Choosing a Pivot From a Sampling Distribution

A sampling distribution can provide more than a description of how a statistic varies from sample to sample. If the statistic can be standardized so that its distribution no longer depends on the unknown parameter, the result is a pivot. Inverting probability statements about that pivot gives confidence limits. The central choices are which sampling distribution applies and how the pivot changes as the parameter changes.

This tutorial focuses on selecting and rearranging pivots from known sampling distributions. The calculations are exact when the stated sampling model and distributional result hold; a pivot based on an approximation instead gives approximate coverage. The method is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: A pivotal quantity is a function of the sample and the unknown parameter whose probability distribution does not depend on any unknown parameter. To use it for interval estimation, choose pivot quantiles with known probabilities, write the resulting probability statement, and solve it for the parameter.

From a Sampling Distribution to Limits

Write $q_p$ for the lower-tail $p$-quantile of a continuous pivot $Q$, so $P(Q\le q_p)=p$. For a central confidence level $1-\alpha$, the equal-tailed quantiles satisfy $P(q_{\alpha/2}\le Q\le q_{1-\alpha/2})=1-\alpha$. The quantiles are numerical constants if the distribution of $Q$ is parameter-free.

The event between the quantiles is not yet an interval for the unknown parameter. It must be rearranged. A transformation that increases with the parameter preserves the order of the limits; a decreasing transformation reverses it. The endpoints must be checked after solving, rather than copied in the order in which the quantiles first appeared.

P\!\left(q_{\alpha/2}\le Q(X,\theta)\le q_{1-\alpha/2}\right)=1-\alpha \quad\Longrightarrow\quad P\!\left(L(X)\le \theta\le U(X)\right)=1-\alpha.

Here $X$ denotes the observed sample and $\theta$ the parameter of interest. The expression on the right is a repeated-sampling coverage statement: before observing the sample, the random limits cover the fixed parameter with probability $1-\alpha$, under the model. After observing the sample, the endpoints are numbers.

State the model.
Record the distributional assumptions, the parameter of interest, and any quantities treated as known.

Find a parameter-free statistic.
Use a sampling distribution that is known under the stated assumptions, then standardize or scale the statistic to remove the unknown parameter.

Choose pivot quantiles.
For an equal-tailed interval, use the lower-tail quantiles at $\alpha/2$ and $1-\alpha/2$.

Invert and check.
Solve the probability inequality for the parameter, respecting monotonicity and the parameter space. Confirm the lower endpoint does not exceed the upper endpoint.

Worked Applications

Worked Example: A Normal Mean With Known Standard Deviation

Method: standardize the sample mean using its normal sampling distribution. A hypothetical portfolio analyst models annual per-policy claim-processing cost as independent observations from a normal distribution with unknown mean $\mu$ dollars and known standard deviation $\sigma=4$ dollars. A random sample of $n=16$ policies has $\bar{x}=\$12.40$. Find an equal-tailed 90% confidence interval for $\mu$. The normal model and known standard deviation are assumptions for this example.

For independent normal observations, the sample mean has distribution $N(\mu,\sigma^2/n)$. Thus $Z=(\bar{X}-\mu)/(\sigma/\sqrt{n})$ has the standard normal distribution, which does not depend on $\mu$. The 90% equal-tailed standard normal quantiles are $-1.6449$ and $1.6449$.

Since $\sigma/\sqrt{n}=4/\sqrt{16}=1$, the pivot event is $-1.6449\le(\bar{X}-\mu)/1\le1.6449$. Rearranging the left inequality gives $\mu\le\bar{X}+1.6449$; rearranging the right gives $\mu\ge\bar{X}-1.6449$. Therefore the parameter limits are $\bar{X}-1.6449$ and $\bar{X}+1.6449$.

12.40-1.6449=10.7551,\qquad 12.40+1.6449=14.0449.

The interval is $[\$10.7551,\$14.0449]$, with endpoints rounded to four decimal places. The order is correct, and the interval is centered at the observed sample mean. Common mistake avoided: using $\sigma$ instead of the standard error $\sigma/\sqrt{n}$ would give an unnecessarily wide interval. Here the standard error is $4/4=1$, not 4.

Worked Example: A Normal Variance From a Chi-Square Pivot

Method: use the chi-square distribution and invert a decreasing function of the variance. Suppose six independent observations come from a normal population with known mean $\mu_0=0$ and unknown variance $\theta=\sigma^2>0$. The observed sum of squared deviations from the known mean is $T=\sum_{i=1}^6(X_i-0)^2=25$. Find a 90% equal-tailed confidence interval for $\theta$. The observations are assumed to be normal and independent.

Under these assumptions, $T/\theta$ has a chi-square distribution with 6 degrees of freedom. Let $q_{.05}=1.6354$ and $q_{.95}=12.5916$ be the lower-tail quantiles of that distribution. The pivot statement is $P(q_{.05}\le T/\theta\le q_{.95})=0.90$.

