Conditional Versus Marginal Treatment Probabilities

Why the Distinction Matters

At each decision time, a treatment probability can answer different questions. One probability describes treatment assignment for a person with a particular observed history; another averages over histories to describe a broader group. Both are legitimate quantities, but they are not interchangeable. Longitudinal inverse probability weights use the history-conditional probabilities in their denominators. A numerator may instead use probabilities averaged over covariate histories, while retaining dependence on prior treatment.

The distinction matters because treatment assignment can depend strongly on covariates. A marginal probability can look moderate even when, within some histories, treatment is nearly certain or nearly impossible. Averaging first can therefore obscure the very assignment patterns that the weights must account for.

Definitions at a Decision Time

Consider decision times $t=1,\ldots,T$. Let $A_t$ be treatment at time $t$, and let $\bar A_t=(A_1,\ldots,A_t)$ denote treatment history through time $t$. Let $L_t$ be the covariate information available at time $t$, and let $\bar L_t=(L_0,\ldots,L_t)$ include baseline covariates $L_0$. Just before treatment $A_t$ is assigned, the observed history is $H_t=(\bar A_{t-1},\bar L_t)$, where $\bar A_0$ is empty.

Definition: The conditional treatment probability at time $t$ is $g_t(a_t\mid\bar a_{t-1},\bar l_t)=P(A_t=a_t\mid\bar A_{t-1}=\bar a_{t-1},\bar L_t=\bar l_t)$. It gives the probability of the current treatment value among individuals with the specified treatment and covariate history. A marginal treatment probability given prior treatment is $q_t(a_t\mid\bar a_{t-1})=P(A_t=a_t\mid\bar A_{t-1}=\bar a_{t-1})$; it averages over the covariate histories observed among people with that prior treatment history.

The word “marginal” here means marginal with respect to the covariates being averaged out. The quantity $q_t$ can still depend on prior treatment. In particular, at $t>1$, it is generally not the unconditional probability $P(A_t=a_t)$. If a numerator retains baseline covariates $V$, a corresponding choice is $P(A_t=a_t\mid\bar A_{t-1},V)$: it is marginal over the other covariates, not necessarily over every covariate.

These probabilities are defined by the observed treatment-assignment distribution. Estimating them is distinct from establishing a causal interpretation for a weighted analysis. For that interpretation, assumptions such as sequential exchangeability and positivity are needed. In particular, sequential positivity requires that each treatment value under consideration have positive conditional probability at histories where that treatment is to be compared.

How Marginal Probabilities Are Obtained

A marginal probability is not usually obtained by taking an unweighted average of fitted probabilities across the entire sample. It averages over the covariate histories conditional on the specified prior treatment history. For discrete covariate histories, the law of total probability gives the required calculation.

Result: For discrete $\bar L_t$, $$ q_t(a_t\mid\bar a_{t-1}) = \sum_{\bar l_t} g_t(a_t\mid\bar a_{t-1},\bar l_t) P(\bar L_t=\bar l_t\mid\bar A_{t-1}=\bar a_{t-1}). $$ For continuous covariates, the sum is replaced by an integral with respect to the conditional distribution of $\bar L_t$ given $\bar A_{t-1}=\bar a_{t-1}$.

To derive this result, partition the event $A_t=a_t$, conditional on the prior treatment history, according to the possible covariate histories. For each history $\bar l_t$, the joint conditional probability is the conditional treatment probability times the probability of that history. Summing over the partition gives

\begin{aligned} P(A_t=a_t\mid\bar A_{t-1}=\bar a_{t-1}) &=\sum_{\bar l_t}P(A_t=a_t,\bar L_t=\bar l_t\mid\bar A_{t-1}=\bar a_{t-1})\\ &=\sum_{\bar l_t}P(A_t=a_t\mid\bar A_{t-1}=\bar a_{t-1},\bar L_t=\bar l_t) P(\bar L_t=\bar l_t\mid\bar A_{t-1}=\bar a_{t-1}). \end{aligned}

The second equality is the product rule for conditional probabilities. The covariate-history probabilities in this calculation depend on the prior treatment history. Consequently, the same conditional treatment model can yield different marginal probabilities in different prior-treatment groups. This derivation assumes the conditioning events have positive probability; for continuous covariates, it is understood as integration over their conditional distribution. $\square$

Worked Example: Averaging Across a Baseline Covariate

Suppose $L_0$ is binary, $P(L_0=1)=0.5$, and the probability of treatment at the only decision time is $0.8$ when $L_0=1$ and $0.2$ when $L_0=0$. Find the marginal probability of treatment.

