Why Endpoint Order Matters
A confidence interval reports a range of plausible values for an unknown parameter. Sometimes the quantity of interest is not the parameter itself, but a transformation of it: a standard deviation instead of a variance, a rate instead of a mean waiting time, or odds instead of a probability. Transforming the two limits is straightforward once one question is answered: does the function increase or decrease over the interval?
For an increasing function, the lower endpoint remains the lower endpoint. For a decreasing function, the endpoints switch places. This change in ordering is essential; it is not an adjustment to the confidence level. Under the conditions below, the transformed interval has the same coverage as the original interval.
If \(g\) is strictly increasing, \([g(L),g(U)]\). If \(g\) is strictly decreasing, \([g(U),g(L)]\).
The endpoints \(L\) and \(U\) can depend on the sample. The displayed rule applies to each realized pair of endpoints.
The Coverage Argument
A confidence interval is a random interval: its limits vary from sample to sample. Its coverage probability is the probability, under the assumed sampling model, that the interval contains the fixed parameter. To see why the transformed interval retains this coverage, compare the two containment events for a particular sample.
The event that it covers \(\theta\) is \(L\leq\theta\leq U\).
Order is preserved, so \(L\leq\theta\leq U\) is equivalent to \(g(L)\leq g(\theta)\leq g(U)\).
Order reverses, so \(L\leq\theta\leq U\) is equivalent to \(g(U)\leq g(\theta)\leq g(L)\).
The original and transformed coverage events are identical for every sample. Therefore, their probabilities under the same sampling model are equal.
For example, with a strictly increasing \(g\), the event equivalence gives:
The equality is not an approximation: the two events are the same event written in different parameter scales. Thus, if the original procedure has coverage \(1-\alpha\), the transformed procedure also has coverage \(1-\alpha\). This reasoning does not require the original interval to be symmetric or to have equal tail probabilities.
For a decreasing function, writing the transformed limits in the wrong order would produce an invalid interval. The lower limit must be the smaller transformed value, which is \(g(U)\), not \(g(L)\). The coverage argument also requires the transformation to be one-to-one on the relevant domain. If a function is not monotone there, its endpoint values alone may not describe the transformed range.
Worked Transformations
Worked Example: Converting a Variance Interval to a Standard-Deviation Interval
Suppose independent observations are sampled from a normal population \(N(\mu,\sigma^2)\), with unknown mean \(\mu\) and positive variance \(\sigma^2\). Assume a valid 95% confidence procedure has produced the interval \([9,25]\) for \(\sigma^2\), in squared measurement units. Find the corresponding interval for the population standard deviation \(\sigma\).
Method. The target is \(\sigma=\sqrt{\sigma^2}\), so use \(g(t)=\sqrt{t}\) for \(t>0\). This function is strictly increasing: if \(0<t_1<t_2\), then \(\sqrt{t_1}<\sqrt{t_2}\). The endpoint order is therefore preserved.
Transform the lower endpoint:
Transform the upper endpoint:
Thus the 95% interval for \(\sigma\) is \([3,5]\) measurement units. The square root converts squared units back to the original measurement units. Since \(\sigma^2\in[9,25]\) exactly when \(\sqrt{\sigma^2}\in[3,5]\), the transformed interval retains the 95% coverage.
Check. Squaring the new limits gives \(3^2=9\) and \(5^2=25\), recovering the original endpoints. Common mistake avoided: the interval is transformed by taking square roots of both limits, not by taking the square root of the interval's midpoint.
Worked Example: Converting a Mean Waiting-Time Interval to a Rate Interval
Suppose independent waiting times follow an exponential distribution with unknown rate \(\lambda>0\), measured per month. The population mean waiting time is \(\theta=1/\lambda\), measured in months. Assume a valid 90% confidence procedure for \(\theta\) gives \([4,10]\) months. Find the corresponding interval for \(\lambda\).
Method. Use \(g(\theta)=1/\theta\). For positive \(\theta\), this function is strictly decreasing: if \(0<a<b\), then \(1/a>1/b\). Therefore, transform both limits and reverse their order.
The transformed value at the upper endpoint is the lower rate limit:
The transformed value at the lower endpoint is the upper rate limit:
The 90% interval for \(\lambda\) is \([0.1,0.25]\) per month. The units reverse appropriately: a mean measured in months becomes a rate measured per month. The coverage is unchanged because, for every positive \(\theta\), \(\theta\in[4,10]\) if and only if \(1/\theta\in[0.1,0.25]\).
