Tutorials › Bioavailability and Bioequivalence › Constructing the FDA Upper Confidence Bound

Reference-Scaled Approaches for Highly Variable Drugs · Tutorial 254 of 399

Constructing the FDA Upper Confidence Bound

Learn to calculate the FDA upper confidence bound, account for both sources of uncertainty, and decide whether the bound is below zero.

Advanced 12 min read

What You'll Learn

  • Define the linearized criterion and the quantities used to construct its upper bound.
  • Derive the bound using a delta-method variance approximation.
  • Include the factor of two in the reference-variance uncertainty contribution.
  • Calculate one-sided upper bounds and interpret their decisions in context.
  • Recognize how degrees of freedom and imprecise estimates can change a decision.
  • Diagnose common calculation and interpretation errors.

Why an Upper Bound Is Needed

The FDA linearized reference-scaled criterion compares the squared log-scale treatment difference with a term based on within-subject reference variability. A calculation using only the estimated treatment difference and estimated reference variance gives a point estimate of that criterion. But both quantities are estimated from finite data. The upper confidence bound addresses this uncertainty: it asks whether the data support keeping the criterion below zero even after allowing for estimation error.

This tutorial focuses on constructing and interpreting that bound. It is not enough to calculate the plug-in criterion and note that it is negative. The bound also depends on how precisely the treatment difference and reference variance have been estimated.

Definition: Let \(D\) be the population log-scale mean difference between test and reference, and let \(\sigma_{WR}^2\) be the within-subject reference variance on the log scale. The FDA linearized criterion is \(D^2-\theta\sigma_{WR}^2\). Its sample estimate is \(\hat D^2-\theta S_{WR}^2\), where \(\hat D\) estimates \(D\), \(S_{WR}^2\) estimates \(\sigma_{WR}^2\), and \(\theta\) is the prespecified scaling constant. The upper-bound decision is to conclude that this criterion is below zero when its one-sided upper confidence bound is less than zero.

For the conventional FDA scaling constant, \(\theta=[\ln(1.25)/\sigma_{W0}]^2\), with \(\sigma_{W0}=0.25\) on the log scale; thus \(\theta\) is approximately \(0.7967\). The criterion, its estimate, and its bound are dimensionless because they are formed from log-scale quantities and variances on the log scale.

The inputs must refer to the same estimand and analysis model: \(\hat D\) is the estimated test-minus-reference log mean difference, \(SE_D\) is its standard error, \(S_{WR}^2\) is the estimated within-subject reference variance, and \(df\) is the degrees of freedom associated with estimating that variance for this bound. Do not substitute a coefficient of variation or a standard deviation where a variance is required.

Deriving the Upper Confidence Bound

Write the estimated criterion as \(g(\hat D,S_{WR}^2)=\hat D^2-\theta S_{WR}^2\). To approximate its uncertainty, use a first-order Taylor expansion around the estimates. The partial derivative with respect to the treatment difference is \(2\hat D\); the partial derivative with respect to the reference variance is \(-\theta\). Under the working approximation that the two estimates are independent, the delta method gives the variance of the estimated criterion as the sum of the two derivative-weighted variance contributions.

$$ \widehat{\operatorname{Var}}\{g(\hat D,S_{WR}^2)\} \approx (2\hat D)^2 SE_D^2+\theta^2\widehat{\operatorname{Var}}(S_{WR}^2) $$

For a variance estimate based approximately on a normal chi-square model, \(df\,S_{WR}^2/\sigma_{WR}^2\) is treated as approximately chi-square with \(df\) degrees of freedom. A chi-square variable with \(df\) degrees of freedom has variance \(2df\). Therefore,

$$ \operatorname{Var}(S_{WR}^2) \approx \frac{2\sigma_{WR}^4}{df} \quad\Longrightarrow\quad \widehat{\operatorname{Var}}(S_{WR}^2) \approx \frac{2S_{WR}^4}{df}. $$

Substituting this approximation and taking the square root gives the estimated standard error of the criterion. Using the one-sided \(t\) critical value \(t_{0.95,df}\), the FDA upper confidence bound is

$$ UCB = \hat D^2-\theta S_{WR}^2 + t_{0.95,df} \sqrt{ (2\hat D\,SE_D)^2+ \frac{2\theta^2 S_{WR}^4}{df} }. $$

The factor of 2 in the reference-variance contribution is essential. It comes from the variance of a chi-square-based variance estimate, not from an optional adjustment. Omitting it understates the estimated uncertainty in \(S_{WR}^2\) and can make the upper bound too low.

