Why Ratios Need Special Care
A ratio estimate can look straightforward: divide an estimate of a numerator by an estimate of a denominator. But that calculation can be unstable when the denominator is poorly determined or could plausibly be near zero. In that situation, a symmetric interval around the estimated ratio can give a misleading account of uncertainty.
Fieller confidence sets address the problem by testing each possible ratio value \(r\). If the true ratio were \(r\), the numerator parameter would equal \(r\) times the denominator parameter. The test therefore concerns a linear contrast, not a direct approximation to the distribution of a quotient. Inverting those tests can produce a bounded interval, two unbounded pieces, a half-line, or even the entire real line.
Constructing the Fieller Set
Suppose the estimates are jointly asymptotically normal, and write their estimated covariance matrix as \( \widehat V=\begin{pmatrix}\hat v_{XX}&\hat v_{XY}\\\hat v_{XY}&\hat v_{YY}\end{pmatrix} \). The entries are variances and covariance of the estimates themselves. For each fixed \(r\), the estimated contrast is \(\hat{\mu}_X-r\hat{\mu}_Y\), and its estimated variance is \(\hat v_{XX}-2r\hat v_{XY}+r^2\hat v_{YY}\). Provided this variance is positive, the corresponding studentized contrast is
For a two-sided confidence level approximately \(1-\alpha\), retain \(r\) when \(|T(r)|\le z_{1-\alpha/2}\), where \(z_{1-\alpha/2}\) is the standard normal critical value. This is an asymptotic procedure when the estimates are only asymptotically normal and their covariance estimate is consistent. In a model with an exact joint normal distribution and known covariance matrix, the same inversion can instead be exact, using the appropriate normal critical value.
Derivation. The acceptance condition \(|T(r)|\le c\), after squaring both nonnegative sides and multiplying by the positive estimated contrast variance, is $$ (\hat{\mu}_X-r\hat{\mu}_Y)^2 \le c^2(\hat v_{XX}-2r\hat v_{XY}+r^2\hat v_{YY}). $$ Expand the left side as \(\hat{\mu}_X^2-2r\hat{\mu}_X\hat{\mu}_Y+r^2\hat{\mu}_Y^2\), then subtract the right side. Collecting the constant, linear, and quadratic terms gives \(C\), \(B\), and \(A\), respectively. Thus the inequality is precisely the set of contrast tests that do not reject. \(\square\)
This inversion explains the coverage logic. At the true ratio \(\theta=\mu_X/\mu_Y\), with \(\mu_Y\ne0\), the null contrast satisfies \(\mu_X-\theta\mu_Y=0\). Under the stated joint asymptotic normality and consistent covariance estimation, the studentized contrast at this fixed value converges in distribution to \(N(0,1)\), when its limiting contrast variance is positive. Consequently, the probability that the true ratio is retained tends to \(1-\alpha\). The confidence set is the collection of retained values, not necessarily a single interval.
How the Shape Is Determined
The sign of \(A\) tells whether the quadratic opens upward or downward. Let \(D=B^2-4AC\) denote its discriminant. The familiar quadratic-root argument gives the following classifications when \(A\ne0\).
- If \(A>0\) and \(D>0\), the set is the bounded interval between the two real roots. If \(A>0\) and \(D=0\), it is the single root. If \(A>0\) and \(D<0\), it is empty.
- If \(A<0\) and \(D>0\), the set lies outside the two roots: it is disconnected and unbounded. If \(A<0\) and \(D\le0\), the set is all real numbers.
If \(A=0\), the inequality reduces to \(Br+C\le0\). When \(B\ne0\), its solution is a half-line. When \(B=0\), the set is all real numbers if \(C\le0\), and empty if \(C>0\). These cases matter in practice: the set may be unbounded precisely when the data do not rule out a denominator close to zero.
Worked Example: A Bounded Fieller Interval
Suppose \(\hat{\mu}_X=2\), \(\hat{\mu}_Y=3\), \(\hat v_{XX}=\hat v_{YY}=0.04\), and \(\hat v_{XY}=0\). Use \(c=1.96\), so \(c^2=3.8416\). The point estimate of the ratio is \(2/3\), but the Fieller set follows from the contrast inequality.
The quadratic coefficients are \(A=3^2-3.8416(0.04)=9-0.153664=8.846336\), \(B=-2(2\cdot3-3.8416\cdot0)=-12\), and \(C=2^2-3.8416(0.04)=4-0.153664=3.846336\). Thus \(A>0\), and \(D=(-12)^2-4(8.846336)(3.846336)=144-136.1039225\approx7.8960775>0\).
