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Given a linear regression model:

$$ Y_i = \beta_0+\beta_1X_{i1}+\epsilon_i $$

I have seen two interpretations of $\beta_1$:

  1. "$\beta_1$ is the amount $Y_i$ increases by when $X_{i1}$ increases by 1 unit"

  2. "$\beta_1$ is the amount $E(Y_i\mid X_{i1})$ increases by when $X_{i1}$ increases by 1 unit"

Which of the two is technically the correct interpretation?

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1 Answer 1

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Here is what $E[Y|X]$ changes when $X$ changes by one unit (using that the expectation operator is linear and that $\epsilon$ is independent of $X$ with $E[\epsilon] = 0$): $$ \begin{align} E[Y|X = x_0+1] - E[Y|X=x_0] &= E[\beta_0 + \beta_1X + \epsilon|X=x_0+1] - E[\beta_0 + \beta_1X + \epsilon|X=x_0]\\ &= \beta_0 + \beta_1 E[X|X=x_0+1] + E[\epsilon|X=x_0+1] - (\beta_0 + \beta_1 E[X|X=x_0] + E[\epsilon|X=x_0])\\ &= \beta_0 + \beta_1(x_0+1)+ 0 - (\beta_0 + \beta_1x_0 + 0)\\ &= \beta_1, \end{align} $$ so the second interpretation is correct.

However, when people use the first version, they mean the second, it is just a bit of a lax way of expressing themselves.


In the comments, @RichardHardy makes the point that one should distinguish between a purely probabilistic and causal interpretation of those two statements.

In the answer above, I presumed those statements to be meant probabilistically, noncausally. Of course, there is also a causal interpretation along the lines: "If $X$ is intervened upon and set to $x_0$ and to $x_0+1$, how would $Y$ change as a result? I.e., what is the average causal effect (ACE) of $X$ on $Y$?" Of course, there are many situations where $\beta_1$ cannot be interpreted as the ACE, e.g. if $Y$ would be causing $X$.

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    $\begingroup$ +1 That last paragraph is an important insight. $\endgroup$
    – Dave
    Jul 31, 2022 at 10:56
  • $\begingroup$ Consider making a clearer distinction between causal and noncausal interpretations. The OPs formulation is essentially causal and is only correct when the regression can be interpreted causally. $\endgroup$ Jul 31, 2022 at 12:36
  • $\begingroup$ @RichardHardy I updated the answer. $\endgroup$
    – frank
    Jul 31, 2022 at 14:51

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