In Section 9.2.3 of Casella's Statistical Inference, they base their confidence interval construction for a parameter $\theta$ on a real-valued statistic $T$ with cdf $F_T(t| \theta)$.

  1. They first assume that $T$ is a continuous random variable. Then by the Probability Integral Transformation, the continuous random variable $F (T | \theta)$ is uniformly distributed over $(0, 1)$, and therefore is a pivot which can be used to construct a confidence region. enter image description here

  2. They say the situation where $T$ is discrete is similar but has a few additional technical details to consider.

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    My question is when $T$ has a discrete distribution. In such case, $F (T | \theta)$ takes countably many values and is not uniformly distributed over $(0, 1)$. Does the distribution of $F (T | \theta)$ depend on $\theta$, and therefore it is not a pivot? In such case, why can Casella still think it as a pivot, or am I missing something?

Thanks and regards!


1 Answer 1


It is not a pivot. However the resulting interval is guaranteed to have at least the stated coverage. For the discrete distribution the coverage could be a lot higher than stated.

  • 3
    $\begingroup$ I'm sure you did not mean to use the word "diapered" (look it up). What do you mean? $\endgroup$
    – whuber
    Dec 7, 2014 at 2:38

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