Discrete-Time Dropout Hazards and Response Indicators

Response Through Time

In a longitudinal study, dropout is not just a single missing-data indicator: it has a time. A participant may provide outcomes at early visits and then stop responding, so the probability of still being observed changes as follow-up proceeds. Discrete-time dropout hazards describe this change visit by visit and give a direct way to calculate the probability of each possible dropout time.

The setup here is monotone dropout: once a participant stops responding, no later scheduled outcomes are observed. This assumption is appropriate when nonresponse represents permanent departure from follow-up, rather than an isolated missed visit followed by a return.

Definition: For participant $i$ and scheduled visits $t=1,\ldots,T$, let $R_{it}=1$ if the outcome at visit $t$ is observed and $R_{it}=0$ otherwise. Set $R_{i0}=1$, indicating that the participant is observed at the start of follow-up. Monotone response means $R_{it}\leq R_{i,t-1}$ for every visit $t$. Define $D_i$ as the first visit with $R_{it}=0$, and set $D_i=T+1$ if the participant is observed at all $T$ visits. Thus, $D_i=t$ means the participant is observed through visit $t-1$, then drops out at visit $t$.

This convention treats the response status at visit $t$ as the event of interest: dropout at $t$ means the outcome scheduled for that visit is not observed. Someone who completes every visit has $D_i=T+1$, not $D_i=T$. Stating the convention matters, because some settings instead label dropout by the last completed visit.

Visit-Specific Dropout Hazards

At visit $t$, the risk set consists of participants who were still responding through visit $t-1$. The discrete-time dropout hazard is the conditional probability of dropout at $t$ among those at risk. It is a conditional probability for one interval, not the probability of having dropped out by that point.

Definition: Let $H_{i,t-1}$ denote information available before the response decision at visit $t$, such as baseline covariates and outcomes observed through visit $t-1$. The visit-specific hazard is $$ h_{it}(H_{i,t-1})=P(R_{it}=0\mid R_{i,t-1}=1,\ H_{i,t-1}). $$ The corresponding conditional probability of remaining observed at visit $t$, given that the participant is still responding at $t-1$, is $1-h_{it}(H_{i,t-1})$.

A hazard may be constant across participants within a specified group, or may differ according to the information in $H_{i,t-1}$. The conditioning is essential: the hazard at visit $t$ concerns participants still under observation just before that visit. Participants who have already dropped out are not in that visit’s risk set.

The definition uses information available before the visit-specific response status is determined. It does not require a value for an outcome that is unobserved because of dropout at that visit. If hazards vary with observed history, the probability calculations below apply along each history path; obtaining an overall probability may additionally require averaging over histories.

From Hazards to Dropout-Time Probabilities

For clarity, first consider a fixed covariate group or setting in which the hazard at each visit is a specified value $h_t$. To drop out at visit $t$, a participant must remain observed at visits $1,\ldots,t-1$, then drop out at $t$. The probability is therefore the product of the probabilities of those successive events.

P(D=t)=\left\{\prod_{j=1}^{t-1}(1-h_j)\right\}h_t,\qquad t=1,\ldots,T.

For $t=1$, the product over no earlier visits is defined to equal 1, so $P(D=1)=h_1$. To remain observed through all visits, the participant must avoid dropout at every visit:

P(D=T+1)=\prod_{j=1}^{T}(1-h_j).

These formulas also express a general sequential rule: multiply the conditional probabilities along the response path. When hazards depend on observed history, evaluate each hazard at the history reached along that path. If the target is a marginal probability across participants with different histories or covariates, average the resulting conditional probabilities over the relevant distribution; multiplying marginally averaged hazards need not give that answer.

