Independence and Instrument Assignment Mechanisms

Why the Assignment Mechanism Matters

An instrument $Z$ is useful only if its variation can support a causal comparison, not merely because it predicts treatment. The earlier material on relevance describes the instrument–treatment contrast. Independence addresses a different question: how was the instrument assigned in relation to the potential treatment and outcome responses? If assignment tends to place people with different potential responses into different instrument groups, then those groups may differ even apart from any effect of the instrument or treatment.

This tutorial considers a binary instrument and binary treatment. The central idea applies more broadly, but the binary case makes the assignment mechanism and its implications easy to see. Independence is a condition on potential outcomes, which are not all observed for any one person; it is therefore not the same as checking whether observed outcomes look similar across instrument groups.

Definition: Let $Z\in\{0,1\}$ be the instrument, $D(z)\in\{0,1\}$ the treatment that would be received if the instrument were set to $z$, and $Y(d)$ the outcome that would occur if treatment were set to $d$. The independence assumption is $$ Z\ \perp\!\!\!\perp\ \bigl(D(0),D(1),Y(0),Y(1)\bigr). $$ It says that instrument assignment is statistically independent of the joint vector of treatment and outcome potential outcomes. In particular, the instrument is independent of each component of that vector.

The joint-vector formulation matters. It rules out dependence between assignment and any combination of a person's potential treatment choices and potential outcomes, not just dependence with one selected potential outcome. For a binary instrument with both assignment levels occurring, an equivalent distributional statement is that the potential-outcome vector has the same distribution in the $Z=1$ and $Z=0$ groups.

\mathcal{L}\!\left(D(0),D(1),Y(0),Y(1)\mid Z=1\right) = \mathcal{L}\!\left(D(0),D(1),Y(0),Y(1)\mid Z=0\right).

Here $\mathcal{L}$ denotes a probability distribution. Independence does not require that every individual have identical potential outcomes; it requires that the instrument groups do not differ in the distribution of those potential outcomes. Nor does the assumption say that $Z$ is unrelated to the observed treatment: relevance is specifically about whether assignment changes treatment. The independence condition concerns potential responses rather than the treatment actually observed.

What Independence Implies

A useful consequence follows directly from the definition. If $Z$ is independent of the potential-outcome vector, then conditioning on $Z$ does not change the distribution of any function of that vector. In particular, the distributions and means of $D(0)$, $D(1)$, $Y(0)$, and $Y(1)$ are the same in the two instrument groups.

Implication: Under independence, for any potential-outcome component $W$ and any set of possible values $A$, $$ P\bigl(W\in A\mid Z=1\bigr)=P\bigl(W\in A\mid Z=0\bigr)=P(W\in A), $$ whenever the conditioning probabilities are defined. If $W$ has a finite mean, it follows that $E(W\mid Z=1)=E(W\mid Z=0)=E(W)$.

For example, if $Y(0)$ is binary, its mean is its probability of equaling 1. Independence therefore implies $P(Y(0)=1\mid Z=1)=P(Y(0)=1\mid Z=0)$. This conclusion concerns the counterfactual outcome under no treatment. It is not generally possible to calculate these probabilities from observed outcomes alone.

Worked Example: Reading the Independence Condition

Suppose assignment $Z$ is independent of the full potential-outcome vector. In the population, 30% would have $Y(0)=1$, and 70% would have $D(1)=1$. Find these probabilities within each instrument group.

By independence, conditioning on either instrument value leaves the distribution of each potential-outcome component unchanged: $$ P(Y(0)=1\mid Z=z)=P(Y(0)=1)=0.30 $$ and $$ P(D(1)=1\mid Z=z)=P(D(1)=1)=0.70,\qquad z\in\{0,1\}. $$

Thus, among people with $Z=1$, the proportions with $Y(0)=1$ and $D(1)=1$ are 0.30 and 0.70, respectively; the same proportions hold among people with $Z=0$. These are implications of the assumed independence, not estimates from observed data. In particular, the example does not assume that $Y(0)$ and $D(1)$ are independent of each other.

Why Randomized Assignment Supports Independence

In a randomized encouragement study, the assignment rule uses a random mechanism rather than a person's potential responses to determine $Z$. If the randomization is implemented as specified, and the random draw is independent of the participants' potential treatment and outcome responses, then the assignment variable is independent of the potential-outcome vector. This is the design-based reason random assignment can support the independence assumption.