Check. The rate corresponding to 10 months is smaller than the rate corresponding to 4 months, consistent with a slower event occurring at a lower rate. Common mistake avoided: writing \([0.25,0.1]\) puts the decreasing transformation's endpoint values in descending order; an interval must be written from its smaller limit to its larger limit.
Worked Example: Transforming a Probability Interval to an Odds Interval
Suppose a binomial model has an unknown event probability \(p\), with \(0<p<1\). A procedure with 90% coverage under this model gives the interval \([0.20,0.50]\) for \(p\). Find the corresponding interval for the odds \(g(p)=p/(1-p)\).
Method. For \(0<p<1\), the odds transformation is strictly increasing. For instance, its derivative is \(1/(1-p)^2\), which is positive throughout this domain. Thus, retain the endpoint order.
At the lower endpoint:
At the upper endpoint:
The 90% interval for the odds is \([0.25,1]\). These odds are ratios, so they have no measurement units. The lower odds, 0.25, correspond to one event for every four non-events; odds of 1 correspond to equal event and non-event probabilities.
Check. The probabilities recovered from odds \(o\) using \(p=o/(1+o)\) are \(0.25/1.25=0.20\) and \(1/2=0.50\). Common mistake avoided: odds are not the same quantity as probability; the transformation is \(p/(1-p)\), not \(p\) itself.
What Changes and What Stays the Same
A monotone transformation can change the interval's numerical width, symmetry, and units. It does not, by itself, change the coverage probability. The intervals in the examples illustrate these distinctions: standard deviation has different units from variance, a reciprocal rate reverses the endpoints, and odds are on a different scale from probabilities.
The transformation must be applied to the limits themselves. Transforming the point estimate at the center of an interval and then adding or subtracting the old half-width is generally incorrect, because a nonlinear function need not preserve distances. For example, a square root compresses larger values differently from smaller ones, and a reciprocal changes scale more sharply near zero. The correct transformed interval comes from evaluating \(g\) at both endpoints and ordering the results.
| Transformation on the interval's domain | Endpoint rule | Coverage |
|---|---|---|
| Strictly increasing \(g\) | \([g(L),g(U)]\) | Same as the original interval |
| Strictly decreasing \(g\) | \([g(U),g(L)]\) | Same as the original interval |
| Not monotone over the interval | Endpoint values alone may not suffice | Do not apply the monotone rule without further analysis |
Common Exam Traps
The central check is the direction of the transformation over the whole interval, not merely whether the formula contains a reciprocal, square, or logarithm. Determine the domain first, then establish whether the function increases or decreases on that domain.
- Forgetting to reverse limits. With \(g(\theta)=1/\theta\) on positive values, \(g(U)<g(L)\), so the transformed interval is \([g(U),g(L)]\).
- Assuming every transformation is increasing. A reciprocal is decreasing on positive values, while a square is increasing only on a domain where the input does not cross zero.
- Claiming a new confidence level. A one-to-one monotone transformation preserves the original coverage; it does not turn a 90% interval into a 95% interval.
- Ignoring units or the domain. A transformed limit must be meaningful for the parameter and use the corresponding transformed units. A reciprocal of a time, for example, is a rate, not another time.
- Transforming only a center and width. For a nonlinear function, this generally does not give the endpoints obtained by transforming the original limits.
Check Your Understanding
For each problem, identify whether the transformation increases or decreases on the stated interval, then transform and order the limits.
- A valid 95% interval for a positive variance parameter \(\sigma^2\) is \([16,36]\). What is the corresponding interval for \(\sigma\)? (Answer: \([4,6]\), with 95% coverage.)
- Under an exponential waiting-time model, a valid 90% interval for the mean waiting time is \([3,12]\) hours. Find the interval for the rate \(\lambda\), in events per hour. (Answer: \([1/12,1/3]\) per hour, with 90% coverage.)
- A valid 95% interval for a binomial event probability is \([0.10,0.40]\). Find the corresponding odds interval. (Answer: \([1/9,2/3]\), with 95% coverage.)
- A valid 90% interval for \(\theta\) is \([1,4]\), and the target is \(\psi=8-2\theta\). Transform and order the endpoints. (Answer: \([0,6]\), with 90% coverage.)
- A valid 95% interval for a positive parameter \(\theta\) is \([e^{-2},e]\). Find the interval for \(\psi=\log(\theta)\), and state why its endpoint order is preserved. (Answer: \([-2,1]\), with 95% coverage; the logarithm is strictly increasing for positive \(\theta\).)