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How do I interpret a log-linear interaction term, is it possible?

My model: $Y= B_1 + B_2\log X_1 + B_3X_2 + B_4(\log X_1 X_2) + u $

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    $\begingroup$ A true "interaction term" would be $X_2\log(X_1)$ rather than $\log X_1X_2 = \log(X_1) + \log(X_2)$ (when the $X_i$ are both positive, anyway). Which one did you mean? $\endgroup$ – whuber Apr 8 at 18:42
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Correct Model

As @whuber suggested, your interaction term should consist of the product of $\log X_1$ and $X_2$. That means your model should be specified as:

$Y= B_1 + B_2\log X_1 + B_3X_2 + B_4(\log X_1)(X_2) + u $

In this model, the effect of $\log X_1$ on $Y$ is assumed to depend on $X_2$. Conversely, the effect of $X_2$ on $Y$ is assumed to depend on $\log X_1$.

What does the effect of of $\log X_1$ on $Y$ look like, given $X_2$?

To see how the effect of $\log X_1$ on $Y$ depends on $X_2$, you can re-write the model as follows:

$Y= B_1 + (B_2 + B_4X_2)\log X_1 + B_3X_2 + u $

From this new expression of the model, you can see that the effect of $\log X_1$ on $Y$ is quantified by the slope $B_2 + B_4X_2$. In particular, given $X_2$, every 1-unit increase in the value of $\log X_1$ is associated with a change of $B_2 + B_4X_2$ units in the expected value of $Y$. This change clearly depends on $X_2$.

What does the effect of of $X_2$ on $Y$ look like, given $X_1$?

To see how the effect of $X_2$ on $Y$ depends on $\log X_1$, you can re-write the model as follows:

$Y= B_1 + B_2\log X_1 + (B_3 + B_4 \log X_1)X_2 + u $

From this alternative formulation of the original model, you can surmise that the effect of $X_2$ on $Y$ is quantified by the slope $B_3 + B_4 \log X_1$. Thus, given $X_1$, every 1-unit increase in the value of $X_2$ is associated with a change of $B_3 + B_4 \log X_1$ units in the expected value of $Y$.

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