Deriving Matched-Set Assignment Bounds

Why Matched-Set Assignment Bounds Are Needed

Matching makes treated and control units similar on measured covariates, but hidden covariates may still affect who receives treatment. A sensitivity model represents the possible impact of such a covariate by limiting how different treatment odds can be within a matched set. The resulting bounds describe how much the assignment probabilities could vary while respecting that limit.

The key is to condition on the set’s fixed number of treated units. The probability of an assignment is then not obtained by treating each unit’s treatment status as an unrelated probability: assignments must have the required treatment count. This conditioning changes the denominator and is essential to the bounds.

Definition: Consider a matched set of $n$ units with exactly $m$ treated units, where $1\leq m<n$. Let $Z_i=1$ if unit $i$ is treated and $Z_i=0$ otherwise, so $\sum_{i=1}^n Z_i=m$. Under the Gamma model, each unit has treatment odds proportional to a positive weight $w_i$, and the ratio of the largest to smallest weight in the set is at most $\Gamma$, with $\Gamma\geq1$. Conditional on the treatment count, the probability of a treated subset $A$, with $|A|=m$, is $$ P(Z_i=1\text{ exactly for }i\in A\mid \textstyle\sum_i Z_i=m) = \frac{\prod_{i\in A} w_i} {\displaystyle\sum_{B:\,|B|=m}\prod_{j\in B} w_j}. $$

This formulation follows from a specific assignment model: conditional on the matched set and its covariates, units’ treatment indicators are independent before conditioning on the total count, with odds proportional to their weights. Conditioning on exactly $m$ treated units makes the probability of a particular assignment proportional to the product of the odds weights for its treated units. A common scale factor in all the weights cancels, so the Gamma constraint allows the weights to be normalized to $1\leq w_i\leq\Gamma$.

The model does not assert that treatment was randomized in the actual study. Rather, it specifies a family of possible assignment mechanisms: $\Gamma=1$ gives equal weights, while larger $\Gamma$ permits greater hidden-odds imbalance. The bounds below are sharp under this model: their endpoints are attained by allowable choices of weights.

Bounds for a Specified Assignment

Fix a particular treated subset $A$. Its probability is largest when its treated units receive the largest permitted weights and the units outside it receive the smallest weights. To see why, compare any alternative assignment $B$ with $A$. If $B$ replaces $k$ units of $A$ with $k$ units outside $A$, the ratio of its assignment weight to the weight of $A$ is the product of the $k$ incoming weights divided by the product of the $k$ removed weights. Setting all weights in $A$ to $\Gamma$ and all weights outside $A$ to $1$ makes each such ratio as small as allowed. This maximizes the share of total probability assigned to $A$. Reversing those extreme weights minimizes its share.

There are $\binom{m}{k}\binom{n-m}{k}$ alternatives that replace exactly $k$ members of $A$. Relative to $A$, their assignment weights are $\Gamma^{-k}$ at the maximizing configuration and $\Gamma^k$ at the minimizing configuration. Consequently, the sharp bounds are:

\frac{1}{\displaystyle\sum_{k=0}^{\min(m,n-m)} \binom{m}{k}\binom{n-m}{k}\Gamma^k} \ \leq\ P(Z=A\mid \textstyle\sum_i Z_i=m) \ \leq\ \frac{1}{\displaystyle\sum_{k=0}^{\min(m,n-m)} \binom{m}{k}\binom{n-m}{k}\Gamma^{-k}}.

The left endpoint uses weights $1$ for units in $A$ and $\Gamma$ for units outside $A$; the right endpoint uses the reverse. When $\Gamma=1$, both endpoints equal $1/\binom{n}{m}$, the uniform probability of each assignment among the $\binom{n}{m}$ possible assignments.

Worked Example: A Matched Pair

Suppose a matched pair contains one treated unit and one control, and take $\Gamma=2$. There are two possible assignments. Let $A$ denote the assignment in which the first unit is treated. Here $n=2$ and $m=1$, so the sum in the bounds has terms for $k=0$ and $k=1$.

For the lower bound, the denominator is $1+\binom{1}{1}\binom{1}{1}(2)=3$, giving $P(Z=A)\geq 1/3$. For the upper bound, it is $1+\binom{1}{1}\binom{1}{1}(1/2)=1.5$, giving $P(Z=A)\leq 1/1.5=2/3$. Thus either unit’s probability of being the treated member lies between $1/3$ and $2/3$. The bounds are attained when the two units’ weights are respectively $1,2$ or $2,1$.

Worked Example: A Two-Treated Assignment in a Four-Unit Set

Consider a set with $n=4$, $m=2$, and $\Gamma=2$. Focus on one specified pair $A$ as the treated subset. The possible numbers of replacements are $k=0,1,2$.

For the lower bound, the denominator is $\binom{2}{0}\binom{2}{0}(1)+\binom{2}{1}\binom{2}{1}(2)+\binom{2}{2}\binom{2}{2}(4)=1+8+4=13$. Thus the probability of this particular assignment is at least $1/13=0.07692$.

For the upper bound, the denominator is $1+4(1/2)+1(1/4)=3.25$, so the probability is at most $1/3.25=0.3077$. Both are probabilities for one specific assignment, not for either member’s marginal treatment status. There are six possible assignments in this set; when $\Gamma=1$, each has probability $1/6$.

Bounds for One Unit’s Treatment Probability

A different question asks for the marginal probability that a particular unit $i$ is treated, allowing any of the other $m-1$ treated positions to vary. This probability is the sum of the probabilities of all assignments that include $i$. Its sharp bounds are:

\frac{m}{m+\Gamma(n-m)} \ \leq\ P(Z_i=1\mid \textstyle\sum_j Z_j=m) \ \leq\ \frac{\Gamma m}{\Gamma m+n-m}.

