# Implications of strict exogeneity for OLS in time series

Zero Conditional Mean (ZCM), or Strict Exogeneity, is given by:
$$E[u|X]=0$$
Equivalently,
$$E[u_t|X]=0, t=1,...,T$$

Is it true that this implies:

• Zero Unconditional Mean: $$E[u_t]=0, \forall t$$
• Contemporaneous Exogeneity: $$E[u_t|x_t]=0, \forall t$$ (Where $$x_t$$ is a vector of explanatory variables)

• $$E[x_su_t]=0, \forall t,s$$

And are there any other things that ZCM imply that are particularly useful?

## 1 Answer

Well consider the Law of Iterated expectation $$\mathbb E[ \mathbb E[Y\lvert W]]= \mathbb E[Y]$$

and apply it with $$Y=\mathbf u$$ and $$W = \mathbf X$$, to get $$\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]]= \mathbb E[\mathbf u]$$ and $$\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]] = \mathbb E = 0$$, hence $$\mathbb E[\mathbf u]=0$$.

Consider a generalized version of Law of Iterated Expectation mentioned in the post A generalization of the Law of Iterated Expectations which states that

$$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$

and let $$W_1 = \mathbf X\setminus \mathbf x_t$$ (read $$\mathbf X$$ except the vector $$\mathbf x_t$$) and let $$W_2=\mathbf x_t$$ then

$$\mathbb E[\mathbb E[u_t\lvert \mathbf X]\lvert \mathbf x_t] = \mathbb E[\mathbb E[u_t\lvert \mathbf X\setminus \mathbf x_t,\mathbf x_t]\lvert \mathbf x_t] = \mathbb E[u_t \lvert \mathbf x_t]$$ and as before it must be 0 because $$\mathbb E[u_t\lvert \mathbf X]=0$$.

Use the same argument to get $$\mathbb E[u_t \lvert \mathbf x_s] = 0$$ for all $$s,t = 1,...,T$$ and use this to conclude that

$$\mathbb E[u_t \mathbf x_s] = \mathbb E[\mathbb E[u_t \mathbf x_s\lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \mathbb E[u_t \lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \cdot 0 ] = 0$$

So the answer must be yes they are all valid statements and further more $$\mathbb E[u_t \lvert \mathbf x_s] = 0$$ for all $$s,t = 1,...,T$$ as was used.

An intuitive statement of the statement $$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$ is that the least information set always dominates so you also have $$\mathbb E[\mathbb E[Y\lvert W_2] \lvert W_1, W_2] = \mathbb E[Y\lvert W_2]$$ again this is mentioned in the post referred to above.

• I think the variables in the first 3 lines are mixed up, but I get the point, thanks. – DQd Dec 28 '18 at 17:50
• pls accept the answer if it was useful. I have edited the mix up. – Jesper for President Dec 28 '18 at 18:13
• Shouldn't the first equation be $\mathbb E[\mathbb E[Y\mid W]]=\mathbb E[Y]$? – Cm7F7Bb Feb 21 '19 at 9:11
• yes offcourse thx – Jesper for President Feb 22 '19 at 0:45