Proving the LATE Theorem of Angrist and Imbens 1994

Assume we have a binary instrument $Z_i$ which can be used to estimate the effect of the endogenous variable $D_i$ on the outcome $Y_i$. Suppose the instrument has a significant first stage, it is randomly assigned, it satisfies the exclusion restriction, and it satisfies monotonicity as outlined in Angrist and Imbens (1994). http://www.jstor.org/discover/10.2307/2951620?uid=3738032&uid=2&uid=4&sid=21104754800073

They state that the probability of being a complier ($C_i$) is $$\text{Pr}(C_i) = \text{Pr}(D_i = 1|Z_i = 1) - \text{Pr}(D_i = 1 - Z_i = 0)$$ and difference in potential outcomes for the subpopulation of compliers is $$E(Y_{i1} - Y_{i0}|C_i) = \frac{E(Y_i|Z_i=1)-E(Y_i|Z_i=0)}{E(D_i|Z_i=1)-E(D_i|Z_i=0)}$$

Can somebody shed some light on how they get these two expressions and more importantly how they combine them? I try to understand this from their journal article but I cannot make sense of it. Any help on this would be very much appreciated.

For the first part you stated that you have a “valid” instrument. This implies for a binary treatment and instrument that $Cov(D_i,Z_i) \neq 0$ is equivalent to $P(D_i = 1|Z_i = 1) \neq P(D_i = 1|Z_i = 0)$, i.e. the instrument has an effect on whether the treatment is chosen or not. This observation which also should be stated in the Angrist and Imbens paper is key for the rest of their proof. For the first stage they assume that $P(D_i = 1|Z_i = 1) > P(D_i = 1|Z_i = 0)$, meaning that the number of compliers ($C_i)$ is larger than that of defiers ($F_i$).

Using the exclusion restriction (for every $z \in$ {$0;1$} we have that $Y_{iz} = Y_{i0z} = Y_{i1z}$, i.e. the instrument does not have a direct effect on the outcome) you can write the difference in the share of compliers and defiers in the population as \begin{align} P(D_i = 1|Z_i = 1) - P(D_i = 1|Z_i = 0) &= P(D_{i1} = 1|Z_i = 1) - P(D_{i0} = 1|Z_i = 0) \newline &= P(D_{i1} = 1) - P(D_{i0} = 0) \newline &= \left[ P(D_{i1} = 1, D_{i0} = 0) + P(D_{i1} = 1, D_{i0} = 1) \right] - \left[ P(D_{i1} = 0, D_{i0} = 1) + P(D_{i1} = 1, D_{i0} = 1) \right] \newline &= P(C_i) – P(F_i) \end{align} where the second step uses independence to get rid of the conditioning on $Z_i$ because potential outcomes are independent of the instrument. The third step uses the law of total probability. In the last step you then only need to use monotonicity which basically assumes that defiers do not exist, so $P(F_i) = 0$ and you get $$P(C_i) = P(D_i = 1|Z_i = 1) - P(D_i = 1|Z_i = 0).$$ This would be your first stage coefficient in a 2SLS regression. The monotonicity assumption is crucial for this and one should think hard about possible reasons for why it might be violated (however, monotonicity can be relaxed, see for instance de Chaisemartin (2012) “All you need is LATE”).

The second part of the proof follows a similar path. For this you need to remember that the observed treatment status is $$D_i = Z_iD_{i1} + (1-Z_i)D_{i0}$$ because you cannot observe both potential outcomes for the same individual. In this way you can relate the observed outcome to the potential outcome, the treatment status, and the instrument as $$Y_i = (1-Z_i)(1-D_i)Y_{i00} + Z_i(1-D_i)Y_{i10} + (1-Z_i)D_iY_{i01} + Z_iD_iY_{i11}$$ For the second part of the proof take the difference in the expected outcome with the instrument being switched on and on, and use the previous representation of observed outcomes and the exclusion restriction in the first step to get: \begin{align} E(Y_i|Z_i = 1) – E(Y_i|Z_i=0) &= E(Y_{i1}D_i + Y_{i0}(1-D_i)|Z_i=0) \newline &- E(Y_{i1}D_i + Y_{i0}(1-D_i)|Z_i=1)\newline &= E(Y_{i1}D_{i1} + Y_{i0}(1-D_{i1})|Z_i=1) \newline &- E(Y_{i1}D_{i0} + Y_{i0}(1-D_{i0})|Z_i=0) \newline &= E(Y_{i1}D_{i1} + Y_{i0}(1-D_{i1})) \newline &- E(Y_{i1}D_{i0} + Y_{i0}(1-D_{i0})) \newline &= E((Y_{i1}-Y_{i0})(D_{i1}-D_{i0})) \newline &= E(Y_{i1}-Y_{i0}|D_{i1}-D_{i0}=1)P(D_{i1}-D_{i0} = 1) \newline &- E(Y_{i1}-Y_{i0}|D_{i1}-D_{i0}=-1)P(D_{i1}-D_{i0} = -1) \newline &= E(Y_{i1}-Y_{i0}|C_i)P(C_i) - E(Y_{i1}-Y_{i0}|F_i)P(F_i) \newline &= E(Y_{i1}-Y_{i0}|C_i)P(C_i) \end{align}