One derivation uses the elementary symmetric sums of the weights excluding $i$. Write $e_r(w_{-i})$ for the sum of all products of $r$ distinct weights among the other $n-1$ units, with $e_0=1$. The total assignment weight for assignments including $i$ is $w_i e_{m-1}(w_{-i})$; for assignments excluding $i$, it is $e_m(w_{-i})$. Therefore the marginal treatment odds are $w_i e_{m-1}(w_{-i})/e_m(w_{-i})$.

For any $m$-element subset of the other units, its product appears $m$ times in the sum that counts each $(m-1)$-element subset multiplied by the sum of weights outside it. Hence $$ e_m(w_{-i}) = \frac{1}{m}\sum_{S:\,|S|=m-1} \left(\prod_{j\in S}w_j\right) \left(\sum_{j\notin S,\,j\ne i}w_j\right). $$

There are $n-m$ weights in each inner sum, and each lies between $1$ and $\Gamma$. It follows that $(n-m)e_{m-1}(w_{-i})/m\leq e_m(w_{-i})\leq\Gamma(n-m)e_{m-1}(w_{-i})/m$. Using $1\leq w_i\leq\Gamma$, the marginal treatment odds are therefore between $m/[\Gamma(n-m)]$ and $\Gamma m/(n-m)$. Converting odds $o$ to probability $o/(1+o)$ gives the stated bounds. The endpoints are attainable: set $w_i=1$ and all other weights to $\Gamma$ for the lower endpoint, or set $w_i=\Gamma$ and all others to $1$ for the upper endpoint.

Worked Example: A Unit’s Marginal Probability in a Five-Unit Set

Suppose five units form a matched set with exactly two treated units, and $\Gamma=1.5$. For a particular unit $i$, the marginal probability bounds are $$ \frac{2}{2+1.5(3)}=\frac{2}{6.5}=0.3077 \quad\text{and}\quad \frac{1.5(2)}{1.5(2)+3}=\frac{3}{6}=0.5000. $$

Under equal weights, the probability is $m/n=2/5=0.4000$. The Gamma model permits the unit’s probability to range from $0.3077$ to $0.5000$, depending on its weight relative to the other four units. These are bounds on one unit’s marginal probability, not the probability of one specified pair of treated units.

Reading the Bounds and Checking the Model

The distinction between a complete assignment and one unit’s treatment status matters in interpretation. A specific assignment probability concerns all $m$ treated positions at once; a marginal probability sums over many assignments. Their bounds are not interchangeable. The two-unit example happens to make the marginal and assignment probabilities coincide because each possible assignment determines exactly which unit is treated.

Useful checks include verifying that $0\leq m\leq n$, that $\Gamma\geq1$, and that the bounds collapse to the uniform probabilities when $\Gamma=1$. For the marginal bounds, substituting $\Gamma=1$ gives $m/n$ at both endpoints. For assignment bounds, the denominator at $\Gamma=1$ is $\binom{n}{m}$, by counting all assignments according to their number $k$ of replacements from $A$. A bound that fails these checks usually reflects a count error or a confusion between odds and probabilities.

The fixed-count assumption must also match the design being analyzed. If sets have different treated counts, use the relevant $m$ for each set; if the treatment count is not fixed, the conditional denominator above does not describe the assignment mechanism. Further, the Gamma restriction is a sensitivity assumption, not something established merely by matching. It limits relative hidden treatment odds but does not prove that hidden bias is absent or that a particular value of $\Gamma$ is realistic.

Common pitfall: A common error is to assign each unit probability $m/n$ even when $\Gamma>1$, because there are $m$ treated units among $n$. That value is correct under equal weights, $\Gamma=1$, but hidden-odds imbalance can shift an individual’s marginal probability within the bounds above. A related error is to use the marginal bounds as bounds for a specified full assignment. The full-assignment probability requires the sum over alternative subsets and is generally much smaller.

The main alternative is to work with an unconditional independent Bernoulli assignment model, in which the treatment count can vary. That model answers a different probability question and should not be substituted for a fixed-count matched-set model. For a pair with one treated unit, the general assignment bound reduces to the familiar interval $1/(1+\Gamma)$ to $\Gamma/(1+\Gamma)$. For larger sets, the combinatorial terms account for all ways that treatment positions can be exchanged. These bounds provide the assignment-probability inputs to sensitivity analyses; they do not by themselves calculate a test statistic or establish a causal effect.

Check Your Understanding

Use the definitions and formulas above. Solutions follow each question.

With $\Gamma=1$, what is the probability of each assignment when $n=5$ and $m=2$? Solution: Equal weights make all $\binom{5}{2}=10$ assignments equally likely, so each has probability $1/10=0.1000$.
For $n=3$, $m=1$, and $\Gamma=2$, find the lower and upper bounds for one specified assignment. Solution: The sum has $k=0,1$. The lower bound is $1/[1+\binom{1}{1}\binom{2}{1}(2)]=1/5=0.2000$. The upper bound is $1/[1+2(1/2)]=1/2=0.5000$.
For the set in question 2, find the marginal bounds for a particular unit. Solution: They are $1/[1+2(2)]=1/5=0.2000$ and $2/[2+2]=1/2=0.5000$. These equal the assignment bounds here because $m=1$: each assignment consists of naming the single treated unit.
For $n=6$, $m=3$, and $\Gamma=2$, state the lower and upper formulas for a particular unit’s treatment probability. Solution: The lower bound is $3/[3+2(3)]=1/3=0.3333$; the upper bound is $2(3)/[2(3)+3]=2/3=0.6667$.

Deriving Matched-Set Assignment Bounds

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