The conditional probabilities are $g_1(1\mid L_0=1)=0.8$ and $g_1(1\mid L_0=0)=0.2$. The marginal probability averages them using the covariate distribution:

$$ P(A_1=1)=0.8(0.5)+0.2(0.5)=0.4+0.1=0.5. $$

Thus the marginal probability is $0.5$, even though neither covariate-specific probability is $0.5$. For someone with $L_0=1$, the relevant conditional probability remains $0.8$; substituting the marginal $0.5$ would not describe that person's treatment assignment probability.

Which Probabilities Enter Longitudinal Weights?

For a treatment history $\bar A_T$, the denominator of the ordinary inverse probability of treatment weight is built from the probability of each observed treatment conditional on the history available at that decision. For a discrete treatment, its form through time $T$ is

W_T=\prod_{t=1}^{T}\frac{1}{g_t(A_t\mid\bar A_{t-1},\bar L_t)}.

The factor at time $t$ uses the probability of the treatment actually received, evaluated at that individual's prior treatments and covariates. This is a history-specific probability, not $P(A_t)$ and not $q_t(A_t\mid\bar A_{t-1})$. Replacing it with a marginal probability would generally fail to account for differences in assignment across covariate histories.

A stabilized construction uses a numerator probability that omits some or all of the covariates from the denominator. A common numerator factor is $q_t(A_t\mid\bar A_{t-1})$, the marginal probability conditional on prior treatment. The corresponding product has the form

SW_T=\prod_{t=1}^{T}\frac{q_t(A_t\mid\bar A_{t-1})}{g_t(A_t\mid\bar A_{t-1},\bar L_t)}.

This expression identifies the distinct roles of the two probabilities: the denominator conditions on the treatment and covariate history, whereas the numerator averages over covariates according to their distribution within the prior-treatment group. A numerator may retain selected covariates by design; it need not be marginal over every measured variable.

Result: In a finite-support population with positive conditional treatment probabilities, weighting by the inverse conditional probability makes treatment values equally likely within each fixed observed history at a decision time. For binary treatment, each value then has conditional probability $1/2$, wherever both values are possible.

To verify this, fix a history $H_t=h$ with positive probability and suppose there are $K$ possible treatment values, each with positive conditional probability $g_t(a\mid h)$. Conditional on $H_t=h$, the original probability of treatment $a$ is $g_t(a\mid h)$. Multiplying each observation's contribution by $1/g_t(A_t\mid h)$, the unnormalized contribution for treatment $a$ is

g_t(a\mid h)\frac{1}{g_t(a\mid h)}=1.

Summing these contributions over the $K$ possible treatment values gives $K$. Therefore, after normalization, the weighted conditional probability of each value is $1/K$. In the binary case $K=2$, it is $1/2$. With weights accumulated over earlier times, the earlier weight is fixed once the earlier history is fixed, so it cancels in this same conditional calculation. This result concerns conditional treatment assignment at a fixed history; it does not by itself establish a causal effect or guarantee that the weights are stable. $\square$

Worked Example: A Marginal Probability Depends on Prior Treatment

At time 2, suppose $L_1$ is binary. Among people with $A_1=1$, let $P(L_1=1\mid A_1=1)=0.75$. Their conditional probabilities of $A_2=1$ are $0.8$ when $L_1=1$ and $0.2$ when $L_1=0$. Find $P(A_2=1\mid A_1=1)$.

Average the conditional probabilities over the covariate distribution in the $A_1=1$ group. The probability of $L_1=0$ in this group is $1-0.75=0.25$, so

P(A_2=1\mid A_1=1)=0.8(0.75)+0.2(0.25)=0.60+0.05=0.65.

Suppose instead that $P(L_1=1\mid A_1=0)=0.25$. The marginal probability in that prior-treatment group is

P(A_2=1\mid A_1=0)=0.8(0.25)+0.2(0.75)=0.20+0.15=0.35.

