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I have solved a number of exercises where I am asked to prove that a particular quantity is pivotal. The most popular example is the $Z$-score. If $Y\sim N(\mu, \sigma^2)$, then $Z=(Y-\mu)/\sigma$ is pivotal since $Z\sim N(0, 1)$, which does not depend on $\mu$ and $\sigma$. However, in these exercises, often the quantity is given and I only need to prove. I am curious about the procedure of how people found those quantities so that students only need to prove.

My question is that, in general, do we have any strategy to find a pivotal statistic?

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  • $\begingroup$ In a time series context you can look up « self-normalisation ». $\endgroup$ Commented Jan 20, 2021 at 23:07
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    $\begingroup$ I think you mean to ask about a pivotal quantity --- a pivotal quantity that is a statistic is an ancillary statistic. $\endgroup$
    – Ben
    Commented Jan 21, 2021 at 2:20

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First, a reasonably coherent definition of pivotal -quantity._ not "statistic", quoted from Wikipedia:

In statistics, a pivotal quantity or pivot is a function of observations and unobservable parameters such that the function's probability distribution does not depend on the unknown parameters (including nuisance parameters).[1] A pivot quantity need not be a statistic—the function and its value can depend on the parameters of the model, but its distribution must not. If it is a statistic, then it is known as an ancillary statistic.

Often pivotal quantities are used to find confidence intervals. It is convenient to have a quantity that involves a sufficient statistic and the parameter to be estimated, such that the quantity has a known distribution for a knows value of the parameter.

As you say, when $\sigma$ and $n$ are known and data are normal with unknown mean $\mu,$ we have the quantity $Z=\frac{\bar X - \mu}{\sigma/\sqrt{n}}\sim \mathsf{Norm}(0,1).$

This quantity contains one unknown parameter $\mu$ and the statistic $\bar X,$ which is a reasonable estimator of $\mu.$ Then we can write $$ P\left(-1.96 <\frac{\bar X - \mu}{\sigma/\sqrt{n}} <1.96\right) = 0.95,$$ because we know the quantity $Z \sim \mathsf{Norm}(0,1).$

Then with a little algebra 'called pivoting' we can isolate $\mu$ to get $$P\left( \bar X - 1.96\frac{\sigma}{\sqrt{n}} < \mu < \bar X + 1.96\frac{\sigma}{\sqrt{n}} \right) =0.95.$$

Thus, a 95% confidence interval for $\mu$ is of the form

$$ \left(\bar X - 1.96\frac{\sigma}{\sqrt{n}},\; \bar X + 1.96\frac{\sigma}{\sqrt{n}}\right).$$


Now for an example you may not have seen before. Suppose you have $n$ random observations $X_i$ from the exponential distribution $\mathsf{Exp}(\mathrm{rate}=\lambda),$ so that $E(X_i) = 1/\lambda = \mu.$ One can show that $\frac{\bar X}{\mu}\sim \mathsf{Gamma}(\mathrm{shape}=n,\mathrm{rate}=n).$

The MLE of $\mu$is the sample mean $\bar X.$ Also, $\bar X/\mu$ is a pivotal quantity, which can be used to get a confidence interval for $\mu.$

One can find constants $L$ and $U$ that cut respective probabilities $0.025$ from the lower and upper tails of $\mathsf{Gamma}(n,n)$ so that $$P\left(L < \frac{\bar X}{\mu} < U\right) = 0.95.$$

Upon pivoting. we have $P(\bar X/U <\mu < \bar X/L) =0.96$ and thus the 95% confidence interval $(\bar X/U, \bar X/L)$ for $\mu.$

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Each pivoting will depend on the details, the key step that all share is finding the sampling distribution of your statistic. From there you can compute probabilities, isolate the parameter of interest either explicitly as shown in the other answer or numerically if that turns out to not be possible.

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