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Quantitative Benefit-Risk Assessment · Tutorial 357 of 596

Probability That Benefits Exceed Harms

Estimate the probability of positive net clinical benefit from joint outcome uncertainty, and interpret it alongside—rather than in place of—expected net benefit.

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What You'll Learn

  • Define the probability of positive net clinical benefit for a specified treatment comparison and uncertainty distribution.
  • Estimate that probability from joint draws of benefit and harm.
  • Calculate it under a bivariate normal model and account for dependence between outcomes.
  • Distinguish a probability of positive net benefit from the expected net benefit.
  • Check whether the uncertainty distribution and model support the reported probability.

From an Average Trade-Off to a Probability

A treatment can have positive expected net clinical benefit while still having substantial uncertainty about whether its benefits outweigh its harms. For a development decision, that distinction matters: the average trade-off describes the center of the uncertainty distribution, while the probability of a positive trade-off describes how often the net benefit is favorable under that distribution. Neither quantity alone summarizes the full decision.

The calculation must preserve the joint uncertainty in benefits and harms. If a treatment’s benefit and harm contributions tend to move together, treating them as independent can change the estimated probability that benefits exceed harms. The result also depends on what the uncertain quantities represent: uncertainty about average treatment effects is not the same as variability in outcomes among individual patients.

Definition: Let \(B\) be the uncertain benefit contribution and \(H\) the uncertain harm contribution, expressed on a common utility scale and oriented so that larger \(B\) is favorable and larger \(H\) is unfavorable. Define net clinical benefit as \(N=B-H\). The probability that benefits exceed harms is \(P(N>0)=P(B>H)\), under a specified joint probability distribution for \(B\) and \(H\).

For a comparison with a treatment alternative, \(B\) and \(H\) should represent the benefit and harm contributions for that comparison, using the same direction and preference weights as the net clinical benefit model. The zero threshold is meaningful only after those choices have been made. If the decision instead requires a clinically meaningful margin \(c\), the relevant probability is \(P(N>c)\), not \(P(N>0)\).

The probability distribution also needs a clear interpretation. For example, it may represent posterior uncertainty about treatment-level benefit and harm summaries, conditional on the available evidence and model. If it represents future individual outcomes instead, the probability answers a different question. In either case, state the target, the evidence or conditioning information, and which sources of uncertainty are included. The earlier material on joint uncertainty in benefits and harms motivates retaining the dependence between \(B\) and \(H\); the calculation here uses that joint distribution to evaluate the sign of net benefit.

Estimation From Joint Uncertainty

A direct approach is to draw paired values \((B_m,H_m)\) from the specified joint distribution, calculate \(N_m=B_m-H_m\) for every draw, and count the proportion for which \(N_m>0\). Pairing matters: each benefit draw must remain with its corresponding harm draw. Shuffling or independently resampling the two columns changes their dependence and can therefore change the result.

$$ \widehat{P}(N>0)=\frac{1}{M}\sum_{m=1}^{M} I(B_m-H_m>0) $$

Here, \(M\) is the number of joint draws and \(I(\cdot)\) equals 1 when its condition is true and 0 otherwise. This proportion estimates the probability under the distribution from which the draws were generated. With a finite number of simulation draws, it also has Monte Carlo error. That error reflects numerical estimation from a finite simulation, not necessarily all uncertainty in the underlying evidence or model.

A convenient parametric result applies when \(B\) and \(H\) have a joint bivariate normal distribution. Write their means as \(\mu_B\) and \(\mu_H\), standard deviations as \(\sigma_B\) and \(\sigma_H\), and covariance as \(\operatorname{Cov}(B,H)\). Then \(N=B-H\) is normally distributed, with mean equal to the difference in means and variance determined by both marginal variances and their covariance.

$$ \mu_N=\mu_B-\mu_H,\qquad \sigma_N^2=\sigma_B^2+\sigma_H^2-2\operatorname{Cov}(B,H) $$

If \(\sigma_N>0\), standardizing \(N\) gives \((N-\mu_N)/\sigma_N\sim N(0,1)\). Therefore, with \(\Phi\) denoting the standard normal cumulative distribution function,

$$ P(N>0) =P\left(\frac{N-\mu_N}{\sigma_N}>-\frac{\mu_N}{\sigma_N}\right) =1-\Phi\left(-\frac{\mu_N}{\sigma_N}\right) =\Phi\left(\frac{\mu_N}{\sigma_N}\right) $$

This formula requires the normal model for the joint distribution, not merely normal-looking marginal distributions considered separately. If \(\sigma_N=0\), net benefit is constant under the model: the probability is 1 if that constant is positive and 0 otherwise. If there is probability mass exactly at zero, the strict event \(N>0\) excludes that mass; it should not be silently counted as favorable.

