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In the case of a single endogenous variable and a single instrumental variable, the IV estimator is given by

$b_{IV} = \frac{cov(z,y)}{cov(z,x)}$

It is often mentioned that "the instrument should not affect the dependent variable directly, but only through the endogenous variable". But if the instrument should not affect the dependent variable directly,

$cov(z,y) = 0$

would hold, making

$b_{IV} = 0$.

So what is happening?

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Short/Technical Answer:

It's untrue that $cov(y,z) = 0$ whenever $z$ does not directly affect $y$. The key is the word "directly". We hope to find an instrument such that "z influences x which influences y". This would mean that $cov(y,z|x) = 0$ (that is, if we observe the value of $x$, knowing $z$ tells us nothing incremental about $y$), but it does not mean that $cov(y,z) = 0$ unconditional on $x$.

Long/Intuitive Answer:

The distinction you need to make here is the difference between an associative (or correlative) effect and a causal one.

Let's first look at an example (linked at bottom). Suppose you want to know how the size of a political protest ($x$) causally affects local political outcomes ($y$).

However, if you simply look at the association between $x$ and $y$, you may note a number of potential issues. For example, the size of the protest ($x$) could itself be correlated with other factors like the importance of the issue or the likelihood of the protest causing change. These factors are also correlated with $y$. Hence, they confound the relationship between $x$ and $y$.

So how can an instrument variable help us? Consider the instrument of how good the weather was on the day of the protest ($z$). The intuition is that weather ($z$) could directly impact crowd size ($x$); however weather on the day of a protest ($z$) should not directly influence a local political outcome ($y$). The only way you would expect $z$ and $y$ to be related is through $x$.

Example Source: https://dash.harvard.edu/bitstream/handle/1/13457753/TeaParty_Protests.pdf

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    $\begingroup$ Thanks for this very good answer. So $cov(y,z|x) = 0$ means that z does not affect y conditional on x. This then implies that $cov(y,z) \neq 0$ because z can affect x and x can affect y. $\endgroup$ – Snoopy Aug 11 '18 at 13:23
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    $\begingroup$ Meanwhile, perhaps we should add that cov(z, y | x) = 0 also implies that z does not affect y through u. $\endgroup$ – Snoopy Aug 11 '18 at 14:05

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