Interference and exposure mappings

Let $n\ge 2$, let $x_1,\dots ,x_n$ and $y_1,\dots ,y_n$ be fixed real numbers, and choose a subset $A$ uniformly from the $m$-element subsets of $\{1,\dots ,n\}$, where $1\le m\le n$. Put $\bar{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i,{\bar{x}}_A=\frac{1}{m}{\sum }_{i\in A}{x}_i,$ and define $\bar{y},{\bar{y}}_A$ analogously. Define the finite-population covariance $S_{xy}=\frac{1}{n-1}{\sum }_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})\mathrm{.}$ Then $\mathbb{E}[{\bar{x}}_A]=\bar{x},\mathrm{Cov}({\bar{x}}_A,{\bar{y}}_A)=(1-\frac{m}{n})\frac{S_{xy}}{m}\mathrm{.}$ When $m\ge 2$, define the sample covariance $s_{xy,A}=\frac{1}{m-1}{\sum }_{i\in A}(x_i-{\bar{x}}_A)(y_i-{\bar{y}}_A)\mathrm{.}$ Then $\mathbb{E}[s_{xy,A}]=S_{xy}$. Taking $y_i=x_i$ gives the corresponding variance identities.

Let $(\mathrm{\Omega },w)$ be a finite probability space and $X:\mathrm{\Omega }\to \mathbb{R}$. For any event $B\subseteq \mathrm{\Omega }$ with $\mathbb{P}(B)>0$, define $\mathbb{E}[X\mid B]:=\frac{1}{\mathbb{P}(B)}{\sum }_{\omega \in B}X(\omega )w(\omega )\mathrm{.}$ If $A\subseteq \mathrm{\Omega }$ has $0<\mathbb{P}(A)<1$, then the function equal to $\mathbb{E}[X\mid A]$ on $A$ and $\mathbb{E}[X\mid A^c]$ on $A^c$ is a version of $\mathbb{E}[X\mid \sigma (A)]$ under the integral characterization. More generally, if $A_1,\dots ,A_r$ is a finite partition of $\mathrm{\Omega }$ into events of positive probability, then $\mathbb{E}[X]={\sum }_{q=1}^r\mathbb{P}(A_q)\mathbb{E}[X\mid A_q]\mathrm{.}$

Let $\mathcal{I}$ be a finite set of treatment-assignment units partitioned into $N\ge 1$ nonempty groups ${\mathcal{I}}_1,\dots ,{\mathcal{I}}_N$, with $n_i=\mathrm{\mid }{\mathcal{I}}_i\mathrm{\mid }$. Label the members of each group ${\mathcal{I}}_i$ by $j=1,\dots ,n_i$. Let $\mathcal{Z}\subseteq \{0,1{\}}^{\mathcal{I}}$ be the nonempty finite set of allowed global assignments in a finite randomized treatment-assignment model (Finite randomized treatment-assignment models, Finite probability spaces, outcome weights, events, and event probabilities). Write ${\mathbf{z}}_i$ for the restriction of $\mathbf{z}\in \mathcal{Z}$ to group $i$, and $Z_{ij}$ for unit $j$'s coordinate in its group.

A fixed finite-population potential-outcome schedule is a family of real-valued functions $Y_{ij}:\mathcal{Z}\to \mathbb{R}$, one for every unit $(i,j)$. The observed outcome at assignment $\mathbf{Z}$ is $Y_{ij}^{\mathrm{o}\mathrm{b}\mathrm{s}}=Y_{ij}(\mathbf{Z})$. The schedule is fixed; only the declared assignment is random. This notation permits arbitrary dependence of $Y_{ij}(\mathbf{z})$ on every component of $\mathbf{z}$.

The schedule satisfies partial interference when, for every unit $(i,j)$ and all allowed $\mathbf{z},{\mathbf{z}}^{\mathrm{\prime }}\in \mathcal{Z}$, equality of their within-group assignments ${\mathbf{z}}_i={\mathbf{z}}_i^{\mathrm{\prime }}$ implies $Y_{ij}(\mathbf{z})=Y_{ij}({\mathbf{z}}^{\mathrm{\prime }})$. In that case the common value is well-defined as $Y_{ij}({\mathbf{z}}_i)$: outcomes may depend arbitrarily on assignments within group $i$, but not on assignments outside that group. Partial interference is a substantive model/design restriction, not a consequence of randomization.

