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?


1 Answer 1


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 $X$.

  • 1
    $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.