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I have the following model:

$\log(y)=\beta_0 + \beta_1 x_1 + \beta_2 \log(x_2) + \beta_3 x_1 \log(x_2) $

In interpreting the % change of $y$ that corresponds with a 1% increase in $x_2$ at a specific value of $x_1$ (.68), is the following correct?

% change in y =

\begin{equation} \bigg[\big[\exp\big(\beta_1.68\ + \beta_2 \log\left(1.01\right) + \beta_3 .68\log(1.01)\big) \times \log(1.01)\big] - 1\bigg] \times 100 \end{equation}

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Mathematically:

$$ \log \left(\frac{ E[Y | X_1 = 0.68, X_2 = 1.01x_2 ] } { E[Y | X_1 = 0.68, X_2 = x_2] }\right) $$ \begin{array}{cll} \\ &=& E[\log Y | X_1 = 0.68, X_2 = 1.01x_2 ]\; -\; E[\log Y | X_1 = 0.68, X_2 = x_2] \\ \\ &=& \beta_0 + \beta_1 0.68 +\beta_2 (\log x_2 + \log 1.01) + \beta_3 0.68 (\log x_2 + \log 1.01) - \\ & & \beta_0 - \beta_1 0.68 - \beta_2 \log x_2 \qquad\qquad\quad- \beta_3 0.68 \log x_2 \\ \\ &=& \beta_2 \log 1.01 + \beta_3 0.68 \log 1.01 \end{array}

Exponentiating the final value gives you what you are after.

$$ \exp (\beta_2) \exp (\beta_3 0.68)$$

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  • $\begingroup$ I edited this to make the $\TeX$ easier to read. If you don't like it, roll it back w/ my apologies. $\endgroup$ – gung Jul 13 '15 at 22:26
  • $\begingroup$ @gung I appreciate it. It was exceptionally difficult to write out the long equations. $\endgroup$ – AdamO Jul 13 '15 at 22:33

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