y_n/\theta$, then the pdf of $Z$ is

$$f_Z(z)=n\,z^{n-1}, \, 0 \leq z \leq 1.$$

We see that $Y_n/\theta$ is a suitable pivotal quantity with the 2 characteristic properties referred to earlier. So we have

$$P(a < Y_n/\theta < b)=.90.$$

Noting that the cdf of Z is $F_Z(z)=z^n,\,0 \leq z \leq 1$, values of $a$ and $b$ may be found as follows.

$$F_Z(a)=.05, \mbox{ and } F_Z(b)=.95$$

$$a^n=.05 \mbox{ and } b^n=.95, \mbox{ giving } a=\sqrt[n]{.05} \mbox{ and } b=\sqrt[n]{.95}.$$

So we may write

$$P(\sqrt[n]{.05} < Y_n/\theta < \sqrt[n]{.95})=.90.$$

Rearranging, the confidence interval for $\theta$ is

$$(Y_n/\sqrt[n]{.95}, \, Y_n/\sqrt[n]{.05}).$$

Comment. Note that there is some arbitrariness in the choice of a confidence interval in a given problem. There are usually several statistics $T=t(X_1, \ldots,X_n)$ that could be used, and it is not really necessary to allocate equal probability to the two tails of the distribution, as was done in the above example. However, it is customary to do this, as this often leads to the shortest confidence interval (for the same confidence coefficient), another property considered desirable.