In the finite-population model of Finite-population potential outcomes under partial interference, an exposure mapping for unit $(i,j)$ is a function $f_{ij}:{\mathcal{Z}}_i\to {\mathcal{D}}_{ij}$, where ${\mathcal{Z}}_i$ is the finite set of allowed assignments to group $i$ and ${\mathcal{D}}_{ij}$ is a finite set of exposure labels. At assignment $\mathbf{Z}$, the unit's exposure is the random label $D_{ij}=f_{ij}({\mathbf{Z}}_i)$. Its design probability is ${\pi }_{ij}(d)=\mathbb{P}\{f_{ij}({\mathbf{Z}}_i)=d\}={\sum }_{\mathbf{z}\in \mathcal{Z}:f_{ij}({\mathbf{z}}_i)=d}p(\mathbf{z}),$ where $p(\mathbf{z})=\mathbb{P}(\mathbf{Z}=\mathbf{z})$ is the declared assignment law.

The mapping is sufficient for the potential outcome of unit $(i,j)$ on a set of labels if, whenever $f_{ij}({\mathbf{z}}_i)=f_{ij}({\mathbf{z}}_i^{\mathrm{\prime }})=d$, the fixed outcomes satisfy $Y_{ij}({\mathbf{z}}_i)=Y_{ij}({\mathbf{z}}_i^{\mathrm{\prime }})$. For each label in the image of $f_{ij}$ this condition defines a unique value $Y_{ij}(d)$ equal to the common outcome over its preimage. The observed-outcome consistency identity is then $Y_{ij}^{\mathrm{o}\mathrm{b}\mathrm{s}}=Y_{ij}(D_{ij})$. An exposure-specific finite-population target may include unit $(i,j)$ at label $d$ only when ${\pi }_{ij}(d)>0$; otherwise its inverse-probability estimator is undefined. Sufficiency is an assumption about the schedule and map, not a consequence of knowing the assignment probabilities.

Let $N>1$ groups have fixed sizes $n_i\ge 2$, and let the fixed potential outcomes satisfy partial interference as in Finite-population potential outcomes under partial interference. Let $\varphi$ and $\psi$ be two group-specific treatment-allocation strategies. Under strategy $s\in \{\varphi ,\psi \}$, group $i$ has a fixed treated count $K_i^s\in \{1,\dots ,n_i-1\}$. Fix $C\in \{1,\dots ,N-1\}$. The two-stage design first chooses uniformly a set of exactly $C$ groups for $\psi$, assigning the other $N-C$ groups to $\varphi$. Conditional on these strategy labels, it independently chooses in each group a uniformly random $K_i^s$-element subset for treatment, where $s$ is that group's strategy. Here $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ denotes the number of $k$-element subsets of an $n$-element set. Thus, for any strategy vector $\mathbf{s}$ with exactly $C$ entries equal to $\psi$ and any assignment vectors ${\mathbf{z}}_i$ with ${\sum }_j{z}_{ij}=K_i^{s_i}$, $\mathbb{P}(\mathbf{S}=\mathbf{s},\mathbf{Z}=\mathbf{z})={\left(\genfrac{}{}{0px}{}{N}{C}\right)}^{-1}{\prod }_{i=1}^N{\left(\genfrac{}{}{0px}{}{n_i}{K_i^{s_i}}\right)}^{-1};$ all other pairs have probability zero. In particular, $\mathbb{P}(Z_{ij}=1\mid S_i=s)=K_i^s\mathrm{/}{n}_i$ and $\mathbb{P}(Z_{ij}=0\mid S_i=s)=(n_i-K_i^s)\mathrm{/}{n}_i$, both positive.