Why the Formula Works

The derivation follows from the conditional probability rule. The event $D=t$ requires response at visits $1$ through $t-1$ and nonresponse at $t$. At each earlier visit $j$, conditional on still responding, the probability of continuing is $1-h_j$. Conditional on continuing through $t-1$, the probability of dropout at $t$ is $h_t$. Multiplying these successive conditional probabilities gives the formula.

\begin{aligned} P(D=t) &=P(R_1=1,\ldots,R_{t-1}=1,R_t=0)\\ &=\prod_{j=1}^{t-1}P(R_j=1\mid R_1=\cdots=R_{j-1}=1)\, P(R_t=0\mid R_1=\cdots=R_{t-1}=1)\\ &=\left\{\prod_{j=1}^{t-1}(1-h_j)\right\}h_t. \end{aligned}

The possible dropout times are mutually exclusive and exhaustive: each participant either first drops out at exactly one visit or completes all visits. The probabilities consequently sum to 1. This can also be checked by writing $S_t=\prod_{j=1}^{t}(1-h_j)$, with $S_0=1$. Then $P(D=t)=S_{t-1}-S_t$, so the sum telescopes:

\sum_{t=1}^{T}P(D=t)+P(D=T+1) =\sum_{t=1}^{T}(S_{t-1}-S_t)+S_T =S_0=1.

Worked Calculations

Worked Example: Calculating Each Dropout-Time Probability

Hypothetical setting. Suppose a study has three visits and the dropout hazards, within the study group considered, are $h_1=0.10$, $h_2=0.20$, and $h_3=0.25$. Assume these hazards apply to each participant still responding before the corresponding visit. Calculate the probability of each dropout time and of completing follow-up.

At visit 1, dropout occurs before any previous visit has been missed: $$ P(D=1)=h_1=0.10. $$

To drop out at visit 2, a participant must first remain observed at visit 1: $$ P(D=2)=(1-h_1)h_2=(1-0.10)(0.20)=0.90(0.20)=0.18. $$

To drop out at visit 3, the participant must remain observed at both earlier visits: $$ P(D=3)=(1-h_1)(1-h_2)h_3=(0.90)(0.80)(0.25)=0.18. $$

The probability of completing all three visits is $$ P(D=4)=(0.90)(0.80)(1-0.25)=(0.90)(0.80)(0.75)=0.54. $$ The probabilities sum to $0.10+0.18+0.18+0.54=1.00$. In this hypothetical setting, 18% drop out at visit 3, while 54% are observed at every visit.

Worked Example: Estimating Hazards From Risk Sets

Hypothetical data. Among 100 participants responding before visit 1, 10 fail to provide the visit-1 outcome. Of the 90 still responding before visit 2, 18 drop out at visit 2. Of the 72 still responding before visit 3, 12 drop out at visit 3. Assume the data follow monotone dropout and that the counts are observed without other reasons for loss from the risk sets. Calculate the empirical hazards and the probability of completing all three visits estimated from them.

At each visit, the empirical hazard is the number dropping out divided by the number at risk just before that visit: $$ \hat h_1=\frac{10}{100}=0.10,\qquad \hat h_2=\frac{18}{90}=0.20,\qquad \hat h_3=\frac{12}{72}=0.1667. $$

The estimated probability of completing all visits is $$ (1-\hat h_1)(1-\hat h_2)(1-\hat h_3) =(0.90)(0.80)(0.8333)\approx0.600. $$ Indeed, 60 of the original 100 participants remained after the third visit, so the observed completion fraction is $60/100=0.600$. The equality here follows from these nested risk sets and empirical hazards. The visit-3 hazard is $12/72$, not $12/100$: its denominator is the participants still responding just before visit 3.

Worked Example: Why Averaging Hazards Can Mislead

Hypothetical setting. Suppose two equally common covariate groups have two-visit dropout hazards. Group A has hazards $0.05$ and $0.10$; Group B has hazards $0.20$ and $0.25$. Assume each group’s hazards apply to its participants who remain at risk. Calculate the overall probability of dropout at visit 2.

