<p>It is very unclear what you are asking. But in most regressions it is assumed that the error term (I assume "e" is the error term) is assumed to be iid for all values of x. In particular, in the discrete case <span class="math-container">$E[\epsilon|d=1]=E[\epsilon|d=0]$</span>, but this can be generalized by saying <span class="math-container">$E[\epsilon|x]=E[\epsilon|x']$</span> for all <span class="math-container">$x, x'$</span>. Now if your regression also includes a constant term, it is without loss of generality to assume that the expected value of <span class="math-container">$\epsilon$</span> (which is the same for all values of x) is equal to zero. Thus, the importance of the assumption <span class="math-container">$E[\epsilon|x]=0$</span> is not so much the "equal to zero" part, but the fact that is the same for all values of x. </p>

CC BY-SA 4.0
2019-07-03T00:44:46.427Z
So just to make sure I understand, if the exogeneity assumption is satisfied, that means in my population (or on average across many samples), when increasing  x the expectation of errors is the same? so basically there is no systematic relationship between the errors and the values of x?    

also, if I dont include an intercept, does that mean that E[ϵ|x]=/=0? but the exogeneity means that the conditional expectation is basically a horizontal line?
Steve
CC BY-SA 4.0
2019-07-03T04:43:58.680Z
The conditional expectation of the errors is the same, yes. Exactly no systematic relationship between errors and x. If you don't have an intercept, but there is still no systematic relationship between errors and x, then $E[\epsilon|x]=c$ for some constant $c$. I am not sure what you mean by a horizontal line. It is always a constant, equal to zero if you include an intercept, or equal to some other value otherwise.
Regio
CC BY-SA 4.0
2019-07-03T14:03:57.880Z
makes sense, thank you! by horizontal I meant if I were to graph the conditional expectation as a continuous function of x (assuming its just one variable), it would just be a flat line. Which I think makes sense if its constant
Steve
CC BY-SA 4.0
2019-07-03T00:01:22.297Z
2019-07-03T00:01:22.297Z
Regio
<p>in the discrete case, if assignemt is random, then i can express E[y|d=1]-E[y|d=0] = B + E[e|d=1]-E[e|d=0], where the expectation of the errors are the same for both groups and become zero. Where I am confused is for a linear regression/continuous variable, we say this condition is expressed by E[e|x]=0, and I struggle to see how they are the same. its not the same as saying E[e|d]=0, right?</p>

<p>If I take a partial derivative of E[y|x] wrt x, I get B+ dE[e|x]/dx. cant I have this last term be zero without having E[e|x]=0? is E[e|x1,x2]=E[e|x2] the same thing as E[e|x]=0</p>

<microeconomics><econometrics><statistics><applied-econometrics>
is there a difference between E[e|x]=0 and E[e|d=1]-E[e|d=0] in continuous vs discrete case in regressions?
Steve
# is there a difference between E[e|x]=0 and E[e|d=1]-E[e|d=0] in continuous vs discrete case in regressions?

(Asked by: username_0 on 2019-07-02)

in the discrete case, if assignemt is random, then i can express E[y|d=1]-E[y|d=0] = B + E[e|d=1]-E[e|d=0], where the expectation of the errors are the same for both groups and become zero. Where I am confused is for a linear regression/continuous variable, we say this condition is expressed by E[e|x]=0, and I struggle to see how they are the same. its not the same as saying E[e|d]=0, right?

If I take a partial derivative of E[y|x] wrt x, I get B+ dE[e|x]/dx. cant I have this last term be zero without having E[e|x]=0? is E[e|x1,x2]=E[e|x2] the same thing as E[e|x]=0

**Tags:** microeconomics, econometrics, statistics, applied-econometrics


---

## 1 answers

### Answer by username_1 on 2019-07-03. Score: 0 | Accepted answer

It is very unclear what you are asking. But in most regressions it is assumed that the error term (I assume "e" is the error term) is assumed to be iid for all values of x. In particular, in the discrete case $E[\epsilon|d=1]=E[\epsilon|d=0]$, but this can be generalized by saying $E[\epsilon|x]=E[\epsilon|x']$ for all $x, x'$. Now if your regression also includes a constant term, it is without loss of generality to assume that the expected value of $\epsilon$ (which is the same for all values of x) is equal to zero. Thus, the importance of the assumption $E[\epsilon|x]=0$ is not so much the "equal to zero" part, but the fact that is the same for all values of x.

> **Comments:**
> **username_0**: So just to make sure I understand, if the exogeneity assumption is satisfied, that means in my population (or on average across many samples), when increasing  x the expectation of errors is the same? so basically there is no systematic relationship between the errors and the values of x?      also, if I dont include an intercept, does that mean that E[ϵ|x]=/=0? but the exogeneity means that the conditional expectation is basically a horizontal line?  
> **username_1**: The conditional expectation of the errors is the same, yes. Exactly no systematic relationship between errors and x. If you don't have an intercept, but there is still no systematic relationship between errors and x, then $E[\epsilon|x]=c$ for some constant $c$. I am not sure what you mean by a horizontal line. It is always a constant, equal to zero if you include an intercept, or equal to some other value otherwise.  
> **username_0**: makes sense, thank you! by horizontal I meant if I were to graph the conditional expectation as a continuous function of x (assuming its just one variable), it would just be a flat line. Which I think makes sense if its constant  

---

