Uncertainty In The Wald Ratio

Why Uncertainty in a Ratio Needs Care

For a single genetic variant, the Wald ratio estimates a causal effect by dividing the estimated outcome association by the estimated exposure association. This simple calculation can be unstable: if the exposure association is close to zero, small changes in its estimate can produce large changes in the ratio. Uncertainty is therefore not a detail added after computing the ratio. It depends on how precisely the denominator is estimated and on whether that denominator is sufficiently separated from zero.

The earlier material on the Wald ratio for a single variant establishes the ratio estimate and its causal interpretation under the instrumental-variable assumptions. Here, assume those causal assumptions are under consideration separately. A narrow statistical interval cannot establish instrument validity, and an interval method cannot repair violations of relevance, independence, or exclusion restriction.

Definition: Let $\hat{\beta}_X$ and $\hat{\beta}_Y$ be estimated associations of the same genetic variant with the exposure and outcome, respectively. The Wald ratio estimate is $\hat{\theta}=\hat{\beta}_Y/\hat{\beta}_X$, provided $\hat{\beta}_X\ne0$. Let their standard errors be $s_X$ and $s_Y$, and let their estimated covariance be $s_{XY}=\widehat{\operatorname{Cov}}(\hat{\beta}_X,\hat{\beta}_Y)$. The delta method approximates the sampling variance of this ratio using a first-order expansion. Fieller’s method instead obtains a confidence set by inverting a test of each candidate ratio value.

Delta-Method Standard Error

Suppose the vector of association estimates is approximately jointly normal around $(\beta_X,\beta_Y)$, with variances estimated by $s_X^2$ and $s_Y^2$, and covariance estimated by $s_{XY}$. The ratio function is $g(x,y)=y/x$, defined for $x\ne0$. The delta method approximates the variance of a smooth function by applying its gradient to the covariance matrix of its inputs. Here the gradient, evaluated at the estimated associations, is $(-\,\hat{\beta}_Y/\hat{\beta}_X^2,\ 1/\hat{\beta}_X)$. Thus the estimated variance is

\widehat{\operatorname{Var}}(\hat{\theta}) \approx \frac{s_Y^2}{\hat{\beta}_X^2} +\frac{\hat{\beta}_Y^2s_X^2}{\hat{\beta}_X^4} -\frac{2\hat{\beta}_Y s_{XY}}{\hat{\beta}_X^3}, \qquad SE(\hat{\theta})= \sqrt{\widehat{\operatorname{Var}}(\hat{\theta})}.

If the two association estimates are independent, then $s_{XY}=0$, and the covariance term drops out. This is often a reasonable working assumption for associations estimated in non-overlapping exposure and outcome samples. With overlapping samples, or other sources of dependence, setting the covariance to zero without justification can misstate uncertainty. The covariance must be estimated or otherwise addressed.

A conventional approximate $100(1-\alpha)\%$ confidence interval is $\hat{\theta}\pm z_{1-\alpha/2}SE(\hat{\theta})$, where $z_{1-\alpha/2}$ is the corresponding standard-normal quantile. A 95% interval uses approximately 1.96. This is a local, approximately normal interval: it relies on the ratio being sufficiently well behaved around the estimated denominator. It is not a guarantee of good coverage when $\hat{\beta}_X$ is near zero.

Worked Example: A Relatively Precise Exposure Association

Suppose hypothetical, harmonized association estimates are $\hat{\beta}_X=0.20$ exposure units per allele, with $s_X=0.04$, and $\hat{\beta}_Y=0.06$ outcome units per allele, with $s_Y=0.02$. Assume independent estimation, so $s_{XY}=0$. The Wald ratio is $0.06/0.20=0.300$ outcome units per exposure unit. For this illustration, the estimates are treated as approximately jointly normal and the exposure estimate is several standard errors from zero: $0.20/0.04=5$. These conditions support using the delta approximation as a working interval.

Substituting into the variance formula gives

\widehat{\operatorname{Var}}(\hat{\theta}) = \frac{0.02^2}{0.20^2} +\frac{0.06^2(0.04^2)}{0.20^4} = \frac{0.0004}{0.04} +\frac{0.0036(0.0016)}{0.0016} = 0.0100+0.0036=0.0136.

Therefore $SE(\hat{\theta})=\sqrt{0.0136}=0.1166$ outcome units per exposure unit. The approximate 95% interval is

0.300\pm1.96(0.1166) = 0.300\pm0.2285 \approx(0.0714,\ 0.5286).

Under the stated approximation, the interval gives a range of values compatible with these data at the 95% confidence level for the ratio parameter. Its units are outcome units per exposure unit. The calculation quantifies sampling uncertainty; it does not show that the variant satisfies the causal assumptions.

Why the Delta Approximation Can Fail

The ratio function changes rapidly near a zero denominator. A first-order Taylor expansion can describe the ratio well when the denominator estimate is concentrated away from zero, but poorly when plausible denominator values include zero. In that setting, the ratio’s sampling distribution can be highly asymmetric or have heavy tails. A symmetric interval centered on $\hat{\theta}$ may then give misleading coverage and conceal the instability.