Assumptions and scope: This construction is a large-sample, first-order approximation. It assumes that the treatment difference estimator is approximately normal with standard error \(SE_D\); that the reference-variance estimator is adequately approximated by a scaled chi-square distribution with \(df\) degrees of freedom; and that the estimates are independent, or that any covariance is negligible under the fitted model. The bound must use the variance estimator, standard error, and degrees of freedom appropriate to the specified reference-scaled analysis. If those approximations or the independence condition are not defensible, this formula may not adequately represent uncertainty.

The decision rule is strict: if \(UCB<0\), the upper bound lies below zero and supports the linearized reference-scaled criterion. If \(UCB\geq 0\), this bound does not establish that the criterion is below zero. A negative plug-in estimate alone is not enough.

Worked Calculations

Worked Example: A Bound Below Zero

Suppose a hypothetical analysis gives \(\hat D=0.100\), \(SE_D=0.020\), \(S_{WR}^2=0.150\), and \(df=40\). Use \(\theta=0.7967\) and \(t_{0.95,40}=1.684\). The estimates and degrees of freedom are assumed to satisfy the working conditions stated above.

First calculate the estimated criterion:

$$ \hat D^2-\theta S_{WR}^2 = (0.100)^2-(0.7967)(0.150) = 0.01000-0.119505 = -0.109505. $$

The treatment-difference contribution to the variance is \((2\hat D\,SE_D)^2=(2(0.100)(0.020))^2=(0.00400)^2=0.00001600\). The reference-variance contribution, including the factor of two, is

$$ \frac{2\theta^2S_{WR}^4}{df} = \frac{2(0.7967)^2(0.150)^2}{40} = \frac{2(0.63473)(0.02250)}{40} \approx 0.0007141. $$

Thus the estimated standard error of the criterion is \(\sqrt{0.00001600+0.00071407225125}\approx0.02701985\), and

$$ UCB=-0.109505+(1.684)(0.02701985)\approx-0.06400. $$

Because the upper bound is below zero, this hypothetical result supports the FDA linearized reference-scaled criterion. The conclusion is about the bound for this criterion, not a claim that the population value is known with certainty.

Worked Example: A Negative Estimate but a Bound Above Zero

Consider hypothetical estimates \(\hat D=0.200\), \(SE_D=0.040\), \(S_{WR}^2=0.100\), and \(df=20\), with \(\theta=0.7967\) and \(t_{0.95,20}=1.725\). Assume the model-based estimates meet the working conditions for the bound.

The point estimate of the criterion is negative:

$$ (0.200)^2-(0.7967)(0.100)=0.04000-0.07967=-0.03967. $$

The two variance contributions are \((2(0.200)(0.040))^2=(0.01600)^2=0.0002560\) and

$$ \frac{2(0.7967)^2(0.100)^2}{20} = \frac{2(0.63473)(0.01000)}{20} \approx0.0006347. $$

Therefore, the standard error is \(\sqrt{0.0002560+0.00063473089}\approx0.0298451\), and the upper bound is

$$ UCB=-0.03967+(1.725)(0.0298451)\approx0.01181. $$

The bound is above zero, so this upper-bound procedure does not establish that the criterion is below zero. Notice that this example has a smaller reference variance than the previous one, so the reference-variance uncertainty contribution is smaller; here that contribution is still larger than the treatment-difference contribution.

Worked Example: Greater Variance Uncertainty Changes the Decision

Suppose a third hypothetical analysis gives \(\hat D=0.150\), \(SE_D=0.025\), and \(S_{WR}^2=0.120\). First use \(df=50\), \(\theta=0.7967\), and \(t_{0.95,50}=1.676\). Assume the same point estimates and standard error are used only to illustrate the effect of changing the variance degrees of freedom.

The estimated criterion is \(0.150^2-(0.7967)(0.120)=0.02250-0.095604=-0.073104\). The treatment contribution is \((2(0.150)(0.025))^2=(0.00750)^2=0.00005625\). The reference-variance contribution is

$$ \frac{2(0.7967)^2(0.120)^2}{50} = \frac{2(0.63473)(0.01440)}{50} \approx0.0003656. $$

The estimated standard error is \(\sqrt{0.00005625+0.0003656}\approx0.02054\), giving \(UCB=-0.073104+(1.676)(0.02054)\approx-0.03868\). This bound is below zero.

Now consider \(df=8\) with the same point estimates and standard error, using \(t_{0.95,8}=1.860\). The reference contribution becomes \(2(0.63473)(0.01440)/8\approx0.0022850\). The standard error is \(\sqrt{0.00005625+0.0022850312}\approx0.04839\), so

$$ UCB=-0.073104+(1.860)(0.04839)\approx0.01690. $$

This upper bound is above zero, so it does not support the criterion. The comparison isolates the role of variance-estimation uncertainty: fewer degrees of freedom increase both the estimated variance of \(S_{WR}^2\) and the one-sided critical value. In an actual analysis, the estimates and degrees of freedom arise from the fitted model together; they should not be altered independently merely to obtain a preferred decision.

