1
$\begingroup$

I have been reading The Perils of Peer Effects paper by Josh Angrist http://www.nber.org/papers/w19774

On page 4, he transforms a condition expectation function:

$E(y|x,z)=\beta\mu_{(y|z)}+\gamma x $ where $\mu_{(y|z)}=E(y|z)$

to "a reduced form relation" by "iterating over x"

$E(y|z)=\frac{\gamma}{1-\beta} E(x|z) $

(All variables are mean zero)

I'm not quite sure how it is done and what he exactly means by "iterating over x". I have tried applying the law of total expectation with reference to x. In particular, I don't see how $E(x|z)$ enters the equation.

$\endgroup$
1
  • $\begingroup$ Typo, should it be $E(y | z) = \frac{\gamma}{1 - \beta} E(x \mid z)$? $\endgroup$ May 19 '16 at 6:18
1
$\begingroup$

I should have written this more succinctly but thought I'd try to be explicit...

For every value $x$ and $z$ of the random variables $X$ and $Z$, the following relation holds:

$$ E[Y \mid X=x, Z=z] = \beta E[Y \mid Z=z] + \gamma x $$

Let $P(X=x \mid Z=z)$ be the probability random variable $X$ takes the scalar value $x$ given that $Z=z$. Multiply both sides: $$ E[Y \mid X=x, Z=z] P(X=x\mid Z=z) = \beta E[Y \mid Z=z] P(X=x\mid Z=z) + \gamma x P(X=x \mid Z = z)$$ $$ \left(\sum_y y P(Y=y \mid X=x, Z=z)\right) P(X=x\mid Z=z) = \beta E[Y \mid Z=z] P(X=x\mid Z=z) + \gamma x P(X=x \mid Z = z)$$ Bayes rule: $$ \sum_y y P(Y=Y, X=x \mid Z=z) = \beta E[Y \mid Z=z] P(X=x\mid Z=z) + \gamma x P(X=x \mid Z = z)$$

Take the summation of both sides over all possible values of x: $$ \sum_x \sum_y y P(Y=Y, X=x \mid Z=z) = \sum_x \beta E[Y \mid Z=z] P(X=x\mid Z=z) + \sum_x \gamma x P(X=x \mid Z = z)$$ $$ \sum_y y P(Y=Y \mid Z=z) = \beta E[Y \mid Z=z] + \sum_x \gamma x P(X=x \mid Z = z)$$ $$ E[Y \mid Z=z] = \beta E[Y \mid Z=z] + \gamma E[X \mid Z=z]$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.