Pivot

Definition taught in Confidence intervals, part of Statistical inference

Pivot

Type	Definition
Taught in	Confidence intervals
Part	Statistical inference
Properties	4
Worked examples	2
Questions using it	2

A pivot is a function h({\boldsymbol{X}};\theta) of the data and the parameter whose distribution does not depend on any unknown parameter.

Properties

One-sample t pivot

Main article: One-sample t pivot

If X_1,\ldots,X_n are independent and \operatorname{N}(\mu,\sigma^2), then

T=\frac{\overline X-\mu}{s/\sqrt n}

has a Student t distribution with n-1 degrees of freedom. A Student variable with m degrees of freedom is Z/\sqrt{V/m} where Z is standard normal, V is chi-squared with m degrees of freedom, and Z and V are independent.

Two-sample t pivot

Main article: Two-sample t pivot

For two independent normal samples with a common variance,

T=\frac{(\overline X_1-\overline X_2)-(\mu_1-\mu_2)}{s_p\sqrt{1/n_1+1/n_2}}

has a t distribution with n_1+n_2-2 degrees of freedom.

Variance-ratio F pivot

Main article: Variance-ratio F pivot

For two independent normal samples,

\frac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}

has the F distribution with (n_1-1,\,n_2-1) degrees of freedom.

Slope t pivot

Main article: Slope t pivot

Under simple linear regression,

\frac{\widehat\beta_1-\beta_1}{S_1}=\frac{\widehat\beta_1-\beta_1}{S/\sqrt{S_{xx}}}

has the t_{n-2} distribution.

Examples

Which of these are statistics?

Question page: Which of these are statistics?

Let X_1,\ldots,X_n be a random sample from \operatorname{N}(\mu,\sigma^2) with both parameters unknown. Which of the following are statistics: \overline X; \overline X-\mu; \max_i X_i-\min_i X_i; (\overline X-\mu)/(s/\sqrt n); \sum_i X_i^2? For each one that is not a statistic, say whether its distribution still depends on unknown parameters.

Worked answer (click to show)

\overline X, \max_iX_i-\min_iX_i and \sum_iX_i^2 are statistics: each can be computed from the data alone.

\overline X-\mu is not a statistic, because it contains \mu. Its distribution, \operatorname{N}(0,\sigma^2/n), still depends on the unknown \sigma^2.

(\overline X-\mu)/(s/\sqrt n) is not a statistic either, for the same reason. Its distribution, however, is t_{n-1} whatever \mu and \sigma^2 are. That combination, a function of data and parameter with a known distribution, is a pivot, and it is what makes the t interval possible.

The interval for a normal mean

Question page: The interval for a normal mean

Let X_1,\ldots,X_n be independent and \operatorname{N}(\mu,\sigma^2) with \sigma unknown. Starting from the t pivot (\overline X-\mu)/(s/\sqrt n), derive the confidence interval for \mu with confidence 1-\alpha.

Worked answer (click to show)

T\sim t_{n-1}, so \mathbb{P}(-t_{n-1;\,1-\alpha/2}\le T\le t_{n-1;\,1-\alpha/2})=1-\alpha. Substituting T=(\overline X-\mu)/(s/\sqrt n) and solving for \mu gives

\overline X\pm t_{n-1;\,1-\alpha/2}\,\frac{s}{\sqrt n}.

Used in

Taught in Confidence intervals, where the derivation and examples are.
Also used in Estimation · Hypothesis tests · Linear regression
Questions that use it (2) — Which of these are statistics? (answer) · The interval for a normal mean
All 6 questions on Confidence intervals