What is a pivotal statistic?

Question

I'm currently reading "Computer Age Statistical Inference" by Efron and Hastie.

In section 2.1, they talk about some of the mechanisms that frequentist inference uses to circumvent the defect of having to calculate properties of estimators $\hat{\Theta} = t(\mathbf{X})$ of an unknown distribution $F$.

One of these mechanisms, they say, is the use of pivotal statistics, which they define as not depending upon the underlying probability of $F$. They give an example of a two-sample $t$ statistic as one that does not depend on the underlying distribution.

More specifically, they say that:

1. given two normally distributed i.i.d. samples $\mathbf{x_1} = (x_{1,1}, \ldots x_{1,n})$ and $\mathbf{x_2} = (x_{2,1}, \ldots x_{2,n})$.

2. given the null hypothesis $H_0: \mu_1 = \mu_2$

3. a test statistic $\hat{\theta} = \bar{\mathbf{x_2}} - \bar{\mathbf{x_1}}$ would have distribution $\sim N\Big(0,\sigma^2(\frac{1}{n_1} + \frac{1}{n_2})\Big)$ under the null hypothesis. In this case $\sigma$ would be non-pivotal

4. Then he proposes that the way we do $t = \frac{\bar{\mathbf{x_2}} - \bar{\mathbf{x_1}}}{\hat{\sigma}(\frac{1}{n_1} + \frac{1}{n_2})^{1/2}}$, is pivotal.

I really didn't understand the difference between these two, and thus, as I didn't really understand how $t$ doesn't rely on the sample.

Answer 1

The population standard deviation $\sigma$ depends on the (unknown) distribution, $E$. The sample standard deviation $\hat{\sigma}$ depends only on the (known) data, $\boldsymbol{x}$.

Because it is a consistent estimator, $\hat{\sigma}\to\sigma$ as sample size goes to infinity. But with a finite sample, only $\hat{\sigma}$ is knowable.

Answer 2

To start off, there is a typo in statement 3: the population variance $\sigma^2$ is a parameter (unknown but fixed value), not a random variable, so it cannot be a pivot. It's the statistic $\widehat{\theta}$ which is not pivotal.

So the question boils down to comparing two statistics for the difference in means $\mu_2 - \mu_1$:

$$ \begin{aligned} \widehat{\theta} &= \overline{\mathbf{x}}_2 - \overline{\mathbf{x}}_1 \\ T &= \frac{\overline{\mathbf{x}}_2 - \overline{\mathbf{x}}_1}{\widehat{\sigma}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)^{1/2}} \end{aligned} $$

Since $\mathbf{x}_1$ is iid $N(\mu_1,\sigma^2)$ and $\mathbf{x}_2$ is iid $N(\mu_2,\sigma^2)$, under the null hypothesis $H_0:\mu_1 = \mu_2$:

$$ \begin{aligned} \widehat{\theta} &\sim N\left(0,\sigma^2\left(\frac{1}{n_1} + \frac{1}{n_2}\right)\right) \\ T &\sim t_{n_1+n_2-2} \end{aligned} $$

$\widehat{\theta}$ is not pivotal because its distribution depends on an unknown parameter; $T$ is pivotal because its distribution depends only on the sample sizes, $n_1$ and $n_2$, which are known.

Here we are interested in comparing the two means, $\mu_1$ and $\mu_2$, not in estimating the variance: $\sigma^2$ is a nuisance parameter. So it's convenient that the distribution of $T$ doesn't depend on the variance. There is no need to plug in an estimator $\widehat{\sigma}$ for $\sigma$ or use any of the other frequentist "devices" described in Section 2.1, in order to test the null hypothesis $\mu_1=\mu_2$. And our analysis (confidence intervals & p-values) will be equally valid for all values of $\sigma^2$.