Because $T=25>0$ and $\theta>0$, multiplying by $\theta$ gives $q_{.05}\theta\le T\le q_{.95}\theta$. Solving the first inequality gives $\theta\le T/q_{.05}$; solving the second gives $\theta\ge T/q_{.95}$. The function $T/\theta$ decreases as $\theta$ increases, which is why the smaller chi-square quantile produces the upper variance limit.

\frac{25}{12.5916}\le\theta\le\frac{25}{1.6354}, \qquad 1.985\le\theta\le15.287.

The interval is approximately $[1.985,15.287]$, with endpoints rounded to three decimal places. The variance parameter is positive, and the lower endpoint is below the upper endpoint. Common mistake avoided: pairing the lower pivot quantile with the lower variance limit. Since $T/\theta$ is decreasing in $\theta$, the quantile order reverses when solving for the variance.

Worked Example: An Exponential Rate From the Sample Sum

Method: scale a gamma-distributed sum into a chi-square pivot, then divide by a positive observed statistic. In a hypothetical study, two independent claim waiting times $X_1$ and $X_2$, measured in years, are modeled as exponential with unknown rate $\lambda>0$ per year. The observed waiting times are 1 year and 2 years, so $S=X_1+X_2=3$ years. Find a 90% equal-tailed confidence interval for $\lambda$. The exponential model and independence are assumptions.

If $X_i$ is exponential with rate $\lambda$, then $2\lambda S$ has a chi-square distribution with $2n=4$ degrees of freedom. This follows because $\lambda S$ has a gamma distribution with shape 2 and scale 1, while twice a gamma variable with integer shape 2 has a chi-square distribution with 4 degrees of freedom. Thus the pivot is parameter-free.

The lower-tail chi-square quantiles for 4 degrees of freedom are $q_{.05}=0.7107$ and $q_{.95}=9.4877$. Since $2S=6>0$, dividing the pivot inequality by 6 preserves its direction:

0.7107\le 2\lambda(3)\le9.4877 \quad\Longrightarrow\quad \frac{0.7107}{6}\le\lambda\le\frac{9.4877}{6}.

The endpoints are approximately $0.1185$ and $1.5813$, so the interval is $[0.1185,1.5813]$ per year, rounded to four decimal places. Both endpoints are positive, as required for a rate. Common mistake avoided: dividing by $S=3$ rather than by $2S=6$; the pivot is $2\lambda S$, not $\lambda S$.

Why the Pivot Choice Matters

A statistic is not a useful pivot merely because its formula contains the parameter. Its distribution must be known and free of unknown parameters after the proposed standardization. For example, the raw sample mean has a distribution centered at $\mu$, so it is not parameter-free; subtracting $\mu$ and dividing by the known standard error produces a standard normal pivot. Likewise, in the normal variance example, the chi-square distribution applies because the observations are normal and the mean is known. Changing those assumptions can change the appropriate sampling distribution.

Quantile conventions matter as well. A lower-tail $p$-quantile satisfies $P(Q\le q_p)=p$. If a table instead labels an upper-tail probability, its entry must be interpreted accordingly before it is used. Confusing these conventions can swap endpoints or produce a confidence level different from the one intended.

Key takeaway: Start with a sampling distribution justified by the model, remove the unknown parameter to form a pivot, and then solve the pivot-quantile inequality. Check monotonicity, endpoint order, units, and the allowed parameter space before reporting the interval.

Check Your Understanding

For each problem, identify the pivot and use it to obtain the requested limits. Quantiles supplied in a question are lower-tail quantiles.

Independent normal observations have unknown mean $\mu$ and known standard deviation $\sigma=5$. A sample of $n=25$ has $\bar{x}=8.60$. Use $z_{.975}=1.96$ to find a 95% equal-tailed interval for $\mu$. (Answer: $[6.64,10.56]$.)
Four independent observations are normal with known mean 0 and unknown variance $\theta$. Their sum of squared deviations from 0 is 9. For 80% confidence, use the chi-square quantiles $q_{.10}=1.0636$ and $q_{.90}=7.7794$ with 4 degrees of freedom. (Answer: approximately $[1.156,8.462]$.)
Three independent waiting times follow an exponential distribution with unknown rate $\lambda$ per year. Their observed sum is 4 years. For a 90% interval, use the chi-square quantiles $q_{.05}=1.6354$ and $q_{.95}=12.5916$ with 6 degrees of freedom. (Answer: approximately $[0.2044,1.5740]$ per year.)
For a normal sample with known standard deviation, explain why $(\bar{X}-\mu)/(\sigma/\sqrt{n})$ is a more useful pivot for $\mu$ than $\bar{X}$ alone, and state what assumption about $\sigma$ is needed for its standard normal distribution. (Answer: Standardizing removes the unknown mean and gives a parameter-free standard normal distribution; $\sigma$ must be known, and the observations must be independent normal observations.)

Choosing a Pivot From a Sampling Distribution

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