Randomization does not guarantee exact equality of potential-outcome distributions in a particular finite sample. Chance imbalances can occur, especially in a small study. The assumption concerns the assignment mechanism: across repetitions of the randomization, assignment does not systematically favor people with particular potential responses. If a fixed number of participants is assigned to each group, different participants' assignments may be dependent on one another, while each participant's assignment can still be randomized independently of that participant's potential outcomes.

Worked Example: A Randomized Encouragement

Consider a hypothetical study in which each participant is assigned $Z=1$ with probability 0.5 using a random draw independent of that participant's potential responses. Suppose 40% of the population would take treatment if assigned $Z=1$, so $P(D(1)=1)=0.40$, and 25% would have outcome 1 under no treatment, so $P(Y(0)=1)=0.25$. What do independence and random assignment imply?

Because the random draw is independent of the potential-outcome vector, $$ P(D(1)=1\mid Z=1)=P(D(1)=1\mid Z=0)=0.40 $$ and $$ P(Y(0)=1\mid Z=1)=P(Y(0)=1\mid Z=0)=0.25. $$

For instance, the joint probability of $Z=1$ and $Y(0)=1$ is $$ P(Z=1,Y(0)=1)=P(Z=1)P(Y(0)=1)=0.50(0.25)=0.125. $$ The probability of $Z=1$ and $D(1)=1$ is likewise $0.50(0.40)=0.20$. These calculations use the stated independence; they do not assert that the observed outcome under the assigned instrument has probability 0.25. That observed outcome may depend on the treatment received, and other assumptions are needed to relate it to the potential outcomes.

How Observational Assignment Can Create Dependence

In an observational setting, instrument status may result from choices, institutions, geography, eligibility rules, or other processes that are related to participants' characteristics. If those characteristics also predict potential treatment or outcome responses, instrument status can be associated with the potential-outcome vector. This is a problem for independence even when the instrument strongly predicts treatment.

The following example makes the required assumptions explicit. It shows one mechanism: a measured severity variable affects both the chance of receiving the instrument and the distribution of a potential outcome. For the mixture calculations to be valid, we specify that within each severity category, assignment is independent of $Y(0)$. The assignment and outcome probabilities alone would not establish that condition.

Worked Example: Severity-Related Instrument Assignment

In a hypothetical population, 40% of patients have high baseline severity and 60% have low severity. The instrument is assigned with probabilities $P(Z=1\mid\text{high})=0.80$ and $P(Z=1\mid\text{low})=0.20$. The potential outcome probabilities are $P(Y(0)=1\mid\text{high})=0.30$ and $P(Y(0)=1\mid\text{low})=0.05$. Assume, in addition, that $Z$ and $Y(0)$ are conditionally independent within each severity category. Calculate the distribution of $Y(0)$ by instrument group.

First calculate the probability of each instrument group: $$ P(Z=1)=0.40(0.80)+0.60(0.20)=0.32+0.12=0.44, $$ $$ P(Z=0)=1-0.44=0.56. $$ Among those with $Z=1$, the fraction with high severity is $$ P(\text{high}\mid Z=1)=\frac{0.40(0.80)}{0.44} =\frac{0.32}{0.44}\approx0.7273. $$ By the stated conditional-independence assumption, the probability of $Y(0)=1$ within each severity-by-assignment group equals the given severity-specific probability. Therefore, $$ \begin{aligned} P(Y(0)=1\mid Z=1) &=(0.7273)(0.30)+(1-0.7273)(0.05)\\ &\approx0.2318. \end{aligned} $$

For $Z=0$, the high-severity fraction is $$ P(\text{high}\mid Z=0)=\frac{0.40(0.20)}{0.56} =\frac{0.08}{0.56}\approx0.1429. $$ Using the same explicitly stated conditional-independence assumption, $$ \begin{aligned} P(Y(0)=1\mid Z=0) &=(0.1429)(0.30)+(1-0.1429)(0.05)\\ &\approx0.08571. \end{aligned} $$ The difference is $0.2318-0.08571\approx0.1461$, or about 14.61 percentage points. Thus $Y(0)$ has different distributions across the instrument groups in this hypothetical population. The instrument is not independent of $Y(0)$, despite the assumption of independence within severity categories. The difference arises because severity is distributed differently across the two groups.