Worked Examples

Worked Example: Probability Under a Joint Normal Model

Suppose a hypothetical treatment comparison is represented on a utility scale in which net benefit is measured in utility points per patient-year. The model gives \(B\) a mean of 0.120 and standard deviation 0.100, and \(H\) a mean of 0.080 and standard deviation 0.080. Their correlation is 0.500. Assume the joint distribution is bivariate normal. Estimate the probability of positive net clinical benefit.

First calculate the covariance from the correlation: \( \operatorname{Cov}(B,H)=0.500(0.100)(0.080)=0.00400 \) utility-points squared. The mean net benefit is \( \mu_N=0.120-0.080=0.0400 \) utility points per patient-year. Its variance and standard deviation are

$$ \sigma_N^2=(0.100)^2+(0.080)^2-2(0.00400) =0.0100+0.00640-0.00800=0.00840 $$

Thus, \(\sigma_N=\sqrt{0.00840}=0.09165\) utility points per patient-year. The standardized mean is \(0.0400/0.09165=0.4364\), so \(P(N>0)=\Phi(0.4364)\approx0.669\). Under the specified model, the probability of positive net clinical benefit is about 66.9%. The expected net benefit is 0.0400 utility points per patient-year; it is a different summary from the probability.

Worked Example: Estimating the Probability From Paired Draws

Consider ten hypothetical paired draws from a model for treatment-level benefit and harm, in utility points per patient-year. The resulting net benefits, calculated draw by draw as \(B_m-H_m\), are 0.10, 0.06, 0.02, 0.01, −0.03, 0.04, −0.01, 0.08, −0.02, and 0.05. Estimate the probability of positive net benefit and compare it with the mean net benefit.

Seven of the ten net benefits are greater than zero. The estimated probability is therefore \( \widehat{P}(N>0)=7/10=0.700 \), or 70.0%. The sum of the ten net benefits is 0.30 utility points per patient-year, so their mean is \(0.30/10=0.0300\) utility points per patient-year.

The estimate from these draws is a 70.0% probability, while the average net benefit across the draws is 0.0300 utility points per patient-year. They answer different questions. With only ten draws, the probability estimate is also imprecise as a simulation estimate: its approximate Monte Carlo standard error is \( \sqrt{0.700(1-0.700)/10}=0.145 \). More simulation draws can reduce this numerical error, but cannot correct a misspecified model or an inappropriate uncertainty distribution.

Worked Example: Why Dependence Changes the Probability

Suppose two hypothetical joint normal models have the same marginal distributions: \(B\) and \(H\) each have mean 0.100 and standard deviation 0.100 utility points per patient-year. Compare the probability of positive net benefit when their correlation is 0 and when it is 0.800.

In both cases, \(\mu_N=0.100-0.100=0\), and the covariance is the correlation multiplied by \(0.100\times0.100\). With correlation 0, the covariance is zero, giving \(\sigma_N^2=0.0100+0.0100=0.0200\). Symmetry of the centered normal distribution gives \(P(N>0)=\Phi(0)=0.500\).

With correlation 0.800, the covariance is \(0.800(0.100)(0.100)=0.00800\). Hence, \( \sigma_N^2=0.0100+0.0100-2(0.00800)=0.00400 \). The standard deviation of \(N\) is \(\sqrt{0.00400}=0.06325\), but its mean remains zero, so \(P(N>0)=\Phi(0/0.06325)=0.500\). In this particular centered, symmetric example, dependence changes the spread but not the probability of being above zero. If the mean net benefit were nonzero, changing that spread would generally change the probability. The example shows why dependence must be represented correctly, while also illustrating that a change in variance does not guarantee a change in this probability for every distribution and threshold.

Why Expected Net Benefit Is Not the Same Quantity

Expected net benefit is \(E(N)\), the probability-weighted average of net benefit. The probability of positive net benefit is \(P(N>0)\), the probability mass on one side of a threshold. A mean does not determine how probability is distributed around that threshold. For example, a net benefit that is always 0.020 utility points has mean 0.020 and probability of being positive equal to 1. A different distribution that gives net benefit 0.100 with probability 0.60 and −0.100 with probability 0.40 has the same mean, \(0.60(0.100)+0.40(-0.100)=0.020\), but its probability of being positive is only 0.60.

