<blockquote>
<p>My first question is about the first assumption. It, combined with the assumption that <span class="math-container">$Y=\beta_0+\beta_1X$</span>, gives <span class="math-container">$E(Y|X)=\beta_0+\beta_1X$</span>. But isn't this equation the whole starting point of our regression analysis? If not, what are we trying to capture in the first place?
It seems like either <span class="math-container">$E(u_i|X_i)=0$</span> or <span class="math-container">$E(Y|X)=\beta_0+\beta_1X$</span> combined with the assumption that <span class="math-container">$Y_i=\beta_0+\beta_1X_i+u_i$</span> implies the other. Which comes first?</p>
</blockquote>
<p>The two are exchangeable. You can either start from:
<span class="math-container">$$
Y_i = \beta_0 + \beta_1X_i + u_i \tag{1}
$$</span>
and use this together with <span class="math-container">$\mathbb{E}(u_i|X_i)$</span> to get
<span class="math-container">$$
\mathbb{E}(Y_i|X_i) = \beta_0 + \beta_1 X_i. \tag{2}
$$</span>
Vice versa if you start from <span class="math-container">$(2)$</span> you can define:
<span class="math-container">$$
u_i = Y_i - \mathbb{E}(Y_i|X_i).
$$</span>
such that <span class="math-container">$Y_i = \beta_0 + \beta_1 X_i + u_i$</span> and:
<span class="math-container">$$
\begin{align*}
\mathbb{E}(u_i|X_i) &amp;= \mathbb{E}(Y_i|X_i) - \mathbb{E}(\mathbb{E}(Y_i|X_i)|X_i),\\
&amp;= 0
\end{align*}
$$</span>
So it does not matter. Either you say that the expected mean of <span class="math-container">$Y$</span> conditional on <span class="math-container">$X$</span> is linear, or you say that <span class="math-container">$Y$</span> is linear in <span class="math-container">$X$</span> and <span class="math-container">$u_i$</span> together with the assumption that the mean of <span class="math-container">$u_i$</span> conditional on <span class="math-container">$X$</span> is zero.</p>
<blockquote>
<p>My second question is about the expectation of <span class="math-container">$u$</span>. Do we assume that <span class="math-container">$E(u_i)=0$</span> or is it implied by other assumptions? Because in Wooldridge's Introduction to Econometrics, it's an assumption but in Stock and Watson's, it looks like it's implied. I'm confused here.</p>
</blockquote>
<p>In order to see what happens if we do not impose it, notice that in this case, we may write:
<span class="math-container">$$
u_i = \mathbb{E}(u_i) + \underbrace{(u_i - \mathbb{E}(u_i))}_{\delta_i}
$$</span>
Now, <span class="math-container">$\delta_i$</span> has mean zero,
<span class="math-container">$$
\mathbb{E}(\delta_i) = \mathbb{E}(u_i) - \mathbb{E}(u_i) = 0.
$$</span>
We can now write:
<span class="math-container">$$
\begin{align*}
Y_i &amp;= \beta_0 + \beta_1 X_i + u_i,\\
&amp;= \left(\beta_0 + \mathbb{E}(u_i)\right) + \beta_1 X_i + \delta_i
\end{align*}
$$</span>
So if the mean of <span class="math-container">$u_i$</span> is not zero, it simply get's added to the intercept <span class="math-container">$\beta_0$</span>. In other words, if you add an intercept to the regression, assuming that <span class="math-container">$\mathbb{E}(u_i) = 0$</span> is without loss of generality.</p>

CC BY-SA 4.0
2021-09-21T14:24:16.113Z
Thanks! So I guess we can say that there's actually four assumptions instead of just three?
WinnieXi
CC BY-SA 4.0
2021-09-24T11:22:41.707Z
@WinnieXi Not really. The condition $\mathbb{E}(u_i) = 0$ follows from the assumption that $\mathbb{E}(u_i|X_i) = 0$ (using the law of iterated expectations).
tdm
CC BY-SA 4.0
2021-09-20T16:10:24.257Z
2021-09-20T16:10:24.257Z
tdm
<p>Here is my understanding of the linear regression model with single regressor:
We suppose that the population regression function takes the form of <span class="math-container">$Y_i=\beta_0+\beta_1X_i+u_i$</span>. Furthermore, for the parameters to have causality meanings, we assume that:</p>
<ol>
<li><span class="math-container">$E(u_i|X_i)=0$</span></li>
<li><span class="math-container">$(X_i,Y_i)$</span> are i.i.d. for <span class="math-container">$i=1,...,n$</span></li>
<li>Large outliers are unlikely</li>
</ol>
<p>My first question is about the first assumption. It, combined with the assumption that <span class="math-container">$Y=\beta_0+\beta_1X$</span>, gives <span class="math-container">$E(Y|X)=\beta_0+\beta_1X$</span>. But isn't this equation the whole starting point of our regression analysis? If not, what are we trying to capture in the first place?<br />
It seems like either <span class="math-container">$E(u_i|X_i)=0$</span> or <span class="math-container">$E(Y|X)=\beta_0+\beta_1X$</span> combined with the assumption that <span class="math-container">$Y_i=\beta_0+\beta_1X_i+u_i$</span> implies the other. Which comes first?</p>
<p>My second question is about the expectation of <span class="math-container">$u$</span>. Do we assume that <span class="math-container">$E(u_i)=0$</span> or is it implied by other assumptions? Because in Wooldridge's Introduction to Econometrics, it's an assumption but in Stock and Watson's, it looks like it's implied. I'm confused here.</p>

