# 2019-03-06 Scratch¶

## The Weak Instruments Problem¶

Suppose: $y = {\beta}x + u$; $x = {\pi}z + v$; $u = {\delta}v + w$

then:

${\beta}_{IV} = \frac{zy}{zx} = \frac{{\beta}xz + uz}{xz}$

${\beta}_{IV} = \frac{{\beta}{\pi}zz + {\beta}vz + uz}{{\pi}zz + vz} = \beta + \frac{uz}{{\pi}zz + vz}$

${\beta}_{IV} = \beta + \frac{{\delta}vz + wz}{{\pi}zz + vz} = \beta + {\delta}\frac{1}{({\pi}zz/(vz)) + 1} + \frac{wz}{{\pi}zz + vz}$

And if $\pi = 0$, then:

${\beta}_{IV0} = \beta + \frac{{\delta}vz + wz}{vz} = \beta + \delta + \frac{wz}{vz}$

This is the weak instruments problem. As you get more and more data, $\frac{wz}{vz}$ is not heading for zero, and even if it were your estimated ${\beta}_{IV0}$ is not headed for $\beta$ but is rather headed for $\beta + \delta$

Now we would like to apply a file-drawer problem filter: $\pi$ is zero, but in the sample you have the calculated $zv$ is large enough that you are happy to run and report the regression. What can we then say?