Local Average Treatment Effect (LATE) is defined as the causal effect of $D$ on $Y$ in the sub-popuation group whose treatment status is induced by the covariates $Z$.

When both $Z$ and $D$ are a binary variable, this estimand is represented in the following regression:

\[\begin{align} \text{LATE} = \mathbb{E}[Y(1) - Y(0) | D(1) > D(0)] \end{align}\]

This quantity is identified even when the treatment variable is not randomly assigned, as long as covariates $Z$ behave as in LATE theorem, as previously discussed. However, in some settings, the desired property might not hold.

We can still recover LATE by imposing additional assumption with the help of a random vector $\boldsymbol{X}$ that have the following relationship with respect to $D$, $Y$, and $Z$.

  1. (1) Conditional independence: conditional on $\boldsymbol{X}$, $Z$ has the desired property as in LATE theorem;
  2. (2) Overlapping: $0 < P(Z = 1| X) < 1$.

When this property $\boldsymbol{X}$ holds, we can recover LATE through a special case of Abadie’s kappa theorem. This theorem tells that LATE is equal to the following regression:

\[\begin{align*} \mathbb{E}[Y(1) - Y(0) | D(1) > D(0)] &= \frac{\mathbb{E}\bigg[\mathbb{E}[Y(1) - Y(0) | D(1) > D(0), X] \times \kappa\bigg]}{P(D(1) > D(0))} \end{align*}\]

where $\kappa$ is below:

\[\begin{align*} \kappa = 1 - \frac{D(1 - Z)}{P(Z = 0|X)} - \frac{(1 - D)Z}{P(Z = 1|X)} \end{align*}\]

Obviously, this approach is analagous to propensity score.

Proof

Additional assumption in the proof of this equality is that LATE is defined in the population, such that regularity conditions hold, so $\mathbb{E}[g(Y, D, X, Z)] < \infty$ for any measurable function $g$.

Starting from the second term in the kappa equation and apply the Law of Total Expectation, we have the following equality:

\[\begin{align*} \mathbb{E}\bigg[\frac{D(1 - Z)}{P(Z = 0|X)}\bigg] \end{align*} = \mathbb{E}[D| Z = 0, X]\]

Similarly, the third term of kappa equation is below:

\[\begin{align*} \mathbb{E}\bigg[\frac{(1 - D)Z}{P(Z = 1|X)}\bigg] \end{align*} = \mathbb{E}[(1 - D)| Z = 1, X]\]

Combining all these terms we then have the following equality:

\[\begin{align*} \frac{\mathbb{E}[Y(1) - Y(0) | D(1) > D(0), X] }{Pr(D(1) > D(0))} \times \bigg(\mathbb{E}[D| Z = 1, X] - \mathbb{E}[D| Z = 0, X]\bigg) \end{align*}\]

Using the same strategies as dicussed previously on the second term in the numerator above and then applying the Law of Total Expectation, we then have the desired result:

\[\begin{align*} &= \frac{\mathbb{E}[Y(1) - Y(0) | D(1) > D(0), X] \times P(D(1) > D(0)|X )}{P(D(1) > D(0))} \\ &= \frac{\mathbb{E}[Y(1) - Y(0)] }{P(D(1) > D(0))} \\ &= \mathbb{E}[Y(1) - Y(0)|D(1) > D(0)] \\ &= \text{LATE} \end{align*}\]