<regression>
Regarding the starting point and assumptions of linear regression model
WinnieXi
# Regarding the starting point and assumptions of linear regression model

(Asked by: username_0 on 2021-09-20)

Here is my understanding of the linear regression model with single regressor:
We suppose that the population regression function takes the form of $Y\_i=\beta\_0+\beta\_1X\_i+u\_i$. Furthermore, for the parameters to have causality meanings, we assume that:

1. $E(u\_i|X\_i)=0$
2. $(X\_i,Y\_i)$ are i.i.d. for $i=1,...,n$
3. Large outliers are unlikely

My first question is about the first assumption. It, combined with the assumption that $Y=\beta\_0+\beta\_1X$, gives $E(Y|X)=\beta\_0+\beta\_1X$. But isn't this equation the whole starting point of our regression analysis? If not, what are we trying to capture in the first place?  
It seems like either $E(u\_i|X\_i)=0$ or $E(Y|X)=\beta\_0+\beta\_1X$ combined with the assumption that $Y\_i=\beta\_0+\beta\_1X\_i+u\_i$ implies the other. Which comes first?

My second question is about the expectation of $u$. Do we assume that $E(u\_i)=0$ or is it implied by other assumptions? Because in Wooldridge's Introduction to Econometrics, it's an assumption but in Stock and Watson's, it looks like it's implied. I'm confused here.

**Tags:** regression

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## 1 answers

### Answer by username_1 on 2021-09-20. Score: 3 | Accepted answer

> My first question is about the first assumption. It, combined with the assumption that $Y=\beta\_0+\beta\_1X$, gives $E(Y|X)=\beta\_0+\beta\_1X$. But isn't this equation the whole starting point of our regression analysis? If not, what are we trying to capture in the first place?
> It seems like either $E(u\_i|X\_i)=0$ or $E(Y|X)=\beta\_0+\beta\_1X$ combined with the assumption that $Y\_i=\beta\_0+\beta\_1X\_i+u\_i$ implies the other. Which comes first?

The two are exchangeable. You can either start from:
$$
Y\_i = \beta\_0 + \beta\_1X\_i + u\_i \tag{1}
$$
and use this together with $\mathbb{E}(u\_i|X\_i)$ to get
$$
\mathbb{E}(Y\_i|X\_i) = \beta\_0 + \beta\_1 X\_i. \tag{2}
$$
Vice versa if you start from $(2)$ you can define:
$$
u\_i = Y\_i - \mathbb{E}(Y\_i|X\_i).
$$
such that $Y\_i = \beta\_0 + \beta\_1 X\_i + u\_i$ and:
$$
\begin{align\*}
\mathbb{E}(u\_i|X\_i) &= \mathbb{E}(Y\_i|X\_i) - \mathbb{E}(\mathbb{E}(Y\_i|X\_i)|X\_i),\\
&= 0
\end{align\*}
$$
So it does not matter. Either you say that the expected mean of $Y$ conditional on $X$ is linear, or you say that $Y$ is linear in $X$ and $u\_i$ together with the assumption that the mean of $u\_i$ conditional on $X$ is zero.

> My second question is about the expectation of $u$. Do we assume that $E(u\_i)=0$ or is it implied by other assumptions? Because in Wooldridge's Introduction to Econometrics, it's an assumption but in Stock and Watson's, it looks like it's implied. I'm confused here.

In order to see what happens if we do not impose it, notice that in this case, we may write:
$$
u\_i = \mathbb{E}(u\_i) + \underbrace{(u\_i - \mathbb{E}(u\_i))}\_{\delta\_i}
$$
Now, $\delta\_i$ has mean zero,
$$
\mathbb{E}(\delta\_i) = \mathbb{E}(u\_i) - \mathbb{E}(u\_i) = 0.
$$
We can now write:
$$
\begin{align\*}
Y\_i &= \beta\_0 + \beta\_1 X\_i + u\_i,\\
&= \left(\beta\_0 + \mathbb{E}(u\_i)\right) + \beta\_1 X\_i + \delta\_i
\end{align\*}
$$
So if the mean of $u\_i$ is not zero, it simply get's added to the intercept $\beta\_0$. In other words, if you add an intercept to the regression, assuming that $\mathbb{E}(u\_i) = 0$ is without loss of generality.

> **Comments:**
> **username_0**: Thanks! So I guess we can say that there's actually four assumptions instead of just three?  
> **username_1**: @WinnieXi Not really. The condition $\mathbb{E}(u_i) = 0$ follows from the assumption that $\mathbb{E}(u_i|X_i) = 0$ (using the law of iterated expectations).  

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