not sure if this was the right place to ask my question, but I saw some questions regarding linear regression so I'd thought I would try to get some answers here. I just started learning about linear regression so this is the homework posed to me.

I assume that the statement is true since 𝛽1 is the coefficient for 𝑥. And its the coefficient that would determine if the slope (i.e. the relationship) is positive or negative.

Am I missing out anything or what should I expound on?

Thanks for reading and for the guides and opinions.


1 Answer 1


Hi and welcome to the forum. Homework is not so well received here in the forum. But still a fair question in my view.

You have two aspects here. In principle you are right that $\beta_1$ is the slope of $x_1$ (you can say the marginal effect of $x_1$ on $y$) and $\beta_0$ is the intercept. This is simply a linear function of form $f(x)=\beta_0 + \beta_1 x$.

However, to claim "causality", a few more things are required. First, you need to make the assumption that there is a causal relation between $x$ and $y$ and $x$ must be exogenous.

Another important aspect is, that if there are additional variables with a causal influence on $y$, say $x_2$, you cannot really claim that $\beta_1$ is the causal effect on $...$, because you omitted $x_2$, so that your model suffers from the omitted variable bias. To claim for causality you need to make sure that your model reflects the data generating process in a proper way.

  • $\begingroup$ Hi Peter, thanks for taking the time to answer my question. When you say x must be exogenous, do you mean that x must be an independent variable? $\endgroup$
    – Van
    Aug 23, 2019 at 14:44
  • $\begingroup$ This means that x determines y, but y does not determine x, so the „direction“ of causality. If you think about the number of police officers (x) and crime (y) you could not claim that x is exogenous, because cities with higher crime rates might hire more police officers, so in this case there is a „more complex“ interaction between x and y. $\endgroup$
    – Peter
    Aug 23, 2019 at 14:57

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