Conditional And Population-Average Risk

Two Different Risk Questions

A competing-risk prediction can answer a question about one person’s covariate profile or about a population. These are related but distinct targets. For a specified profile, the question is: what is the probability of failure from a particular cause by time \(t\), for individuals with these covariate values? For a population, the question is: what is that probability on average among individuals whose covariates follow a specified distribution?

The distinction matters even when the profile-specific predictions are unchanged. A population containing many people with profiles associated with higher predicted risk can have a higher average risk than a population containing mostly lower-risk profiles. Thus, reporting an average without defining the population leaves an important part of the prediction unspecified.

Definition: Let \(T\) be time to the first event, \(J\) its cause, \(X\) a vector of baseline covariates, and \(t\) a fixed prediction horizon. The conditional cause-specific risk for cause \(k\) at profile \(x\) is \(F_k(t\mid x)=P(T\leq t, J=k\mid X=x)\). For a target population with covariate distribution \(G\), the population-average risk is \(R_k(t;G)=E_G\{F_k(t\mid X)\}\).

Here, conditional means conditional on the specified baseline covariates, not conditional on having survived event-free up to time \(t\). The risk is a probability over the interval from the defined time origin through the horizon \(t\), and it concerns the event of cause \(k\) occurring before or at that horizon. A profile-specific prediction is one value of the function \(F_k(t\mid x)\); a population-average prediction averages that function over the covariates of a defined target population.

The distribution \(G\) might describe a clinic’s eligible patients, a regional population, or a trial population. It must be the distribution relevant to the question. The average is not automatically the risk for a “typical” individual, and it is not necessarily the average risk in the data used to fit a model.

How Averaging Connects the Targets

For a finite set of covariate profiles \(x_1,\ldots,x_m\), suppose profile \(x_j\) occurs with proportion \(w_j\) in the target population. The proportions are nonnegative and sum to one. The population-average risk is the weighted mean of the profile-specific risks:

R_k(t;G)=\sum_{j=1}^{m} w_j F_k(t\mid x_j), \qquad w_j\geq 0,\quad \sum_{j=1}^{m}w_j=1.

For continuous or mixed covariates, the corresponding operation is an expectation, or equivalently an integral with respect to \(G\):

R_k(t;G)=\int F_k(t\mid x)\,dG(x).

The justification is the law of total probability. Let \(A_k(t)\) denote the event that cause \(k\) occurs by time \(t\). Partition the population into covariate profiles. The probability of \(A_k(t)\) is the sum, across profiles, of the probability of the profile multiplied by the probability of \(A_k(t)\) within that profile. Replacing those within-profile probabilities by \(F_k(t\mid x_j)\) gives the weighted-mean formula. With a general covariate distribution, the sum becomes the integral.

P\{A_k(t)\} =\sum_{j=1}^{m}P(X=x_j)P\{A_k(t)\mid X=x_j\} =\sum_{j=1}^{m}w_j F_k(t\mid x_j).

This identity is a statement about probabilities under a common target population and a common definition of the event. It does not require the covariates to have any particular distribution. But the numerical result does depend on that distribution: changing the weights generally changes the average even when every conditional risk stays fixed.

Worked Example: Averaging Risk Across Two Profiles

Suppose a hypothetical model gives a 5-year risk of cause 1 of 0.08 for profile A and 0.20 for profile B. In the target population, 70% have profile A and 30% have profile B. Find the population-average 5-year risk.

The horizon and cause are the same for both profile-specific predictions, so they can be averaged using the target proportions. The weights sum to \(0.70+0.30=1.00\).

$$ R_1(5;G) =(0.70)(0.08)+(0.30)(0.20) =0.056+0.060 =0.116. $$

The target population’s average predicted risk is 0.116, or 11.6%, by 5 years. This is not the risk for every individual: the two profiles have predicted risks of 8% and 20%. Nor is it their unweighted average, because the profiles are not equally common in the target population.

The Population Distribution Is Part of the Answer

The same set of conditional predictions can yield different population averages under different covariate distributions. This is not a contradiction. Each average answers a question about a different population. It is especially important to name the target distribution when comparing groups, regions, or time periods with different covariate compositions.

Worked Example: Why Population Composition Changes the Average

Consider two hypothetical profiles with predicted 3-year cause 1 risks of 0.04 for profile L and 0.25 for profile H. A target population has 80% profile L and 20% profile H. A second population has equal proportions of the two profiles. Calculate both average risks.

For the first population, the target weights are 0.80 and 0.20:

$$ R_1(3;G_1) =(0.80)(0.04)+(0.20)(0.25) =0.032+0.050 =0.082. $$

The average predicted risk is 8.2%. For the second population, the weights are 0.50 and 0.50:

$$ R_1(3;G_2) =(0.50)(0.04)+(0.50)(0.25) =0.020+0.125 =0.145. $$

The second average is 14.5%. The profile-specific predictions did not change; the population composition did. Therefore, the difference between these averages alone does not show that the conditional risk associated with either profile changed. It reflects, at least in this calculation, a different mix of profiles.

For several competing causes, the same averaging operation applies separately to each cause. If the cause-specific risks are defined for the same horizon and profile, the average for each cause uses the same target weights. Since risks across causes add to the risk of any failure by the horizon, averaging preserves that addition: the average of the sum equals the sum of the averages.

\sum_{k=1}^{K}R_k(t;G) =\sum_{k=1}^{K}E_G\{F_k(t\mid X)\} =E_G\left\{\sum_{k=1}^{K}F_k(t\mid X)\right\}.

