Interference and exposure mappings — Examples

The global assignments $(z_{11},z_{21})=(0,0)$ and $(0,1)$ have the same first-group restriction $z_{11}=0$, but their first-group outcomes are respectively $0$ and $1$.

The inequality of those outcomes violates the invariance in [F1]. Therefore no function of the first group's assignment alone can represent this schedule, and partial interference fails.

In every assignment under strategy $s$, the outcome of every unit in that group equals the fixed treated count $K_i^s$, independently of which particular units are treated. Thus all conditional and unconditional group means under $s$ equal $K_i^s$.

Consequently $\bar{Y}(0;\varphi )=\bar{Y}(1;\varphi )=\bar{Y}(\varphi )=1$ and $\bar{Y}(0;\psi )=\bar{Y}(1;\psi )=\bar{Y}(\psi )=2$. Substitution into the definitions gives $DE(\psi )=0$, $IE(\varphi ,\psi )=-1$, $TE(\varphi ,\psi )=-1$, and $OE(\varphi ,\psi )=-1$.

In each realized group, every observed outcome equals its strategy's fixed count. The within-group mean in either treatment arm and the full group mean therefore equal that same count. Averaging over the groups assigned each strategy yields hatted means $1$ under $\varphi$ and $2$ under $\psi$, so all four observed contrasts equal the targets from step 2.1 for every assignment.

Only assignment $(0,1)$ maps either unit to $d$, so ${\pi }_1(d)={\pi }_2(d)=1\mathrm{/}4$. The target is $\mu (d)=(4+10)\mathrm{/}2=7$.

Under assignment $(0,1)$ the estimator equals $\frac{1}{2}(4\mathrm{/}(1\mathrm{/}4)+10\mathrm{/}(1\mathrm{/}4))=28$. Under each of the other three assignments it is zero. Its design expectation is therefore $(28+0+0+0)\mathrm{/}4=7=\mu (d)$, verifying the unbiasedness identity directly.

Unit 1 has $D_1=0$ under both assignments $(0,0)$ and $(0,1)$, but its outcomes there are respectively $0$ and $1$. Thus [F1] fails although ${\pi }_1(0)={\pi }_2(0)=1\mathrm{/}2>0$.

If one nevertheless sets the attempted exposure-zero values to $Y_1^{\ast }(0)=Y_1(0,0)=0$ and $Y_2^{\ast }(0)=0$, their mean is $0$. The attempted Horvitz–Thompson estimator is $\frac{1}{2}\{1(Z_1=0)Y_1\mathrm{/}(1\mathrm{/}2)+1(Z_2=0)Y_2\mathrm{/}(1\mathrm{/}2)\}=1(Z_1=0)Z_2$. It equals $1$ only at assignment $(0,1)$ and zero at the other three assignments, so its expectation is $1\mathrm{/}4\mathrm{\ne }0$. Hence positivity does not repair the misspecified exposure map, and [F2] cannot be applied without its sufficiency premise.

For any fixed group assignment, $Y_{ij}({\mathbf{z}}_i)$ depends only on $z_{ij}$. Hence changing peer assignments while holding the unit's treatment and the treated count fixed leaves its outcome unchanged, so stratified interference holds. The unit direct effects are $d_{ij}=Y_{ij}(0)-Y_{ij}(1)=-a_j=(-1,0,0,0)$, which are not constant.

The finite-population mean direct effect in each group is $-1\mathrm{/}4$, and its finite-population variance is $S_{iD}^2=(3\mathrm{/}4)\mathrm{/}3=1\mathrm{/}4$. Within a group, the estimated direct effect is $-1\mathrm{/}2$ if unit 1 is treated and $0$ otherwise, each with probability $1\mathrm{/}2$. Its variance is therefore $1\mathrm{/}16$. The within-group variance component ${\widehat{v}}_i=s_{i0}^2\mathrm{/}2+s_{i1}^2\mathrm{/}2$ is $1\mathrm{/}4$ when unit 1 is treated and $0$ otherwise, so $\mathbb{E}[{\widehat{v}}_i]=1\mathrm{/}8$.

Two groups are selected for $\psi$, and their within-group assignments are independent. The population direct-effect estimator is their mean. Its values are $-1\mathrm{/}2,-1\mathrm{/}4,0$ with probabilities $1\mathrm{/}4,1\mathrm{/}2,1\mathrm{/}4$, respectively, so its mean is $-1\mathrm{/}4$ and its design variance is $1\mathrm{/}32$. For two selected groups, the between-group sample variance of their direct-effect estimates is $1\mathrm{/}8$ when their outcomes differ and $0$ otherwise; the outcomes differ with probability $1\mathrm{/}2$, giving expected sample variance $1\mathrm{/}16$.

The expected between-group term in the source variance estimator is $(1-C\mathrm{/}N)s_{D,\psi }^2\mathrm{/}C=(1\mathrm{/}4)s_{D,\psi }^2$, hence its expectation is $1\mathrm{/}64$. The expected within-group term is $(CN{)}^{-1}$ times the sum over the two selected groups' ${\widehat{v}}_i$ values, giving $(1\mathrm{/}8)(2\cdot 1\mathrm{/}8)=1\mathrm{/}32$. Therefore the complete variance estimator has expectation $1\mathrm{/}64+1\mathrm{/}32=3\mathrm{/}64$.

The positive bias is $3\mathrm{/}64-1\mathrm{/}32=1\mathrm{/}64$. It also equals the theorem's correction $N^{-2}{\sum }_i{S}_{iD}^2\mathrm{/}{n}_i={16}^{-1}\cdot 4\cdot (1\mathrm{/}4)\mathrm{/}4=1\mathrm{/}64$. Thus the nonadditive individual effects give a strict conservative bias despite stratified interference and all required sample sizes.