Write ${\mathbb{E}}_{i,s}$ for expectation under the finite uniform assignment law within group $i$ under strategy $s$. Conditional means given $Z_{ij}=z$ are the finite conditional weighted sums, defined because that event has probability strictly between zero and one; Conditional expectation on positive-probability events in finite spaces verifies that their piecewise versions satisfy the integral characterization of Conditional expectation given a sigma algebra. Define the strategy-specific group and population mean potential outcomes by ${\bar{Y}}_i(z;s)=\frac{1}{n_i}{\sum }_{j=1}^{n_i}{\mathbb{E}}_{i,s}[Y_{ij}({\mathbf{Z}}_i)\mid Z_{ij}=z],\bar{Y}(z;s)=\frac{1}{N}{\sum }_{i=1}^N{\bar{Y}}_i(z;s),$ for $z\in \{0,1\}$, and ${\bar{Y}}_i(s)=\frac{1}{n_i}{\sum }_{j=1}^{n_i}{\mathbb{E}}_{i,s}[Y_{ij}({\mathbf{Z}}_i)],\bar{Y}(s)=\frac{1}{N}{\sum }_{i=1}^N{\bar{Y}}_i(s)\mathrm{.}$ These are finite-population estimands: the outcomes are fixed, and the expectations average only over the specified within-group assignment strategy. Define $DE(\psi )=\bar{Y}(0;\psi )-\bar{Y}(1;\psi ),IE(\varphi ,\psi )=\bar{Y}(0;\varphi )-\bar{Y}(0;\psi ),$ $TE(\varphi ,\psi )=\bar{Y}(0;\varphi )-\bar{Y}(1;\psi ),OE(\varphi ,\psi )=\bar{Y}(\varphi )-\bar{Y}(\psi )\mathrm{.}$

For an observed design realization, put $m_{i1}^s=K_i^s$ and $m_{i0}^s=n_i-K_i^s$, and define, whenever $S_i=s$, ${\widehat{\bar{Y}}}_i(z;s)=\frac{{\sum }_{j=1}^{n_i}1\{Z_{ij}=z\}{Y}_{ij}^{\mathrm{o}\mathrm{b}\mathrm{s}}}{m_{iz}^s},{\widehat{\bar{Y}}}_i(s)=\frac{1}{n_i}{\sum }_{j=1}^{n_i}{Y}_{ij}^{\mathrm{o}\mathrm{b}\mathrm{s}}\mathrm{.}$ The denominators are the fixed, positive realized arm sizes. With $C_{\psi }=C$ and $C_{\varphi }=N-C$, define the observed population estimators $\widehat{\bar{Y}}(z;s)=\frac{1}{C_s}{\sum }_{i=1}^{N}1\{S_i=s\}{\widehat{\bar{Y}}}_i(z;s),\widehat{\bar{Y}}(s)=\frac{1}{C_s}{\sum }_{i=1}^{N}1\{S_i=s\}{\widehat{\bar{Y}}}_i(s)\mathrm{.}$ The four strategy-contrast estimators $\widehat{DE}(\psi )$, $\widehat{IE}(\varphi ,\psi )$, $\widehat{TE}(\varphi ,\psi )$, and $\widehat{OE}(\varphi ,\psi )$ are obtained by replacing each mean in the corresponding displayed target by its hatted estimator. No stratified-interference or additive-effects assumption is part of this design definition.

Let $\mathcal{U}=\{(i,j):1\le i\le N,1\le j\le n_i\}$ be a fixed population of $M={\sum }_i{n}_i$ units. Let $\mathbf{Z}$ have a known finite assignment law $p$ and fixed potential-outcome schedule satisfying partial interference. Fix an exposure label $d$ for which the mapping in Exposure mappings and exposure consistency is sufficient and ${\pi }_{ij}(d)>0$ for every $(i,j)\in \mathcal{U}$. Then ${\widehat{\mu }}_{HT}(d)=\frac{1}{M}{\sum }_{(i,j)\in \mathcal{U}}\frac{1\{D_{ij}=d\}{Y}_{ij}^{\mathrm{o}\mathrm{b}\mathrm{s}}}{{\pi }_{ij}(d)}$ is an estimator of the finite-population exposure mean $\mu (d)=M^{-1}{\sum }_{(i,j)\in \mathcal{U}}{Y}_{ij}(d)$ and satisfies ${\mathbb{E}}_p[{\widehat{\mu }}_{HT}(d)]=\mu (d)$. Therefore, for two labels $d,e$ with the same positivity and sufficiency conditions, ${\widehat{\mu }}_{HT}(d)-{\widehat{\mu }}_{HT}(e)$ is design-unbiased for $\mu (d)-\mu (e)$. Dependence among units' exposures does not affect these expectation identities.