The roots are \(r=(-B\pm\sqrt D)/(2A)=(12\pm2.8100)/17.692672\), approximately \(0.5194\) and \(0.8371\). Since the quadratic opens upward, the retained values are between the roots. The Fieller confidence set is approximately \([0.5194,0.8371]\), a bounded interval containing the ratio estimate \(0.6667\).
Worked Example: A Disconnected Confidence Set
Now let \(\hat{\mu}_X=3\), \(\hat{\mu}_Y=0\), \(\hat v_{XX}=\hat v_{YY}=1\), and \(\hat v_{XY}=0\), again using \(c=1.96\). Here the denominator estimate is zero, so a ratio estimate by direct division is unavailable.
The coefficients are \(A=0^2-3.8416(1)=-3.8416\), \(B=-2(3\cdot0-3.8416\cdot0)=0\), and \(C=3^2-3.8416(1)=9-3.8416=5.1584\). The inequality is therefore \(-3.8416r^2+5.1584\le0\). Its roots are approximately \(-1.159\) and \(1.159\); equivalently, \(r^2\ge5.1584/3.8416\approx1.3428\).
Because the quadratic opens downward, it is nonpositive outside its roots. The confidence set is approximately \((-\infty,-1.159]\cup[1.159,\infty)\), not a bounded interval. This is a coherent result: with an uncertain denominator centered at zero, large ratios of either sign cannot be excluded.
Worked Example: The Entire Real Line and a Sign Check
Take \(\hat{\mu}_X=0\), \(\hat{\mu}_Y=1\), \(\hat v_{XX}=\hat v_{YY}=1\), \(\hat v_{XY}=0\), and \(c=1.96\). The acceptance inequality before collecting terms is \(r^2\le3.8416(1+r^2)\).
Subtract the right side from the left: \(r^2-3.8416(1+r^2)=(1-3.8416)r^2-3.8416=-2.8416r^2-3.8416\). Both terms are nonpositive for every real \(r\), so the inequality holds for all \(r\). In coefficient form, \(A=1-3.8416=-2.8416\), \(B=0\), and \(C=0-3.8416=-3.8416\), agreeing with the expanded expression. Thus the set is all real numbers.
The result does not say that every ratio is equally plausible. It says that this contrast test, at this confidence level, cannot rule out any candidate value. The example also shows why subtracting the two sides carefully matters: the coefficient of \(r^2\) is \(1-3.8416=-2.8416\), not \(-3.8416\).
Interpretation and a Common Pitfall
The denominator estimate and its uncertainty jointly determine the set's shape. In the quadratic coefficient, \(A=\hat{\mu}_Y^2-c^2\hat v_{YY}\), the first term reflects the squared estimated denominator and the second reflects its uncertainty. If the denominator is estimated precisely enough relative to its distance from zero, \(A\) is positive and a bounded set is possible. If its uncertainty is large, \(A\) can be negative, allowing disconnected or all-real sets.
A common mistake is to divide first and then insist on a symmetric interval around \(\hat{\mu}_X/\hat{\mu}_Y\). That approach is especially fragile when \(\hat{\mu}_Y\) is near zero: the quotient can be unstable, and a symmetric interval cannot represent two separated regions of plausible ratios. Fieller's method instead tests the linear contrast for every candidate \(r\), retaining the possibility that the data provide little information about the ratio.
Fieller sets should be reported as obtained, including their unbounded or disconnected form. Replacing such a set with a convenient finite interval changes the inferential result. The procedure also depends on an appropriate joint covariance estimate; ignoring covariance when the two estimates are dependent changes the contrast variance and can change the confidence set.
Check Your Understanding
Use the contrast inequality and the sign of its quadratic coefficient to answer these questions.
- Why does a Fieller set test \(\mu_X-r\mu_Y=0\) rather than directly test whether \(r\) is close to a quotient estimate?
- For \(Ar^2+Br+C\le0\), what is the solution shape when \(A<0\) and the discriminant is positive?
- If \(A>0\) and the discriminant is negative, what is the solution set?
- What feature of the denominator estimate and its variance is summarized by the coefficient \(A=\hat{\mu}_Y^2-c^2\hat v_{YY}\)?
- In the third worked example, verify the coefficient of \(r^2\) after subtracting the right side of the inequality from the left.