The added conditional-independence statement in that example is essential. The given severity-specific probabilities describe $Y(0)$ by severity, and the assignment rates describe $Z$ by severity; without a condition linking $Z$ and $Y(0)$ within severity, neither probability tells us the distribution of $Y(0)$ among people with a particular value of $Z$. The next example isolates this issue.

Worked Example: Why Marginal Probabilities Are Not Enough

Within one severity category, suppose $P(Z=1)=0.50$ and $P(Y(0)=1)=0.20$. The joint probability is $P(Z=1,Y(0)=1)=0.16$. Calculate the potential-outcome probability in each instrument group and compare it with what independence would require.

The remaining joint probabilities follow from the stated margins: $$ P(Z=1,Y(0)=0)=0.50-0.16=0.34, $$ $$ P(Z=0,Y(0)=1)=0.20-0.16=0.04, $$ and $$ P(Z=0,Y(0)=0)=1-0.16-0.34-0.04=0.46. $$ Thus, $$ P(Y(0)=1\mid Z=1)=\frac{0.16}{0.50}=0.32, \qquad P(Y(0)=1\mid Z=0)=\frac{0.04}{0.50}=0.08. $$

The two conditional probabilities differ, so $Z$ and $Y(0)$ are dependent in this category. If they were independent, the joint probability would instead be the product of the margins: $$ P(Z=1,Y(0)=1)=P(Z=1)P(Y(0)=1)=0.50(0.20)=0.10. $$ The stated margins, 0.50 and 0.20, do not determine the joint probability: the value 0.16 is compatible with those margins but does not satisfy independence. This is why a mixture calculation based only on marginal severity-specific outcome and assignment probabilities needs an additional within-severity condition.

What Independence Does—and Does Not—Establish

Independence is a substantive condition about the assignment mechanism and potential responses. In a randomized design, it is supported by how the instrument is assigned. In an observational setting, it cannot usually be guaranteed just by naming a variable an instrument or showing that it predicts treatment. Assignment may share causes with potential outcomes, or depend directly on characteristics associated with them.

Observed balance in measured baseline variables can be informative about the assignment process, but it does not prove independence from the full potential-outcome vector. Potential outcomes are not jointly observed, and unmeasured characteristics may still generate dependence. Conversely, a chance difference in observed baseline variables in a randomized sample does not by itself show that the randomization mechanism failed; finite samples can be imbalanced by chance.

Independence is also only one part of an instrumental-variable argument. By itself, it does not establish that changing the instrument changes treatment, nor does it say how the instrument may affect outcomes. Those are separate conditions. Keeping these questions distinct prevents a strong first-stage relationship from being mistaken for evidence that assignment is independent of potential responses.

Common Mistake

Wrong reasoning: “The potential outcome probability is 0.30 among high-severity patients, so it is 0.30 among high-severity patients assigned $Z=1$.” This conclusion silently assumes that assignment and the potential outcome are independent within the severity group.
Correction: The within-severity conditional probability is not determined by the severity-specific marginal probability alone. State conditional independence explicitly, or provide the outcome probability separately for each severity-by-assignment group. Otherwise, the mixture calculation is unsupported.
Another error: Treating an instrument–treatment association as evidence of independence. Relevance concerns whether treatment varies with the instrument; independence concerns whether assignment is unrelated to potential responses. One does not imply the other.

Key takeaway: Independence means that instrument assignment is unrelated to the joint distribution of treatment and outcome potential outcomes. Randomization can support this condition by design. For observational assignment, the mechanism may create dependence, and marginal probabilities cannot be combined into conditional probabilities without an explicit assumption or the necessary joint data.

Check Your Understanding

Answer each question using the distinction between an assignment mechanism and a potential-outcome distribution.

Write the independence assumption using $Z$, $D(0)$, $D(1)$, $Y(0)$, and $Y(1)$.
If independence holds and $P(Y(1)=1)=0.42$, what is $P(Y(1)=1\mid Z=0)$? Explain.
Why can randomized assignment support independence even when the two realized groups have different sample averages of a baseline variable?
In the severity example, what additional condition justified using the severity-specific probabilities in the instrument-group mixture calculations?
Can equal marginal probabilities $P(Z=1)$ and $P(Y(0)=1)$ establish independence? State what further probability is needed to check independence.