Within Group A, the probability of reaching visit 2 and then dropping out is $$ (1-0.05)(0.10)=0.95(0.10)=0.095. $$ Within Group B, it is $$ (1-0.20)(0.25)=0.80(0.25)=0.200. $$

With equal group proportions, the marginal probability is $$ P(D=2)=0.5(0.095)+0.5(0.200)=0.1475. $$ For comparison, the initial average hazards are $0.5(0.05)+0.5(0.20)=0.125$ at visit 1 and $0.5(0.10)+0.5(0.25)=0.175$ at visit 2. Multiplying the complements of the first average hazard by the second average hazard gives $0.875(0.175)=0.1531$, not $0.1475$. After visit 1, the group mix in the risk set has changed: the higher-dropout Group B has lost a larger fraction of its members. Thus an overall dropout-time probability must account for group-specific paths or changing risk-set composition.

Interpretation, Checks, and Limitations

A hazard is conditional on still being observed immediately before a visit. It is not a cumulative probability, nor is it generally the fraction of the original cohort that has dropped out by that visit. For example, the probability of having dropped out by visit $t$ is $P(D\leq t)=1-\prod_{j=1}^{t}(1-h_j)$ when the hazards are fixed within the group considered. The hazard at visit $t$ is only the conditional chance of dropout at that visit among those who reached it.

Useful checks include verifying that each risk-set denominator contains only participants still responding before that visit, confirming that estimated hazards lie between 0 and 1, and checking that the dropout-time probabilities plus the completion probability sum to 1. A mismatch in the final check often signals an omitted dropout time or an inconsistent visit convention. Nonmonotone response histories, such as a participant missing visit 2 but returning at visit 3, do not fit this permanent-dropout setup without a different response model.

Hazards summarize the distribution of response patterns. By themselves, they do not specify a model for the outcomes, establish why participants drop out, or determine what unobserved outcomes would have been. The earlier material on identifiability is relevant to interpreting assumptions about unobserved data. Pattern-mixture and selection approaches organize outcome and response distributions differently; the hazards here describe response timing, rather than supplying either approach’s complete outcome model.

A frequent mistake is to treat a visit-specific hazard as the probability of dropout by that visit, or to use the original cohort size as the denominator at every visit. The correction is to condition on the visit-specific risk set and then accumulate the hazards through the product of conditional survival probabilities. Another common error is to multiply hazards averaged across groups as if the risk-set composition stayed fixed. The third worked example shows why that shortcut can fail.

Finally, hazards based on small risk sets can be unstable, especially late in follow-up; report the risk-set sizes alongside estimates. Administrative study closure can also end observation without participant dropout. It should not automatically be counted as dropout: the event and the follow-up window must be defined to match the scientific question.

Check Your Understanding

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

With two visits and hazards $h_1=0.30$ and $h_2=0.10$, calculate the probabilities of dropout at visits 1 and 2 and of completing follow-up.
In a risk set of 80 participants before visit 2, 12 drop out. What is the empirical visit-2 hazard, and what does it mean?
Explain why a hazard of $0.20$ at visit 3 does not necessarily mean that 20% of the original cohort drops out at visit 3.
For two equally sized groups, hazards are $(0.10,0.20)$ in Group A and $(0.30,0.40)$ in Group B. Calculate the marginal probability of dropout at visit 2.

Solutions. (1) $P(D=1)=0.30$, $P(D=2)=(0.70)(0.10)=0.070$, and $P(D=3)=(0.70)(0.90)=0.630$; these sum to 1. (2) $\hat h_2=12/80=0.150$, meaning 15% of those still responding before visit 2 drop out at that visit. (3) The hazard is conditional on being in the risk set immediately before visit 3, which may be smaller than the original cohort; the unconditional probability also depends on the chance of reaching visit 3. (4) Group A contributes $0.5(0.90)(0.20)=0.090$, and Group B contributes $0.5(0.70)(0.40)=0.140$. The marginal probability is $0.090+0.140=0.230$.