IV and LATE — Examples

In the product law, $Z$ is independent of all type-specific potential values, and both assignment atoms have positive probability. The observed treatment and outcome satisfy consistency by construction; $Y(a,z)=Y(a)$ is exclusion. The first-stage contrast is $E[A\mid Z=1]-E[A\mid Z=0]=(0.2+0.4)-(0.2+0.1)=0.3\mathrm{\ne }0$, so relevance holds in both models.

For $Z=0$, both models give the same observed pairs: always-takers have $(A,Y)=(1,5)$, never-takers and compliers have $(0,0)$, and defiers have $(1,3)$. For $Z=1$, both give $(1,5)$ to always-takers, $(0,0)$ to never-takers and defiers, and $(1,1)$ to compliers. Since type probabilities and the law of $Z$ also agree, the entire observed laws of $(Z,A,Y)$ agree.

The reduced-form contrast in either model is ${\mathrm{\Delta }}_Y=(0.2\cdot 5+0.4\cdot 1)-(0.2\cdot 5+0.1\cdot 3)=0.1$. Thus the common Wald ratio is $0.1\mathrm{/}0.3=1\mathrm{/}3$ by [F4]. However, $\mathrm{ATE}(M_0)=0.2\cdot 5+0.3\cdot 0+0.4\cdot 1+0.1\cdot 3=1.7$, whereas $\mathrm{ATE}(M_1)=0.2\cdot (-5)+0.3\cdot 10+0.4\cdot 1+0.1\cdot 3=2.7$. Therefore [F2] has different values at two parameters with the same observed law, and [F3] shows the ATE is not identifiable under only these core conditions.

Since $0.02\mathrm{\ne }0$, the two ratios are defined by [F1] and equal $0.06\mathrm{/}0.02=3$ and $0.07\mathrm{/}0.02=3.5$.

Their difference is $0.5$, exactly the reduced-form change $0.01$ divided by the common first stage $0.02$. The calculation is a property of the ratio only and implies no sampling or causal conclusion.

The product assignment makes $Z$ independent of the type and hence of all potential values; consistency holds by construction. Under $Z=1$, half the population has $A=1$ and the mean outcome is $(3+1)\mathrm{/}2=2$; under $Z=0$, $A=0$ for both types and the outcome mean is $0$. Thus ${\mathrm{\Delta }}_A=1\mathrm{/}2$ and ${\mathrm{\Delta }}_Y=2$.

By [F2], the Wald ratio is $2\mathrm{/}(1\mathrm{/}2)=4$, although the treatment contrast $Y(1,z)-Y(0,z)$ equals $2$ for every unit and each fixed $z$. The discrepancy comes from the violated exclusion restriction, not treatment-effect heterogeneity.

The first-stage contrast is $P(C)-P(D)=0.4-0.1=0.3$. By [F2], the outcome contrast is $E[(A(1)-A(0))(Y(1)-Y(0))]=0.4\cdot 1-0.1\cdot 3=0.1$; always-takers and never-takers contribute zero.

The Wald ratio is $0.1\mathrm{/}0.3=1\mathrm{/}3$ by [F3]. The complier effect is $1$, whereas the defier contribution enters the numerator with a negative sign. Thus the LATE conclusion cannot be used without monotonicity.

On the product of the four equally likely types and the fair assignment, $Z$ is independent of the potential values. The two compliers have effects $1$ and $3$, while the always-taker and never-taker both have treatment effect $4$; thus $P(C)=1\mathrm{/}2$, $\mathrm{LATE}=(1+3)\mathrm{/}2=2$, and $\mathrm{ATE}=(4+4+1+3)\mathrm{/}4=3$.

The treatment rate is $3\mathrm{/}4$ under $Z=1$ and $1\mathrm{/}4$ under $Z=0$, so ${\mathrm{\Delta }}_A=1\mathrm{/}2$. The outcome mean is $2$ under $Z=1$ and $1$ under $Z=0$, so ${\mathrm{\Delta }}_Y=1$. Therefore ${\psi }_W=1\mathrm{/}(1\mathrm{/}2)=2$, agreeing with the LATE in [F2] and differing from the ATE.

The complier event has probability $1\mathrm{/}2$ and its two effects are $1$ and $3$, so $\mathrm{LATE}=2$. Averaging all four effects gives $\mathrm{ATE}=(4+4+1+3)\mathrm{/}4=3$.

The always-taker and never-taker outcomes do not change across instrument arms. The two compliers each have probability $1\mathrm{/}4$, so their treatment effects give reduced-form contrast $\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 3=1$; the first-stage contrast is $1\mathrm{/}2$. The Wald ratio is therefore $2$, equal to the LATE by [F2] but not to the population ATE.