The following example deliberately illustrates a weak exposure association. It shows the arithmetic of the conventional interval, not a recommendation to trust that interval.

Worked Example: A Symmetric Interval Under Weak Relevance

Consider hypothetical estimates $\hat{\beta}_X=0.050$ exposure units per allele, $s_X=0.0705$, $\hat{\beta}_Y=0.100$ outcome units per allele, and $s_Y=0.020$, with independent estimation. The ratio estimate is $0.100/0.050=2.00$ outcome units per exposure unit. The exposure association is only $0.050/0.0705=0.709$ standard errors from zero, so the denominator is not well separated from zero.

Applying the delta formula mechanically gives

\widehat{\operatorname{Var}}(\hat{\theta}) = \frac{0.020^2}{0.050^2} +\frac{0.100^2(0.0705^2)}{0.050^4} = 0.1600+7.9524 = 8.1124, \qquad SE(\hat{\theta})\approx2.848.

The mechanical 95% interval is

2.00\pm1.96(2.848) = 2.00\pm5.58208 \approx(-3.582,\ 7.582).

The large standard error reflects the denominator’s uncertainty, but the symmetric form still comes from a local approximation. In this example a confidence set obtained by Fieller’s method is more informative about the denominator problem: it can be unbounded or split into two separate ranges rather than forming one finite interval.

Fieller Confidence Sets

Fieller’s method tests each proposed value $\theta$ of the ratio by asking whether $\hat{\beta}_Y-\theta\hat{\beta}_X$ is compatible with zero. If the candidate is the true ratio, then the corresponding contrast has expectation zero. Under approximate joint normality, its estimated variance is $s_Y^2+\theta^2s_X^2-2\theta s_{XY}$. Retaining candidate values whose standardized contrast is no larger than the normal critical value in absolute value produces a confidence set:

\mathcal{C}_{1-\alpha} = \left\{ \theta: (\hat{\beta}_Y-\theta\hat{\beta}_X)^2 \le z_{1-\alpha/2}^2 \left(s_Y^2+\theta^2s_X^2-2\theta s_{XY}\right) \right\}.

Expanding and collecting powers of $\theta$ gives $A\theta^2+B\theta+C\le0$, where

A=\hat{\beta}_X^2-z^2s_X^2,\qquad B=-2\hat{\beta}_X\hat{\beta}_Y+2z^2s_{XY},\qquad C=\hat{\beta}_Y^2-z^2s_Y^2, \quad z=z_{1-\alpha/2}.

This inequality explains the possible shapes of a Fieller set. If $A>0$, the quadratic opens upward and, when there are two real roots, the set lies between them. If $A<0$, it opens downward and the set lies outside the roots; that can yield two unbounded ranges. If there are no real roots, the sign of the quadratic determines whether the set is all real numbers or empty. These shapes are not errors: they express that the data do not pin the ratio to one finite range. In particular, $A\le0$ indicates that the usual normal confidence interval for the exposure association includes zero.

Worked Example: A Disconnected Fieller Set

Use hypothetical independent estimates $\hat{\beta}_X=0.050$, $s_X=0.040$, $\hat{\beta}_Y=0.010$, and $s_Y=0.002$, with associations measured in exposure units and outcome units per allele. Assume the joint-normal approximation is reasonable and use $z=1.96$. The exposure association is weak relative to its uncertainty: $0.050/0.040=1.25$. Here $s_{XY}=0$, so

$$ A=0.050^2-3.8416(0.040^2) =0.0025-0.00614656=-0.00364656, $$

$$ B=-2(0.050)(0.010)=-0.001000, \qquad C=0.010^2-3.8416(0.002^2) =0.000100-0.0000153664=0.0000846336. $$

The boundary equation is $-0.00364656\theta^2-0.001000\theta+0.0000846336=0$. Its roots, from the quadratic formula, are approximately $-0.3420$ and $0.06784$. Since $A<0$, the quadratic is nonpositive outside the two roots. The 95% Fieller confidence set is therefore approximately

(-\infty,\,-0.3420]\ \cup\ [0.06784,\,\infty).

The point ratio is $0.010/0.050=0.200$ outcome units per exposure unit and belongs to this set. The set is disconnected and unbounded because values of the exposure association compatible with the data include zero. Reporting only a finite interval would hide this feature of the uncertainty.