Reading the Result and Checking the Calculation

The sign of the upper bound, not the sign of the plug-in estimate, determines this part of the decision. A negative estimated criterion may have a positive upper bound when the uncertainty allowance is large. Conversely, a bound well below zero indicates that the estimated uncertainty, as represented by this method, does not extend the criterion to zero.

The bound is a one-sided confidence procedure. In repeated applications under its assumptions, this first-order approximate procedure is intended to have approximately 95% coverage of the target criterion. It does not mean there is a 95% probability that the fixed population criterion lies below the calculated bound.

A useful calculation sequence is to write down the inputs with their meanings and units, calculate the plug-in criterion, calculate each variance contribution separately, add them before taking the square root, apply the one-sided critical value, and then compare the resulting upper bound with zero. Keeping the two uncertainty contributions visible makes an omitted term or an incorrect scale easier to detect.

1
Check the inputs.
Confirm that \(\hat D\) is a log-scale test-minus-reference mean difference, \(SE_D\) is its standard error, and \(S_{WR}^2\) is the within-subject reference variance on the log scale.
2
Calculate both uncertainty terms.
Use \((2\hat D\,SE_D)^2\) and \(2\theta^2S_{WR}^4/df\). The factor of two belongs in the second term.
3
Form and interpret the bound.
Add the one-sided \(t\)-margin to \(\hat D^2-\theta S_{WR}^2\). A bound below zero supports the criterion; a bound at or above zero does not establish it.

Diagnostics, Alternatives, and Limitations

Several checks help identify a calculation that is going wrong. Since \(S_{WR}^2\) is a variance, it must be nonnegative; its fourth power in the formula is \(S_{WR}^4=(S_{WR}^2)^2\), not the fourth power of a standard deviation entered as though it were a variance. The two contributions inside the square root must be nonnegative, and the square root and resulting standard error must be real. Verify that \(df\) is positive and that \(t_{0.95,df}\) is a one-sided 95th-percentile critical value, not a two-sided 95% critical value. Finally, the confidence margin must be added to the point estimate when constructing an upper bound.

A common mistake is to use \(\theta^2S_{WR}^4/df\) for the reference-variance contribution. That omits the factor of two in \(\operatorname{Var}(S_{WR}^2)\approx2S_{WR}^4/df\). The correction is to use \(2\theta^2S_{WR}^4/df\) before adding the treatment-difference contribution. Omitting the factor makes the standard error too small and can produce an upper bound that is too low, potentially changing the decision.

The plug-in criterion is simpler to calculate, but it does not account for estimation uncertainty and is not a substitute for the upper-bound decision. Conventional average bioequivalence procedures address a different criterion; they should not be treated as interchangeable with this reference-scaled bound. This bound also does not by itself resolve every requirement of a scaled bioequivalence assessment. Its accuracy depends on the specified analysis model and on how well the normal and chi-square approximations describe the estimates, especially when the reference-variance estimate has few degrees of freedom or is unstable. Model fit and variance estimation therefore matter to the validity of the calculation, not just to its arithmetic.

Key takeaway: The FDA upper bound adds a one-sided uncertainty margin to \(\hat D^2-\theta S_{WR}^2\). That margin must include both \((2\hat D\,SE_D)^2\) and \(2\theta^2S_{WR}^4/df\). The bound supports the linearized criterion only when its calculated value is below zero.

Check Your Understanding

Try each question, then check the solution summary below.

  1. Which two estimated quantities contribute uncertainty to the upper bound?
  2. Why does the reference-variance term include a factor of two?
  3. If the plug-in criterion is negative but the upper bound is positive, what is the decision for this criterion?
  4. For \(\hat D=0\), what happens to the treatment-difference contribution to the variance approximation?
  5. Name two checks that can help detect a calculation or input error.

Solutions. (1) The treatment difference estimate \(\hat D\), through its standard error \(SE_D\), and the reference variance estimate \(S_{WR}^2\), through its degrees of freedom \(df\), both contribute. (2) Under the scaled chi-square approximation, \(\operatorname{Var}(S_{WR}^2)\approx2S_{WR}^4/df\); multiplying by the squared derivative \(\theta^2\) preserves that factor. (3) The bound does not establish that the criterion is below zero, because the decision requires \(UCB<0\). (4) The term \((2\hat D\,SE_D)^2\) is zero when \(\hat D=0\), although uncertainty in the reference variance remains. (5) For example, confirm that the variance input is on the log scale and that \(df>0\); also check the factor of two, use a one-sided critical value, and ensure the uncertainty margin is added rather than subtracted.