Under the two-stage design of Two-stage randomized design and strategy-specific effects, for either strategy $s\in \{\varphi ,\psi \}$, treatment value $z\in \{0,1\}$, and group $i$, $\mathbb{E}[{\widehat{\bar{Y}}}_i(z;s)\mid S_i=s]={\bar{Y}}_i(z;s),\mathbb{E}[{\widehat{\bar{Y}}}_i(s)\mid S_i=s]={\bar{Y}}_i(s)\mathrm{.}$ Consequently, $\mathbb{E}[\widehat{\bar{Y}}(z;s)]=\bar{Y}(z;s),\mathbb{E}[\widehat{\bar{Y}}(s)]=\bar{Y}(s)\mathrm{.}$ By taking the displayed contrasts, the design estimators of $DE(\psi )$, $IE(\varphi ,\psi )$, $TE(\varphi ,\psi )$, and $OE(\varphi ,\psi )$ are unbiased. The proof does not require stratified interference or additive effects.

The assignment space and every sum in this proof are finite. The argument is choice-free and uses no conditional-expectation existence theorem on general probability spaces.

In the setting of Two-stage randomized design and strategy-specific effects, stratified interference means that, for each group $i$, unit $j$, and treatment count $k\in \{1,\dots ,n_i-1\}$, the value $Y_{ij}({\mathbf{z}}_i)$ is the same for any two assignments ${\mathbf{z}}_i,{\mathbf{z}}_i^{\mathrm{\prime }}$ with ${\sum }_{\mathrm{\ell }}{z}_{i\mathrm{\ell }}={\sum }_{\mathrm{\ell }}{z}_{i\mathrm{\ell }}^{\mathrm{\prime }}=k$ and $z_{ij}=z_{ij}^{\mathrm{\prime }}$. Thus at a fixed group treatment count an outcome may depend on the unit's own assignment, but not on which particular peers are treated. This assumption is additional to partial interference and is not needed for point-effect unbiasedness.

For strategy $s$ and own treatment $z$, write $Y_{ij}(z;s)$ for the common outcome at count $K_i^s$ and own assignment $z$. Define $S_{iz}^2(s)=\frac{1}{n_i-1}{\sum }_{j=1}^{n_i}\{Y_{ij}(z;s)-{\bar{Y}}_i(z;s){\}}^2,d_{ij}(s)=Y_{ij}(0;s)-Y_{ij}(1;s),$ $S_{i01}(s)=\frac{1}{n_i-1}{\sum }_{j=1}^{n_i}\{Y_{ij}(0;s)-{\bar{Y}}_i(0;s)\}\{Y_{ij}(1;s)-{\bar{Y}}_i(1;s)\},$ and $S_{iD}^2(s)=\frac{1}{n_i-1}{\sum }_{j=1}^{n_i}\{d_{ij}(s)-D_i(s){\}}^2,$ where $D_i(s)={\bar{Y}}_i(0;s)-{\bar{Y}}_i(1;s)$. In an observed group assigned $s$, define the estimated group direct effect by ${\widehat{D}}_i(s):={\widehat{\bar{Y}}}_i(0;s)-{\widehat{\bar{Y}}}_i(1;s)\mathrm{.}$ Let $s_{iz}^2(s)$ be the ordinary sample variance of the $m_{iz}^s$ observed outcomes in arm $z$, with denominator $m_{iz}^s-1$, and let $s_{D,s}^2$ be the sample variance of the observed group direct-effect estimates ${\widehat{D}}_i(s)$ across the $C_s$ groups assigned to $s$.

