Binary Instruments And Wald Estimands

Why a Binary Instrument Produces a Ratio

When an instrument takes only two values, its role in a just-identified IV model can be expressed with a simple comparison: how much does the mean outcome differ between the two instrument groups, and how much does the mean endogenous regressor differ? The ratio of those contrasts is the Wald estimand. Its simplicity makes the algebra transparent, but the ratio identifies a structural coefficient only under the model and instrument conditions that justify it.

Let \(Z\in\{0,1\}\) be a binary instrument, \(D\) a scalar endogenous regressor, and \(Y\) an outcome. Start with the structural equation

Y=\alpha+\beta D+u,

where \(\alpha\) is an intercept, \(\beta\) is the structural coefficient of interest, and \(u\) contains the remaining determinants of \(Y\). The instrument must be relevant, so the mean of \(D\) differs across instrument groups. It must also satisfy the required exogeneity and exclusion condition in this structural equation: the conditional mean of \(u\) must be the same at both values of \(Z\). This is a conditional-mean statement, not a claim that \(Z\) is independent of every feature of the data.

Definition: For a binary instrument, the Wald estimand is the difference in conditional mean outcomes across instrument values divided by the corresponding difference in conditional mean values of the endogenous regressor:

\beta_W = \frac{E[Y\mid Z=1]-E[Y\mid Z=0]} {E[D\mid Z=1]-E[D\mid Z=0]}.

The denominator must be nonzero. Under the structural equation above and \(E[u\mid Z=1]=E[u\mid Z=0]\), this ratio equals \(\beta\).

Deriving the Ratio

The derivation uses the structural equation and the instrument’s conditional-mean restriction. Taking the conditional expectation of the structural equation given \(Z=z\) gives

E[Y\mid Z=z] = \alpha+\beta E[D\mid Z=z]+E[u\mid Z=z], \qquad z\in\{0,1\}.

Subtract the expression for \(z=0\) from the expression for \(z=1\). The common intercept cancels. If the conditional mean of the structural error is also the same in both groups, its difference is zero. Therefore,

E[Y\mid Z=1]-E[Y\mid Z=0] = \beta\big(E[D\mid Z=1]-E[D\mid Z=0]\big).

Relevance means \(E[D\mid Z=1]-E[D\mid Z=0]\ne 0\), so dividing both sides by that difference yields \(\beta_W=\beta\). If the error’s conditional mean differs across instrument groups, the numerator also contains that difference; the ratio need not identify \(\beta\). Thus, a nonzero denominator alone does not establish validity.

This is the binary-instrument form of the just-identified IV ratio. The numerator is the reduced-form difference in mean outcomes; the denominator is the first-stage difference in mean endogenous-regressor values. Both are measured across the same two instrument groups.

Worked Example: Participation and Annual Earnings

Suppose, hypothetically, \(Z=1\) denotes eligibility for a training invitation, \(D\) indicates training participation (one for participation and zero otherwise), and \(Y\) is annual earnings in thousands of dollars. In a sample, 80 people have \(Z=1\), with \(\sum D_i=48\) and \(\sum Y_i=2240\). Among 120 people with \(Z=0\), \(\sum D_i=48\) and \(\sum Y_i=3120\). Calculate the sample Wald estimator.

First compute the group means. For the eligible group, \(\bar{D}_1=48/80=0.60\) and \(\bar{Y}_1=2240/80=28.0\) thousand dollars. For the other group, \(\bar{D}_0=48/120=0.40\) and \(\bar{Y}_0=3120/120=26.0\) thousand dollars. The first-stage difference is \(0.60-0.40=0.20\), and the outcome difference is \(28.0-26.0=2.0\) thousand dollars.

\hat{\beta}_W = \frac{\bar{Y}_1-\bar{Y}_0}{\bar{D}_1-\bar{D}_0} = \frac{28.0-26.0\ \text{thousand dollars}} {0.60-0.40} = 10.0\ \text{thousand dollars}.

The estimate is 10.0 thousand dollars in annual earnings per one-unit change in the participation indicator. The denominator is nonzero, so the sample ratio is defined. This calculation does not establish that eligibility is a valid instrument: the structural model and equal-error-conditional-means condition are assumptions that must be justified separately. Conditional on those assumptions, the ratio estimates the structural coefficient.

From Population Estimand to Sample Estimator

The population Wald estimand uses conditional expectations. In data, replace each conditional mean with the sample mean for the corresponding instrument group. Let \(n_1\) and \(n_0\) be the numbers of observations with \(Z=1\) and \(Z=0\), respectively, and let \(\bar{Y}_z\) and \(\bar{D}_z\) denote the sample means in group \(z\). When both groups are observed and the sample first-stage difference is nonzero, the sample Wald estimator is

\hat{\beta}_W = \frac{\bar{Y}_1-\bar{Y}_0}{\bar{D}_1-\bar{D}_0}.

The group means are calculated separately, so unequal group sizes do not require an additional adjustment in this formula. Equivalently, the sample estimator solves the just-identified IV moment equation with an intercept and instrument \(Z\). To see the equivalence, write the sample moment after removing the intercept as

\sum_{i=1}^{n}(Z_i-\bar{Z})(Y_i-\beta D_i)=0.

For binary \(Z\), \(\bar{Z}=n_1/n\). For any variable \(A\), its centered cross-product with \(Z\) is

\sum_{i=1}^{n}(Z_i-\bar{Z})A_i = \frac{n_1n_0}{n}(\bar{A}_1-\bar{A}_0).

