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Suppose I have a linear regression form of

$$ \log(Y) = \beta_0 + \beta_1X_2 + \beta_2X_3 + \beta_3X_1Z + \beta_4X_2Z + \epsilon $$

where $X_1, X_2, X_3$ are binary and $X_1$ is omitted as a reference variable. Suppose $Z$ is also binary 0-1. I am wondering how we would be interpret $\beta_1$ and $\beta_2$?

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  • $\begingroup$ Note that $\beta_1 = \mathbb{E}[\log Y \mid X_2 = 1, Z = 0] - \mathbb{E}[\log Y \mid X_2 = 0, Z = 0]$ which of course the mean difference in $\log Y$ between a particular treatment ($X_2 = 1$) and control ($X_2 = 0$) when $Z = 0$. If the difference is small then you can interpret it as a percentage change in $Y$ between treatment and control holding Z at 0 ... $\endgroup$ Commented Feb 14, 2023 at 6:22
  • $\begingroup$ Since you have X1 in an interaction you should have it as a main effect. Alternatively, you can take it out of the interaction and add X3. Any good stat software will do this for you. $\endgroup$
    – Peter Flom
    Commented Jan 15 at 17:41

1 Answer 1

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You can interpret $\beta_1$ as the percent change in Y when the treatment effect $X_2$ is applied, when you're holding other variables constant, in this case that is Z.

You can scale this interpretation to other $\beta_s$ as well.

Hope this helps.

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    $\begingroup$ This is incorrect in two ways. (1) Assuming "treatment effect $X_2$ is applied" means that $X_2$ is changed from $0$ to $1,$ then $\log(Y)$ changes by adding $\beta_1+\beta_4 Z,$ not just $\beta_1$ alone. (2) Interpreting that as a proportional change in $\log(Y)$ assumes the change is relatively small: less than $0.1$ in absolute value will do; less than $0.25$ is OK; beyond that it's usually a good idea to distinguish changes in the log from proportional changes. $\endgroup$
    – whuber
    Commented Feb 14, 2023 at 15:37

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