Confidence Intervals - distribution of a Pivot? I am working on some self study and have come across "The Pivot Method". It seems quiet abstract and I can't understand what the usefulness of the following statement is regarding pivots: The distribution of a pivot does not depend on the parameter of interest?
What exactly is a pivot and what is meant by the statement "the distribution of a pivot does not depend on the parameter of interest"?
 A: A pivotal quantity is function of the sample observations and a parameter whose distribution is completely specified (does not contain any unknown parameters). 
(1) For example, if $X_{1},X_{2},\cdots,X_{n}$ is a random sample from $N(\mu,5)$ and let $\bar{X}$ be the sample mean. Then the distribution of $\bar{X}-\mu \sim N(0,5)$, is completely specified, as there are no unknown parameters. The quantity $\bar{X}-\mu$ is a pivotal quantity. 
(2) As another simple example, let $X_{1},X_{2},\cdots,X_{n}$ is a random sample from $N(\mu,\sigma^2)$ and let $\bar{X}$ be the sample mean. Then the distribution of $\frac{ \bar{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1)$. Here $\frac{ \bar{X}-\mu}{\sigma/\sqrt{n}}$ is a pivot. 
(3) If $\bar{X}$ is the sample mean of a random sample of size $n$ obtained from normal population $N(\mu,\sigma^2)$, with $\sigma^{2}$ unknown,  the $t$-statistic defined by $\dfrac{\bar{X}-\mu}{S/\sqrt{n}}$ is a pivot because the $t$-distribution does not depend on the parameters $\mu$ and $\sigma$.
