Covariance Between Cause-Specific Estimates

Why Cause-Specific Estimates Are Dependent

Cumulative incidence estimates for different causes are calculated from the same people and the same estimated event-free survival. They are therefore not independent. At an event time, a person can experience one cause or another, but not both as a first event; this mutual exclusivity contributes to dependence. In addition, a change in estimated event-free survival changes the amount of the risk set available to contribute to every cause-specific cumulative incidence function (CIF).

This covariance matters whenever the question compares risks across causes. For example, the variance of the estimated difference between two CIFs is not generally the sum of their separate variances. It also includes their covariance. The earlier material on the Aalen variance estimate motivates estimating uncertainty from the event-time contributions; here the focus is on the covariance between causes.

Definition: For causes \(k\) and \(m\), the covariance between their estimated cumulative incidences at horizon \(t\) is \(\operatorname{Cov}\{\hat F_k(t),\hat F_m(t)\}\). Its estimate describes how the two estimates vary together under repeated sampling. A negative estimate means that, in the estimated sampling variation, larger values of one tend to accompany smaller values of the other.

Setup and Covariance Formula

Consider discrete event times \(j=1,\ldots,J\) up to horizon \(t\), with \(K\) mutually exclusive causes. Immediately before time \(j\), let \(Y_j\) be the number at risk and let \(d_{kj}\) be the number experiencing cause \(k\). Let \(d_j=\sum_{k=1}^K d_{kj}\). Conditional on being at risk at \(j\), the estimated probability of cause \(k\) is \(\hat p_{kj}=d_{kj}/Y_j\), and the estimated probability of remaining event-free through that time is \(\hat p_{0j}=1-\sum_k\hat p_{kj}\). The Aalen–Johansen estimates are \(\hat S_{j-}=\prod_{\ell<j}\hat p_{0\ell}\) and \(\hat F_k(t)=\sum_{j\leq J}\hat S_{j-}\hat p_{kj}\).

For the formula below, treat the cause and event-free outcomes at each time as multinomial categories among those at risk. Assume independent individuals, correctly constructed risk sets, mutually exclusive first events, and independent censoring conditional on the information used to form the risk sets. In particular, censoring must not selectively remove people with different subsequent event risks in a way unaccounted for by the model. The horizon must be within follow-up support, with positive estimated event-free probability after each time at which a future conditional risk is calculated.

Let \(\hat{\boldsymbol p}_j\) be the \(K\)-vector of cause probabilities at time \(j\). The plug-in covariance matrix of these estimated probabilities is \(\hat{\boldsymbol C}_j=\{\operatorname{diag}(\hat{\boldsymbol p}_j)-\hat{\boldsymbol p}_j\hat{\boldsymbol p}_j^{\,T}\}/Y_j\). For each cause \(k\), define \(\hat G_{kj}\) as the estimated conditional probability of cause \(k\) after time \(j\) through \(t\), among people event-free just after \(j\). Set \(\hat G_{kj}=0\) if there are no later event times through \(t\). Define the \(K\)-vector \(\hat{\boldsymbol g}_{kj}\) with entry \(a\) equal to \(1(a=k)-\hat G_{kj}\). Then the covariance estimate is:

\widehat{\operatorname{Cov}}\{\hat F_k(t),\hat F_m(t)\} = \sum_{j=1}^{J} \hat S_{j-}^{\,2}\, \hat{\boldsymbol g}_{kj}^{\,T} \hat{\boldsymbol C}_j \hat{\boldsymbol g}_{mj}.

The diagonal entries of \(\hat{\boldsymbol C}_j\) reflect variation in each cause’s estimated probability. Its off-diagonal entries are negative because the cause categories are mutually exclusive. The vectors \(\hat{\boldsymbol g}_{kj}\) also account for the shared survival estimate: an event at time \(j\) changes the amount of event-free probability available for later events. Thus the covariance is not just the covariance of the event counts at one time; it accumulates the propagated contributions across times.