Justification and Practical Checks

The delta formula follows directly from the first-order Taylor approximation. For $g(x,y)=y/x$, the partial derivatives are $\partial g/\partial x=-y/x^2$ and $\partial g/\partial y=1/x$. If the covariance matrix of $(\hat{\beta}_X,\hat{\beta}_Y)$ has diagonal entries $s_X^2,s_Y^2$ and off-diagonal entry $s_{XY}$, the gradient-quadratic form is

\left(-\frac{\hat{\beta}_Y}{\hat{\beta}_X^2},\frac{1}{\hat{\beta}_X}\right) \begin{pmatrix} s_X^2&s_{XY}\\ s_{XY}&s_Y^2 \end{pmatrix} \begin{pmatrix} -\frac{\hat{\beta}_Y}{\hat{\beta}_X^2}\\ \frac{1}{\hat{\beta}_X} \end{pmatrix} = \frac{\hat{\beta}_Y^2s_X^2}{\hat{\beta}_X^4} -\frac{2\hat{\beta}_Y s_{XY}}{\hat{\beta}_X^3} +\frac{s_Y^2}{\hat{\beta}_X^2}.

The approximation uses only the first-order terms, which is why it is most defensible when denominator uncertainty is modest relative to the estimated denominator. Fieller’s inequality, by contrast, retains the candidate ratio in the variance of the contrast and does not linearize the ratio around its observed value.

Worked Example: Accounting for Covariance

Suppose hypothetical association estimates are $\hat{\beta}_X=0.20$, $s_X=0.04$, $\hat{\beta}_Y=0.06$, and $s_Y=0.02$, as in the first example, but now their estimated covariance is $s_{XY}=0.00020$. Assume the covariance estimate is justified by the sampling design and that the joint-normal approximation is appropriate. The ratio remains $0.300$, but the covariance term changes its estimated variance:

\widehat{\operatorname{Var}}(\hat{\theta}) = 0.0100+0.0036 -\frac{2(0.06)(0.00020)}{0.20^3} = 0.0136-0.0030 = 0.0106.

Thus $SE(\hat{\theta})=\sqrt{0.0106}=0.1030$ outcome units per exposure unit, and the approximate 95% interval is $0.300\pm1.96(0.1030)$, or approximately $(0.0982,\ 0.5018)$. Positive covariance reduces the variance here because the covariance contribution has a negative sign for a positive ratio and positive $\hat{\beta}_X$. That direction is not universal: the sign of the contribution depends on the associations and covariance.

Diagnostics, Alternatives, and Limitations

Before reporting a ratio interval, check the sign and magnitude of $\hat{\beta}_X$, its standard error, and whether a confidence interval for the exposure association includes zero. Examine whether exposure and outcome estimates came from independent or overlapping samples, and whether a defensible covariance estimate is available. For Fieller’s method, verify the signs of $A$ and the discriminant and check that the reported set satisfies the original inequality. A quadratic’s roots alone do not tell you which side belongs to the confidence set.

The delta interval is convenient and usually gives a simple finite interval when the denominator is well estimated away from zero. Fieller’s set is preferable when denominator uncertainty is material, because it can expose the resulting unbounded or disconnected uncertainty. Neither method is automatically reliable if the association estimates are poorly approximated by a joint normal distribution, their standard errors are wrong, or the covariance is omitted when important. A bootstrap can be considered when the sampling procedure supports it, but resampling does not make a weak denominator stable and must preserve the relevant dependence structure.

These are uncertainty calculations for the single-variant ratio, not a solution to weak-instrument bias or a test of the instrumental-variable assumptions. A very broad Fieller set may be scientifically uninformative, but narrowing it by choosing a more convenient interval does not add information. Report the ratio, its uncertainty method, the association estimates and standard errors, and any relevant covariance assumption so that readers can see what drives the conclusion.

Common pitfall: “The ratio estimate is finite, so its confidence interval should be a finite interval around it.” This is not necessarily true. A nonzero observed denominator can still be statistically compatible with zero. The delta method may then give a fragile symmetric interval, whereas Fieller’s method can correctly return an unbounded or disconnected set. Check the denominator uncertainty and the quadratic inequality before interpreting interval endpoints.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

For $\hat{\beta}_X=0.25$, $s_X=0.05$, $\hat{\beta}_Y=0.10$, and $s_Y=0.03$, with independent estimation, calculate the Wald ratio and delta-method standard error.
Why can a delta-method confidence interval be unreliable when the exposure-association confidence interval includes zero?
For a Fieller quadratic with $A<0$ and two distinct real roots, which values satisfy $A\theta^2+B\theta+C\le0$?
In the covariance example, state the estimated ratio standard error and explain why including the positive covariance reduced it.

Solutions. (1) The ratio is $0.10/0.25=0.400$. Its variance estimate is $0.03^2/0.25^2+0.10^2(0.05^2)/0.25^4=0.0144+0.0064=0.0208$, so the standard error is $\sqrt{0.0208}=0.1442$ outcome units per exposure unit. (2) The ratio changes sharply near a zero denominator, so a first-order approximation around the observed denominator can fail and a symmetric interval can misrepresent uncertainty. (3) The inequality holds outside the roots, including the two unbounded tails, because the quadratic opens downward. (4) The standard error is $\sqrt{0.0106}=0.1030$. With positive associations and positive covariance, the covariance contribution $-2\hat{\beta}_Y s_{XY}/\hat{\beta}_X^3$ is negative, reducing the estimated variance in this case.

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