The source direct-effect variance estimator under strategy $s$ is ${\widehat{V}}_D(s)=(1-\frac{C_s}{N})\frac{s_{D,s}^2}{C_s}+\frac{1}{C_{s}N}{\sum }_{i:S_i=s}\{\frac{s_{i0}^2(s)}{n_i-K_i^s}+\frac{s_{i1}^2(s)}{K_i^s}\}\mathrm{.}$ For the indirect, total, and overall effect contrasts define ${\widehat{V}}_I(\varphi ,\psi )=\frac{s_{0,\varphi }^2}{N-C}+\frac{s_{0,\psi }^2}{C},{\widehat{V}}_T(\varphi ,\psi )=\frac{s_{0,\varphi }^2}{N-C}+\frac{s_{1,\psi }^2}{C},$ ${\widehat{V}}_O(\varphi ,\psi )=\frac{s_{\varphi }^2}{N-C}+\frac{s_{\psi }^2}{C},$ where $s_{z,s}^2$ is the sample variance across groups assigned $s$ of the observed group estimates ${\widehat{\bar{Y}}}_i(z;s)$, and $s_s^2$ is the sample variance across such groups of ${\widehat{\bar{Y}}}_i(s)$.

These are design-randomization variance estimators for the fixed finite-population point estimators. The group-arm sample variances in these formulas are defined only if $C\ge 2$ and $N-C\ge 2$. In addition, ${\widehat{V}}_D(s)$ requires $2\le K_i^s\le n_i-2$ for every group, so both within-group arm sample variances have nonzero denominators. The point estimators themselves remain defined under the broader counts $1\le K_i^s<n_i$.

Let $N\ge 4$ groups be completely randomized, with exactly $C$ groups assigned $\psi$ and $N-C$ assigned $\varphi$, where $2\le C\le N-2$. For each group $i$ and strategy $s\in \{\varphi ,\psi \}$, let $a_i(s)$ be a fixed real value and let $X_i(s)$ be a real-valued random variable from that group's finite second-stage assignment law, with $\mathbb{E}[X_i(s)]=a_i(s)$ and $v_i(s)=\mathrm{Var}(X_i(s))$. Conditional on the first-stage vector, the group-specific second-stage random variables are independent. Let $C_{\psi }=C$, $C_{\varphi }=N-C$, and define ${\widehat{\mu }}_s=\frac{1}{C_s}{\sum }_{i:S_i=s}{X}_i(s),s_{X,s}^2=\frac{1}{C_s-1}{\sum }_{i:S_i=s}(X_i(s)-{\widehat{\mu }}_s{)}^2\mathrm{.}$ For $S_s^2=(N-1{)}^{-1}{\sum }_i(a_i(s)-\bar{a}(s){)}^2$ and $S_{\psi -\varphi }^2=(N-1{)}^{-1}{\sum }_i\{a_i(\psi )-a_i(\varphi )-\overline{a(\psi )-a(\varphi )}{\}}^2$, one has $\mathrm{Var}({\widehat{\mu }}_s)=(1-\frac{C_s}{N})\frac{S_s^2}{C_s}+\frac{1}{C_{s}N}{\sum }_{i=1}^N{v}_i(s),\mathbb{E}[s_{X,s}^2]=S_s^2+\frac{1}{N}{\sum }_{i=1}^N{v}_i(s),$ and $\mathrm{Var}({\widehat{\mu }}_{\psi }-{\widehat{\mu }}_{\varphi })=\frac{S_{\psi }^2}{C}+\frac{S_{\varphi }^2}{N-C}-\frac{S_{\psi -\varphi }^2}{N}+\frac{1}{CN}{\sum }_i{v}_i(\psi )+\frac{1}{(N-C)N}{\sum }_i{v}_i(\varphi )\mathrm{.}$ Consequently, $\mathbb{E}\text{\hspace{0.17em}⁣}[\frac{s_{X,\psi }^2}{C}+\frac{s_{X,\varphi }^2}{N-C}]=\mathrm{Var}({\widehat{\mu }}_{\psi }-{\widehat{\mu }}_{\varphi })+\frac{S_{\psi -\varphi }^2}{N}\mathrm{.}$