Why the Future-Risk Term Appears

A small change in a cause probability at time \(j\) affects the CIF at that time and changes the probability of remaining event-free for later times. To see both effects, hold the probabilities at all other times fixed. Increasing the probability of cause \(a\) by a small amount \(\delta\) adds \(\hat S_{j-}\delta\) to cause \(a\) at time \(j\). It also reduces the event-free probability after \(j\) by \(\delta\), reducing the later CIF for cause \(k\) by \(\hat S_{j-}\hat G_{kj}\delta\). Therefore the derivative of \(\hat F_k(t)\) with respect to \(p_{aj}\) is \(\hat S_{j-}\{1(a=k)-\hat G_{kj}\}\).

The vector of estimated cause probabilities at time \(j\) has multinomial covariance \(\hat{\boldsymbol C}_j\). Applying the multivariate delta method to the derivatives just obtained gives that time’s covariance contribution, \(\hat S_{j-}^{\,2}\hat{\boldsymbol g}_{kj}^{\,T}\hat{\boldsymbol C}_j\hat{\boldsymbol g}_{mj}\). Summing over event times gives the displayed formula. This justification also explains why omitting the future-risk term would miss the uncertainty transmitted through event-free survival.

Worked Example: Covariance at One Event Time

Suppose 10 hypothetical individuals are at risk at one event time. Two experience cause 1, one experiences cause 2, and seven remain event-free through the interval. The horizon is this event time, so there are no later events and both future-risk terms are zero.

The estimated probabilities are \(\hat p_1=2/10=0.2\) and \(\hat p_2=1/10=0.1\). The off-diagonal entry of the probability covariance matrix is \(\hat C_{12}=-\hat p_1\hat p_2/Y=-(0.2)(0.1)/10=-0.002\). Since \(\hat S_-=1\) and there is no future-risk term, the covariance between the two CIF estimates is \(-0.002\).

For a check, each estimate is just its cause probability at this time. Their estimated variances are \(\hat C_{11}=\{0.2-(0.2)^2\}/10=0.016\) and \(\hat C_{22}=\{0.1-(0.1)^2\}/10=0.009\). The negative covariance is consistent with the fact that, among the 10 individuals, a first event cannot be counted in both cause categories.

Worked Example: Covariance Across Two Event Times

In a hypothetical cohort, 10 people are at risk at the first event time: two experience cause 1, one experiences cause 2, and seven remain event-free. At the second event time, those seven are at risk: one experiences cause 1, two experience cause 2, and four remain event-free. There is no censoring before the horizon, and events occur at distinct times.

The estimated survival just before the second time is \(0.7\). The CIF estimates at the horizon are \(\hat F_1=2/10+(7/10)(1/7)=0.3\) and \(\hat F_2=1/10+(7/10)(2/7)=0.3\). After the first time, the conditional future CIFs are \(\hat G_{1,1}=1/7\) and \(\hat G_{2,1}=2/7\). At the second time both future terms are zero.

At the first time the cause-probability covariance matrix is \(\hat{\boldsymbol C}_1=\begin{pmatrix}0.016&-0.002\\-0.002&0.009\end{pmatrix}\). The gradient vectors are \(\hat{\boldsymbol g}_{1,1}=(6/7,-1/7)^T\) and \(\hat{\boldsymbol g}_{2,1}=(-2/7,5/7)^T\). Their covariance contribution is \(1^2\hat{\boldsymbol g}_{1,1}^{\,T}\hat{\boldsymbol C}_1\hat{\boldsymbol g}_{2,1}=-0.006142857\).

At the second time, the cause probabilities are \(1/7\) and \(2/7\), and \(Y_2=7\). The off-diagonal probability covariance is \(-(1/7)(2/7)/7=-2/343\). The survival factor squared is \(0.7^2=0.49\), so the second-time contribution is \(0.49(-2/343)=-0.002857143\).

Adding the two unrounded contributions gives \(\widehat{\operatorname{Cov}}(\hat F_1,\hat F_2)=-0.009000\). In this example, both the direct mutual-exclusivity contribution at each time and the propagated survival contribution are included through the gradient vectors.

