# How to show that the 2SLS estimand is equal to weighted average of two instruments

How can I show that below is true, using the properties of covariances.

We are interested in how a binary treatment variable $$D_i$$ affects an outcome variable $$Y_i$$. We have access to \emph{two valid binary instruments} for $$D_i$$, $$Z_{1i}$$ and $$Z_{2i}$$. Assume that the instruments are mutually exclusive, meaning that $$Cov(Z_{1i},Z_{2i})=0$$.

We assume that treatment effects are heterogeneous across individuals $$i$$, i.e. that

$${ Y_i = \alpha + \beta_i D_i + \varepsilon_i, }$$

which means that the IV estimands when only using either $$Z_{1i}$$ or $$Z_{2i}$$ as the instrument are not generally the same.

Let $${ \beta_1 = \frac{Cov(Y_i,Z_{1i})}{Cov(D_i,Z_{1i})} }$$ denote the IV estimand when only using $$Z_{1i}$$ as the instrument, and define $$\beta_2$$ similarly when only using $$Z_{2i}$$ as the instrument.

Suppose that we use $$Z_{1i}$$ and $$Z_{2i}$$ to instrument for $$D_{i}$$. Then the (population) first-stage equation is given by

$${ D_i = \pi_0 + \pi_1 Z_{1i} + \pi_2 Z_{2i} + v_{i}. }$$

Denote the (population) fitted values from this first-stage equation by $$\tilde{D}_i = \pi_0 + \pi_1 Z_{1i} + \pi_2 Z_{2i}$$. Show that the 2SLS estimand is equal to a weighted average of the two IV estimands $$\beta_1$$ and $$\beta_2$$:

$${ \beta_{2SLS} = \frac{Cov(Y_i,\tilde{D}_i)}{Cov(D_i,\tilde{D}_i)} = \psi \beta_1 + (1-\psi) \beta_2, }$$

where $$\psi = \frac{\pi_1 Cov(D_i,Z_{1i})}{\pi_1 Cov(D_i,Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}$$.

Plugging $$\tilde{D}_i = \pi_0 + \pi_1 Z_{1i} + \pi_2 Z_{2i}$$ into $$\beta_{2SLS} = \frac{Cov(Y_i,\tilde{D}_i)}{Cov(D_i,\tilde{D}_i)}$$ gives $$\beta_{2SLS} = \frac{Cov(Y_i,\pi_0 + \pi_1 Z_{1i} + \pi_2 Z_{2i})}{Cov(D_i,\pi_0 + \pi_1 Z_{1i} + \pi_2 Z_{2i})}$$ or $$\beta_{2SLS} = \frac{\pi_1Cov(Y_i, Z_{1i}) + \pi_2 Cov(Y_i,Z_{2i})}{\pi_1Cov(D_i, Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}$$ or $$\beta_{2SLS} = \frac{\pi_1Cov(Y_i, Z_{1i}) }{\pi_1Cov(D_i, Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}+\frac{\pi_2 Cov(Y_i,Z_{2i})}{\pi_1Cov(D_i, Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}$$ or $$\beta_{2SLS} = \frac{\pi_1Cov(Y_i, Z_{1i})Cov(D_i, Z_{1i})/Cov(D_i, Z_{1i}) }{\pi_1Cov(D_i, Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}+\frac{\pi_2 Cov(Y_i,Z_{2i})Cov(D_i, Z_{2i})/Cov(D_i, Z_{2i}) }{\pi_1Cov(D_i, Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}$$ or $$\beta_{2SLS} = \psi\beta_1+(1-\psi)\beta_2$$ as $$1-\psi = \frac{\pi_1 Cov(D_i,Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})-\pi_1 Cov(D_i,Z_{1i})}{\pi_1 Cov(D_i,Z_{1i}) + \pi_2 Cov(D_i,Z_{2i})}$$
• Thank you! I failed in doing this derivation because I thought I had to take into account the parameters in $D$ when factoring out the covariates. Thank you for this derivation! Feb 18, 2021 at 18:39