The conditional treatment model is the same in both groups, but the marginal probabilities differ because the covariate distributions differ. If $P(A_1=1)=0.5$, then the unconditional probability is $P(A_2=1)=0.5(0.65)+0.5(0.35)=0.50$. That unconditional value is neither prior-treatment-specific marginal probability.

Worked Example: Selecting the Denominator for an Observed History

Suppose a person has $A_1=1$, $L_1=1$, and then receives $A_2=0$. The fitted conditional probabilities are $g_1(1\mid L_0)=0.8$ and $g_2(0\mid A_1=1,L_0,L_1=1)=0.3$. Suppose also that $P(A_1=1)=0.5$ and $P(A_2=0\mid A_1=1)=0.4$. Calculate the two-time unstabilized and stabilized weights for this history.

The unstabilized denominator uses the conditional probabilities of the treatments actually received:

W_2=\frac{1}{0.8(0.3)}=\frac{1}{0.24}=4.1667\text{ (approximately)}.

The stabilized numerator uses the specified marginal probabilities for the same observed treatments, while the denominator is unchanged:

SW_2=\frac{0.5(0.4)}{0.8(0.3)}=\frac{0.20}{0.24}=0.8333\text{ (approximately)}.

The value $0.4$ is conditional on $A_1=1$, not the unconditional probability of $A_2=0$. The value $0.3$ is conditional on the person's covariate and treatment history. Using $0.4$ in the denominator would discard that history-specific assignment information.

Diagnostics, Alternatives, and Common Pitfalls

A practical diagnostic is to compare treatment probabilities across covariate histories, not only their overall averages. For each decision time, inspect the fitted conditional probabilities for the observed treatment and look for histories where they are close to zero. A healthy-looking marginal probability can conceal such histories: averaging $0.8$ and $0.2$, for example, gives $0.5$, but does not remove the low probability $0.2$ in one subgroup. Small conditional probabilities can generate large weights and signal a positivity problem or an inadequately supported treatment comparison.

The main alternative to a history-conditional denominator is a model using a smaller conditioning set, perhaps only prior treatment. That produces a marginal treatment probability, but it is not a substitute for the standard denominator when covariates predict treatment and are needed to address treatment-confounder association. Conversely, using a fully conditional numerator is possible, but it retains more history information in the numerator and changes the weighting scheme. The choice of numerator is a design decision; it does not change which conditional assignment probabilities define the usual denominator.

Common mistake: “The overall treatment rate is the probability needed for everyone.” The overall rate averages over covariate and treatment histories. The denominator requires the probability conditional on the individual's observed history.
Common mistake: averaging over the wrong group. To calculate $P(A_t=a_t\mid\bar A_{t-1})$, average over covariate histories conditional on that prior treatment history, not over the full sample without conditioning.
Common mistake: assuming a moderate marginal probability proves positivity. Positivity concerns conditional probabilities at relevant histories. Marginalization can hide values near zero.

The formulas also have limitations. Estimated conditional probabilities can be unstable when histories are sparse, and a misspecified treatment model can give inaccurate probabilities even if its marginal predictions look plausible. Marginal probabilities require the relevant covariate-history distribution as well as the conditional treatment model. Neither marginalization nor weighting creates information in histories where a treatment is never observed.

Check Your Understanding

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

If $P(L=1)=0.4$, $P(A=1\mid L=1)=0.75$, and $P(A=1\mid L=0)=0.25$, what is $P(A=1)$?
Why is $P(A_t=a_t\mid\bar A_{t-1})$ generally different from $P(A_t=a_t)$ when $t>1$?
State the conditioning history in the denominator probability for treatment $A_t$ in a longitudinal inverse probability weight.
Can a marginal treatment probability of $0.5$ rule out a conditional positivity problem? Explain.

Solutions. 1. $P(A=1)=0.75(0.4)+0.25(0.6)=0.30+0.15=0.45$. 2. The conditional probability averages over covariates among people with a specified prior treatment history; the unconditional probability also averages over prior treatment histories, which can have different covariate distributions and treatment probabilities. 3. The denominator is $P(A_t=a_t\mid\bar A_{t-1},\bar L_t)$, evaluated at the individual's observed prior treatments and covariates. 4. No. A marginal value averages conditional probabilities and may conceal histories where a treatment probability is near zero, so positivity must be assessed conditionally.