Worked Example: Averaging Two Cause-Specific Risks

For a hypothetical 2-year prediction, profile L has cause 1 risk 0.04 and cause 2 risk 0.02. Profile H has cause 1 risk 0.18 and cause 2 risk 0.12. Suppose 75% of the target population has profile L and 25% has profile H. Calculate the average risk for each cause and for any failure.

For cause 1, the weighted average is:

$$ R_1(2;G) =(0.75)(0.04)+(0.25)(0.18) =0.030+0.045 =0.075. $$

For cause 2, it is:

$$ R_2(2;G) =(0.75)(0.02)+(0.25)(0.12) =0.015+0.030 =0.045. $$

The average risk of any failure is \(0.075+0.045=0.120\), or 12.0%. Checking by profile gives the same result: profile L has any-failure risk \(0.04+0.02=0.06\), and profile H has \(0.18+0.12=0.30\). Thus,

$$ (0.75)(0.06)+(0.25)(0.30)=0.045+0.075=0.120. $$

The arithmetic is coherent: cause-specific risks sum within each profile, and the corresponding population averages sum to the population-average risk of any failure.

Choosing and Checking the Averaging Operation

To calculate a population average, first specify the target population and horizon, then obtain profile-specific predictions on that same scale, and finally average them over the target covariate distribution. If the target population is represented by \(n\) individuals with covariate profiles \(X_1,\ldots,X_n\), an empirical average is:

\widehat{R}_k(t) =\frac{1}{n}\sum_{i=1}^{n}\widehat{F}_k(t\mid X_i).

This calculation weights each represented individual equally. If individuals represent different numbers of people or the target population has specified sampling weights \(a_i\), use normalized weights instead:

\widehat{R}_k(t) =\frac{\sum_{i=1}^{n}a_i\widehat{F}_k(t\mid X_i)} {\sum_{i=1}^{n}a_i}.

The result is an average of predictions over the covariate profiles supplied; the choice of profiles and weights determines which population the average describes. The calculation does not by itself correct a model that produces poor conditional predictions, or ensure that the supplied profiles represent the intended target population.

A tempting shortcut is to calculate risk at the average covariate value. In general, this is not the same as averaging the risks. The two operations agree only under special conditions, such as a risk function that is linear in the covariates over the relevant distribution. In competing-risk models, risk functions are generally not guaranteed to have that form. For categorical covariates, an “average profile” may not even be a possible profile.

Common pitfall: “The average risk is the model’s risk for the average patient.” This is generally wrong. The population average is \(E_G\{F_k(t\mid X)\}\), whereas the risk for average covariates is \(F_k(t\mid E_G[X])\). These expressions need not be equal. Use the first when the target is average risk across a population.

Diagnostics, Alternatives, and Limits

Several checks can reveal an averaging error before the result is reported:

Check the target weights. They should be nonnegative and sum to one after normalization. A zero or negative weight, or weights that do not total one, may indicate a coding or scaling problem.
Match the predictions. Every risk being averaged must refer to the same cause, horizon, time origin, and event definition. Mixing 3-year and 5-year risks does not produce a well-defined average for either horizon.
Check probability bounds and coherence. Each cause-specific risk must be between zero and one. At a profile, the sum across mutually exclusive causes must not exceed one; the same must hold for the corresponding averages, apart from numerical rounding.
Check the population represented. A convenient data set or an equal-weight average may not describe the intended target population. Compare the covariate distribution used for averaging with the population named in the question.
Investigate implausible results. An average outside the range of the profile-specific risks being averaged cannot arise from nonnegative weights summing to one. It suggests a mismatch, invalid weights, or an arithmetic error.

A profile-specific prediction and a population average are not competing calculations from which one is universally preferable. They answer different questions. The profile-specific value is useful when a decision concerns a specified covariate profile; the average is useful when the question concerns the expected risk in a target population. A crude estimate from an observed group is another possible summary, but it reflects that group’s covariate mix and observation process. It should not be treated as interchangeable with a model-based average over a separately defined target distribution.

The averaging identity is exact when the conditional risks and target distribution are correctly defined. In practice, both may be estimated. The average then inherits uncertainty from the fitted risk function and, when estimated rather than fixed, from the target covariate distribution. Reporting a point estimate alone does not convey this uncertainty. Also, an average can conceal substantial variation between profiles, so it should not replace profile-specific results when that variation is important to the decision.

Key takeaway: Conditional risk describes a specified covariate profile. Population-average risk is the expectation of that conditional risk over an explicitly defined target covariate distribution. Keep the horizon, cause, and target population aligned, and average predicted risks rather than substituting average covariates.

Check Your Understanding

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

Two profiles have 1-year cause-specific risks 0.03 and 0.15. If their target-population proportions are 0.60 and 0.40, what is the average risk?
Why can two populations have different average risks when their profile-specific predictions are identical?
A target population has 25 people in profile A with predicted risk 0.08 and 75 people in profile B with predicted risk 0.12. Find the empirical average risk.
Is the average of predicted risks generally equal to the predicted risk at average covariate values? Explain.

Solutions: (1) \(0.60(0.03)+0.40(0.15)=0.018+0.060=0.078\), or 7.8%. (2) The populations can have different covariate distributions, so they use different weights in the average. (3) \(\{25(0.08)+75(0.12)\}/100=(2+9)/100=0.11\), or 11.0%. (4) No, not generally: \(E_G\{F_k(t\mid X)\}\) need not equal \(F_k(t\mid E_G[X])\), particularly when the risk function is nonlinear or the average covariate values do not describe a possible profile.