# Confidence intervals from pivotal quantities

I am reading a book somewhere it introduces pivot as a way of constructing a confidence interval for a parameter.

My question is:

it says $q_{0.025} < = \lambda n \bar{X} <= q_{0.975}$ is a 95% confidence interval of $\lambda n \bar{X}$ where $q_{0.025}$ is the 2.5% quantile of a gamma distribution and $q_{0.975}$ is the 97.5% quantile of a gamma distribution.

Then it says to find the confidence interval of $\lambda$, it just divides the above inequality by $n\bar{x}$ to get $\frac{q_{0.025}}{n\bar{X}} < = \lambda <= \frac{q_{0.975}}{n\bar{X}}$ as a 95% confidence interval of $\lambda$.

My question is, what is the justification of saying the above is a 95% confidence interval of $\lambda$?

Is it true that as long as we do the same manipulation (i.e. apply a function to every term on each of the 3 terms in the inequality), then we will arrive at a 95% confidence interval for the middle term?

i.e. say I do the following:

$\cos(q_{0.025})<= \cos(\lambda n \bar{X})<= \cos(q_{0.975})$ , then this will become the 95% confidence interval of $\cos(\lambda n \bar{X})$ ?

and if I do the following:

2.5 + $\cos(q_{0.025})<= 2.5 + \cos(\lambda n \bar{X})<= 2.5+ \cos(q_{0.975})$ , then this is a 95% confidence interval of $2.5 + \cos(\lambda n \bar{X})$?

and if I do this:

$\sqrt{q_{0.025}} <= \sqrt{\lambda n \bar{X}} < = \sqrt{q_{0.975}}$, then $(\sqrt{q_{0.025}}, \sqrt{q_{0.975}})$ will be the 95% confidence interval of $\sqrt{\lambda n \bar{X}}$ ?

• You need to give a clearer context for the circumstances near the start of your question. How are the $X_i$ distributed? What parameterization are you using (/what's $\lambda$?) – Glen_b Mar 15 '15 at 2:12
• Hi Glen, thanks for answering. $\X_i$ is independently distributed as exponential exp($\lambda$). $\lambda$ is the rate in the exponential model. – john_w Mar 15 '15 at 18:13

The question boils down to the purely mathematical consideration -- "when are inequalities preserved?".

If you make changes that preserve ordering, then inequalities will be preserved.

Consider a simple probability statement like $P(a\leq X\leq b)=c$ and a monotonic-increasing (i.e. order-preservng) transformation, $g()$. Then $P(g(a)\leq g(X)\leq g(b))=c$. (With a monotonic decreasing function the probability is preserved but the direction of the inequalities would flip.)

$\qquad$

But if $g$ were not monotonic, such a statement would not be true in general. e.g. consider $g(x)=(x-m)^2$, where $m=\frac{a+b}{2}$ and $X$ continuous, then $P(g(a)\leq g(X)\leq g(b))=0$.

So you can't take a function like $\cos$ and hope that in general the interval with transformed limits will preserve the coverage in the original probability statement. If the random variable on an interval is such that $\cos$ is monotonic increasing over that interval, then the probability statement would be preserved.

Note that if $\bar{X}$ is positive then you can divide through by it while leaving the inequality unaltered, since dividing all terms by a positive constant preserves ordering. (This could be relaxed to almost surely while preserving the probability statement.)

• oh thank you. So as long as the function is 1 to 1 mapping , then the manipulation will preserve the probability and only direction might change for the inequality. So the left and right will still be the 95% confidence interval for the middle term I assumed. – john_w Mar 15 '15 at 18:16