Indeed, the contribution from the \(Z=1\) group is \((1-n_1/n)n_1\bar{A}_1=(n_0/n)n_1\bar{A}_1\); the contribution from the \(Z=0\) group is \(-(n_1/n)n_0\bar{A}_0\). Adding them gives the displayed expression. Apply it first to \(Y\), then to \(D\). Solving the moment equation gives the ratio of the two group-mean differences, provided the denominator is nonzero.

Worked Example: A Negative First Stage

In a hypothetical study, \(Z=1\) indicates assignment to a prescribing protocol. The endogenous regressor \(D\) is a daily medication dose in milligrams, and \(Y\) is a symptom score in points. Suppose the instrument groups have means \(\bar{D}_1=16\) mg and \(\bar{D}_0=20\) mg, while their mean scores are \(\bar{Y}_1=5.8\) points and \(\bar{Y}_0=7.0\) points. Assume the structural model and instrument conditions are maintained for this calculation.

The first-stage difference is \(16-20=-4\) mg, and the reduced-form difference is \(5.8-7.0=-1.2\) points. Both differences are negative, so their ratio is positive:

\hat{\beta}_W = \frac{-1.2\ \text{points}}{-4\ \text{mg}} = 0.300\ \text{points per mg}.

The negative first stage does not invalidate the arithmetic or require changing the order of the groups. Reversing the order in both differences would change both signs and leave the ratio unchanged. Here the ratio is 0.300 symptom-score points per milligram under the stated model and assumptions. The group means show relevance in this sample because their dose difference is nonzero; they cannot, by themselves, verify the instrument’s exogeneity or exclusion.

Units, Group Sizes, and the Meaning of the Ratio

The units of the Wald estimator follow directly from its construction. If \(Y\) is measured in dollars and \(D\) in hours, the ratio is dollars per hour. If \(D\) is an indicator, its group-mean difference is a difference in proportions and has no physical unit; the ratio then has the units of \(Y\). State the units of both contrasts before interpreting their ratio, and retain the same group ordering in numerator and denominator.

A useful check is to ask what the ratio compares. It is not the mean outcome in the \(Z=1\) group divided by the mean outcome in the \(Z=0\) group. It is the difference between group means for \(Y\), divided by the difference between group means for \(D\). Nor is it generally a simple difference in outcomes: the denominator rescales the outcome contrast by the observed first-stage contrast.

Worked Example: Why Group Means Matter When Group Sizes Differ

Consider hypothetical group summaries with \(n_1=50\), \(n_0=100\), \(\bar{Y}_1=14\), \(\bar{Y}_0=12\) points, \(\bar{D}_1=0.35\), and \(\bar{D}_0=0.25\). The sample Wald estimator is

\hat{\beta}_W = \frac{14-12\ \text{points}}{0.35-0.25} = \frac{2}{0.10} = 20\ \text{points per unit of }D.

The totals would be \(\sum_{Z=1}Y_i=50(14)=700\), \(\sum_{Z=0}Y_i=100(12)=1200\), \(\sum_{Z=1}D_i=50(0.35)=17.5\), and \(\sum_{Z=0}D_i=100(0.25)=25\). Dividing differences of totals would give \((700-1200)/(17.5-25)=66.7\), not 20. That calculation is wrong because it compares totals that reflect different group sizes rather than the required mean differences.

The estimator’s denominator is \(0.10\), so it is defined for these sample summaries. As in the other examples, a nonzero sample difference is not proof that the population instrument conditions hold. The value 20 has the intended structural interpretation only under the model and validity conditions used to derive the Wald ratio.

What the Wald Ratio Does—and Does Not—Establish

The ratio packages two observable contrasts into a single IV quantity. Under the stated linear structural equation, conditional-mean exogeneity, exclusion, and relevance, it identifies the coefficient \(\beta\). Without those conditions, it remains a ratio of conditional mean differences, but it need not equal a structural parameter. A large or precisely calculated ratio cannot compensate for a failure of instrument validity.

The arithmetic also depends on the denominator. If the first-stage difference is close to zero, the ratio can be numerically large and sensitive to small changes in either group contrast. That fact is a warning about the calculation, not a validity check or an inference procedure. A sample denominator that is nonzero permits division; it does not establish that the population instrument is sufficiently relevant for reliable estimation.

Key takeaway: For a binary instrument, calculate the outcome difference and endogenous-regressor difference using the same \(Z=1\) versus \(Z=0\) comparison, then divide the former by the latter. The interpretation as a structural coefficient additionally requires the structural model and valid-instrument conditions.

Common Mistake

A frequent error is to divide the two group totals, or their differences, and call the result the Wald estimate. For instance, in the unequal-size example, using differences in totals gives \(66.7\), although the ratio of mean differences is \(20\). The totals combine the variable values with the number of observations in each group, so they do not represent the required conditional-mean contrasts. Calculate each group mean first, subtract within each variable, and only then form the ratio.

A second interpretive error is to assume that a nonzero first-stage difference proves the instrument is valid. Relevance is necessary for the denominator, but validity also requires the appropriate exogeneity and exclusion conditions. Those conditions concern the relationship between \(Z\) and the structural error; they are not established by this ratio’s arithmetic.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

Write the population Wald estimand using conditional means of \(Y\) and \(D\) for a binary instrument.
Starting from \(Y=\alpha+\beta D+u\), state the condition on \(E[u\mid Z]\) that allows differencing the two conditional means to identify \(\beta\).
If the outcome difference is 3 points and the first-stage difference is 0.15 units of \(D\), what is the Wald ratio, including its units?
Why must group means be used rather than differences in raw totals when instrument groups have different sample sizes?
Does a nonzero sample first-stage difference establish that the instrument is valid? Explain the distinction between relevance and the other required conditions.