The normal result gives another way to see the distinction: for a normal net benefit distribution, \(P(N>0)=\Phi(\mu_N/\sigma_N)\). The probability depends on the mean relative to the standard deviation, not on the mean alone. A positive expected net benefit can coexist with a probability below one-half if the distribution is sufficiently skewed; under a normal model specifically, that cannot happen because the normal distribution is symmetric and the sign of the mean determines whether the probability is above or below one-half.

These measures support different questions. Expected net benefit summarizes average utility on the chosen scale. The probability of positive net benefit summarizes the chance, under the chosen uncertainty distribution, that the sign is favorable. A decision-maker may consider both, along with whether the threshold is appropriate and the consequences of unfavorable outcomes. Neither statistic by itself establishes that treatment causes a particular outcome for an individual patient.

Assumptions, Diagnostics, and Limitations

Before reporting a probability, check that the benefit and harm components are defined consistently, that the utility weights and direction of the score are explicit, and that their joint uncertainty distribution represents the intended target. Check that paired simulations preserve dependence. For a normal calculation, verify that the joint normal approximation is defensible and that the resulting variance is nonnegative. An impossible negative variance usually signals an arithmetic error or an inconsistent covariance specification.

For simulation-based estimates, inspect the number of draws, Monte Carlo error, and the distribution of \(N\), especially near zero. If only a small number of draws are positive or negative, the estimated probability can be sensitive to a few draws. Check whether zero values occur and state how ties are handled. Compare the estimated probability across reasonable model specifications or uncertainty inputs when those choices are consequential; a precise simulation estimate from one questionable model is not evidence that the probability itself is known precisely.

The principal alternative is expected net benefit, which emphasizes average utility rather than the proportion above a threshold. A further alternative is a probability of exceeding a prespecified positive margin \(c\), which asks whether net benefit is not merely favorable but larger than that margin. These summaries are not interchangeable: the threshold must be chosen for a clinical or decision reason, and moving it changes the probability. Reporting a mean, a probability, and—when relevant—a clinically motivated threshold can make the trade-off more transparent than relying on one summary.

The probability remains conditional on the utility model, the evidence, and the uncertainty included. It does not resolve uncertainty about whether the chosen preferences represent all relevant stakeholders, whether the outcomes or weights omit an important consequence, or whether the model is appropriate. A probability based on treatment-level parameter uncertainty also should not be described as the proportion of individual patients who will benefit; that interpretation requires a different target distribution.

Common pitfall: “Expected net benefit is positive, so the probability that benefits exceed harms must be high.” This does not follow: the mean alone does not determine the probability of being above zero. Calculate \(P(N>0)\) from the specified joint distribution, and report expected net benefit separately with its units.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A net benefit distribution has mean −0.020 utility points and standard deviation 0.050 utility points. Under a normal model, what is \(P(N>0)\)?

    Solution: Standardize: \(P(N>0)=\Phi(\mu_N/\sigma_N)=\Phi(-0.020/0.050)=\Phi(-0.400)\approx0.345\). Under this normal model, the probability of positive net benefit is about 34.5%.

  2. In 200 paired draws, 124 net benefit values are positive. What is the empirical probability estimate, and what is its approximate Monte Carlo standard error?

    Solution: The estimate is \(124/200=0.620\). Its approximate Monte Carlo standard error is \(\sqrt{0.620(1-0.620)/200}=\sqrt{0.001178}\approx0.0343\), or 3.43 percentage points.

  3. Why is it generally inappropriate to calculate the probability of positive net benefit by independently shuffling the simulated benefit and harm values?

    Solution: Shuffling destroys the dependence structure in the paired draws and creates a different joint distribution. Because \(N=B-H\), its uncertainty and potentially \(P(N>0)\) depend on that joint structure; use the original paired values unless independence is an explicit, justified assumption.

  4. Two models have the same expected net benefit of 0.030 utility points. Can their probabilities of positive net benefit differ? Explain.

    Solution: Yes. The mean does not determine the distribution’s probability mass above zero. Models with the same mean can differ in variance, shape, or tail mass, and therefore in \(P(N>0)\). The probability must be evaluated from each model’s uncertainty distribution.