Using Covariance for a Contrast

For any two estimates, the variance of their difference is the sum of their variances minus twice their covariance:

\widehat{\operatorname{Var}}\{\hat F_k(t)-\hat F_m(t)\} = \widehat{\operatorname{Var}}\{\hat F_k(t)\} + \widehat{\operatorname{Var}}\{\hat F_m(t)\} - 2\widehat{\operatorname{Cov}}\{\hat F_k(t),\hat F_m(t)\}.

Worked Example: Variance of a Difference Between Causes

Continue the two-time example. The separate variance contributions for cause 1 at the first time are \(0.012428571\), and at the second time \(0.008571429\), giving \(\widehat{\operatorname{Var}}(\hat F_1)=0.021000\). For cause 2, the corresponding contributions are \(0.006714286\) and \(0.014285714\), giving \(\widehat{\operatorname{Var}}(\hat F_2)=0.021000\). The covariance is \(-0.009000\).

Substituting without first rounding the intermediate values, \(\widehat{\operatorname{Var}}(\hat F_1-\hat F_2)=0.021000+0.021000-2(-0.009000)=0.060000\). The estimated standard error of the difference is \(\sqrt{0.060000}\approx0.24495\). Since the two point estimates are both 0.3, their estimated difference is zero; the standard error describes sampling variability in that estimated difference, not a probability that the true difference equals zero.

Diagnostics, Alternatives, and Limitations

A useful check is to verify each time’s probability accounting: \(0\leq\hat p_{kj}\leq1\), \(\sum_k\hat p_{kj}\leq1\), and \(\hat p_{0j}=1-\sum_k\hat p_{kj}\geq0\). Confirm that the event counts and risk sets refer to the same time convention and that the risk set decreases consistently with events and censoring. If a future conditional CIF is calculated by dividing by estimated event-free probability, check that this probability is positive. A zero or very small denominator signals that the requested horizon is unsupported or that the estimate may be unstable.

The covariance need not always be negative. Mutual exclusivity contributes negative dependence between causes at a given time, but the shared survival pathway and the future-risk gradients can change the net contribution. A positive estimated covariance is not, by itself, evidence of a coding error; inspect the risk sets, probability vectors, and gradient terms. Conversely, a zero covariance should not be assumed just because the causes are distinct.

Treating the CIF estimates as independent is a tempting shortcut, but it discards their shared risk sets and survival estimate. It gives the wrong variance for a contrast unless the covariance actually is zero. The multinomial delta-method expression is suited to discrete event times and their conditional cause probabilities. Other variance procedures, including influence-function implementations and resampling, can be useful under different formulations or for more complex data structures; they do not justify ignoring dependence. The calculation also relies on valid event classification and an appropriate censoring condition. It does not correct bias from informative censoring, misspecified risk sets, or a horizon with inadequate follow-up.

Key takeaway: Covariance between cause-specific CIF estimates combines mutually exclusive event outcomes with the shared event-free survival pathway. Estimate it from the joint event-time contributions, then include it when calculating uncertainty for a contrast.

Check Your Understanding

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

At a single event time, \(Y=20\), with \(\hat p_1=0.15\), \(\hat p_2=0.10\), and no later events, calculate the covariance contribution.
Why does the covariance matrix of cause-probability estimates have a negative off-diagonal entry under the multinomial model?
At time \(j\), what two effects are represented by the derivative \(\hat S_{j-}\{1(a=k)-\hat G_{kj}\}\)?
If the covariance of two CIF estimates is \(-0.004\), why is it incorrect to calculate the variance of their difference by simply adding their variances?

Solutions. (1) With no later events, the covariance is \(-\hat p_1\hat p_2/Y=-(0.15)(0.10)/20=-0.0007500\). (2) A person’s first event at a time can be assigned to only one cause, so the multinomial covariance is \(-\hat p_1\hat p_2/Y\). (3) The indicator term gives the immediate contribution if the probability for cause \(a\) changes; the future-risk term accounts for the consequent change in event-free probability and later cause-\(k\) incidence. (4) The variance of the difference is \(V_1+V_2-2\operatorname{Cov}\). Here it is \(V_1+V_2+0.008\), not merely \(V_1+V_2\).