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