For a basic linear regression:
$$y = \beta_0 + \beta_1 x + \varepsilon$$
Its clear to see the interpretation of the estimated parameters:
$$y = \beta_0 + \beta_1 x + \varepsilon$$
$$\frac{dy}{dx} = \beta_1$$
A unit increase in $x$ is associated with an estimated $\beta_1$ unit change in $y$.
I am trying to understand the same interoperations for the following models:
Model 1 (log-log): $$\log(y) = \beta_0 + \beta_1 \log(x) + \varepsilon$$
Model 2 (log-linear): $$\log(y) = \beta_0 + \beta_1 x + \varepsilon$$
I tried to derive these myself.
Model 1:
$$\log(y) = \beta_0 + \beta_1 \log(x) + \varepsilon$$
$$\frac{d(\log(y))}{dx} = \beta_1 \frac{d(\log(x))}{dx}$$
$$\frac{1}{y} \cdot \frac{dy}{dx} = \beta_1 \cdot \frac{1}{x}$$
$$\frac{dy}{dx} \cdot \frac{x}{y} = \beta_1$$
$$\beta_1 = \frac{dy}{dx} \cdot \frac{x}{y}$$
$$\beta_1 = \frac{dy/y}{dx/x}$$
From calculus, we know that:
$$\text{Percentage change in }y = \frac{y_{\text{new}} - y_{\text{old}}}{y_{\text{old}}} \times 100\%$$
$$\Delta y = y_{\text{new}} - y_{\text{old}}$$ $$\text{Proportional change in }y = \frac{\Delta y}{y}$$
Thus:
- $\frac{dy}{y} \approx$ percentage change in $y$
- $\frac{dx}{x} \approx$ percentage change in $x$
Therefore $\beta_1$ represents the ratio of these percentage changes:
$$\beta_1 = \frac{\text{percentage change in }y}{\text{percentage change in }x}$$
Model 2:
$$\log(y) = \beta_0 + \beta_1 x + \varepsilon$$
$$\frac{d\log(y)}{dx} = \beta_1$$
$$\frac{1}{y} \cdot \frac{dy}{dx} = \beta_1$$
$$\frac{dy}{dx} = \beta_1 \cdot y$$
$$\frac{dy/y}{dx} = \beta_1$$
Thus, the interpretation becomes:
$$\beta_1 = \frac{\text{percentage change in }y}{\text{unit change in }x}$$
To conclude:
- In Model 1, a percent increase in $x$ is associated with an estimated $\beta_1$ percent change in $y$
- In Model 2, a unit increase in $x$ is associated with an estimated $\beta_1$ percent change in $y$
Are these the correct interpretations?
Based on Greg H's comments, I tried to correct the derivations for Model 2:
$$\log(y) = \beta_0 + \beta_1 x + \varepsilon$$
$$e^{\log(y)} = e^{\beta_0 + \beta_1 x + \varepsilon}$$ $$y = e^{\beta_0} \cdot e^{\beta_1 x} \cdot e^{\varepsilon}$$
Define $A = e^{\beta_0}$ (ignore the error term): $$y = A \cdot e^{\beta_1 x}$$
Define the original value $y_1$ (at $x_1$) and the new value $y_2$ (at $x_2 = x_1 + 1$):
At $x_1$: $$y_1 = A \cdot e^{\beta_1 x_1}$$
At $x_2 = x_1 + 1$: $$y_2 = A \cdot e^{\beta_1 (x_1 + 1)}$$ $$y_2 = A \cdot e^{\beta_1 x_1} \cdot e^{\beta_1}$$ $$y_2 = y_1 \cdot e^{\beta_1}$$
Thus, the ratio of the new value to the original value is: $$\frac{y_2}{y_1} = e^{\beta_1}$$
$$\Delta y = y_2 - y_1 = y_1 \cdot e^{\beta_1} - y_1 = y_1(e^{\beta_1} - 1)$$
And the proportional change is: $$\frac{\Delta y}{y_1} = \frac{y_1(e^{\beta_1} - 1)}{y_1} = e^{\beta_1} - 1$$
$$\text{Percentage change in } y = (e^{\beta_1} - 1) \times 100\%$$
When $x$ increases by one unit, $y$ changes by $(e^{\beta_1} - 1) \times 100\%$.