Now this was quite a bit of work but it's not too bad if you know the steps you need to take. For the second line use again the exclusion restriction to write out the potential treatment states. In the third line use independence to get rid of the conditioning on $Z_i$ as before. In the fourth line you just factor terms. The fifth line uses the law of iterated expectations. The last line arises due to the monotonicity assumption, i.e. $P(F_i)=0$. Then you just need to divide as a last step and you arrive at \begin{align} E(Y_{i1}-Y_{i0}|C_i) &= \frac{E(Y_i|Z_i = 1) – E(Y_i|Z_i=0)}{P(C_i)} \newline &= \frac{E(Y_i|Z_i = 1) – E(Y_i|Z_i=0)}{P(D_i = 1|Z_i = 1) - P(D_i = 1|Z_i = 0)} \newline &= \frac{E(Y_i|Z_i = 1) – E(Y_i|Z_i=0)}{E(D_i|Z_i = 1) - E(D_i|Z_i = 0)} \end{align} since $D_i$ and $Z_i$ are binary. This should show how you combine the two proofs and how they arrive at the final expression.

There are four types of people:

1. Never Takers (NT): $$D = 0$$ for both values of Z
2. Defiers (DF): $$D=0$$ when $$Z =1$$ and $$D=1$$ when $$Z =0$$
3. Compliers (C): $$D=1$$ when $$Z =1$$ and $$D=0$$ when $$Z =0$$
4. Always Takers (AT): $$D =1$$ for both values of $$Z$$.

The formula for the Wald estimator is: $$\Delta_{IV} = \frac{E(Y \vert Z=1)−E(Y \vert Z=0)}{Pr(D=1 \vert Z =1)−Pr(D=1 \vert Z =0)}$$

Using our 4 groups and the basic rules of probability, we can rewrite the two numerator pieces as: $$E(Y \vert Z=1)=E(Y_1 \vert AT)\cdot Pr(AT)+E(Y_1 \vert C)\cdot Pr(C)+E(Y_0 \vert DF) \cdot Pr(DF)+E(Y_0 \vert NT) \cdot Pr(NT)$$ and $$E(Y \vert Z=0)=E(Y_1 \vert AT)\cdot Pr(AT)+E(Y_0 \vert C)\cdot Pr(C)+E(Y_1 \vert DF) \cdot Pr(DF)+E(Y_0 \vert NT) \cdot Pr(NT)$$

The two denominator terms are: $$Pr(D=1 \vert Z =1)=Pr(D=1 \vert Z =1,AT) \cdot Pr(AT)+Pr(D=1 \vert Z =1,C) \cdot Pr(C) \\ =Pr(AT)+Pr(C)$$ and $$Pr(D=1 \vert Z =0)=Pr(D=1 \vert Z = 0,AT) \cdot Pr(AT)+Pr(D=1 \vert Z =0,DF) \cdot Pr(DF) \\ =Pr(AT)+Pr(DF)$$

The first of these corresponds to your first expression.

Coming back to the Wald formula and plugging these in, we see that some of these terms cancel out in the subtraction, leaving

$$\Delta_{IV} =\frac{[E(Y_1 \vert C) \cdot Pr(C)+E(Y_0 \vert D) \cdot Pr(D)]−[E(Y_0 \vert C) \cdot Pr(C)+E(Y_1 \vert DF) \cdot Pr(DF)]}{Pr(C) − Pr(DF)}$$ ￼ This yields some insight. The Wald IV estimator is a weighted average of the treatment effect on the compliers and the negative of the treatment effect on the defiers.

Now we make two assumptions. First, we assume monotonicity, so that the instrument can only increase or decrease the probability of participation. This means that $$Pr(DF) = 0$$. The monotonicity assumption is equivalent to assuming an index function model for treatment. The second assumption is that there are some compliers, which is to say that $$Pr(C) > 0$$. The behavior of some individuals must be altered by the instrument. This should be the case if the instrument is relevant. These two assumptions produce

$$\Delta_{IV} =\frac{E(Y_1 \vert C) \cdot Pr(C)−E(Y_0 \vert C) \cdot Pr(C)}{Pr(C)}=E(Y_1 \vert C)−E(Y_0 \vert C)=LATE.$$

• +1, I think the two answers complement each other very well. This one shows more the intuition of the Wald estimator and where the assumptions come from rather than just utilize them in a purely technical manner
– Andy
Sep 26 '14 at 22:20