I am having trouble following a short derivation that uses the Law of Iterated Expectations that is found in the answer to another question: How to derive a regression formula

I will repeat it here:

Let $E(y|z) = \mu_{y|z}.$ Then it is shown that $E(y \mu_{y|z}) = Var(\mu_{y|z})$ in the following steps:

(1) $E(y \mu_{y|z}) = E(E(y|z, \mu_{y|z}) *\mu_{y|z})$

(2) $~~~~~~~~~~~~~= E(E(y|z) *\mu_{y|z})$

(3) $~~~~~~~~~~~~~= E(\mu^2_{y|z})$

(4) $~~~~~~~~~~~~~= Var(\mu^2_{y|z})$

I don't know all the properties of the LIE, but I do know that in general it gives $E(W) = E_Z(E(W|Z))$. With that said:

Q1. In line 1, what is going on? How is the LIE being applied in this way?

Q2. In going from line 1 to line 2, why do we not condition on $\mu_{y|z}$ anymore?

Q3. In going from line 3 to line 4, why is $(E(\mu_{y|z}))^2 = 0$ so that we get the variance?


1 Answer 1


By the defining property of the conditional expectation,

$$E\big[E(W\mid Q)\big] = E(W)$$

Set $W \equiv y\cdot E(y\mid z)$ and $Q \equiv z$. Substitute to get

$$E\Big(E\big [y\cdot E(y\mid z) \Big|z\big]\Big)= E\big [y\cdot E(y\mid z)\big] $$

The right-hand-side is what you are starting with, so we need to manipulate the left-hand-side.

By the measurability of $E(y\mid z)$ with respect to $z$ ("take out what is known") we have

$$E\Big(E\big [y\cdot E(y\mid z) \Big|z\big]\Big) = E\Big(E(y\mid z)\cdot E\big [y \mid z\big]\Big)$$

$$=E\left(\left[E(y\mid z)\right]^2\right) = {\rm Var}[E(y\mid z)] +\big(E\left[E(y\mid z)\right]\big)^2$$

$$={\rm Var}[E(y\mid z)] +\big(E[y]\big)^2$$

So for the stated result to go through, it must be the case that the unconditional expected value of $y$ is zero. This is of course not a general result, but it is an assumption explicitly made in Angrist's paper (which is used in the CV thread the OP linked to), where just above his eq. $(2)$ the author writes:

"Let $\beta$ denote the population regression coefficient from a regression of (mean zero) $y$ on $μ_{y|z} = E[y|z]$, for any random variables, $y$ and $z$."

  • 1
    $\begingroup$ +1 That's a good point Alecos, I had not mentioned this assumtpion in my previous answer to the OP. I will edit the answer accordingly. $\endgroup$
    – Andy
    Commented Dec 1, 2014 at 8:40

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