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


1 Answer 1


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

  • $\begingroup$ 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! $\endgroup$ Feb 18, 2021 at 18:39

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