# Effect of scaling the dependent variable

I was wondering about this problem while I was running a normal regression. Assume that we have the standard regression setup with one dependent variable $Y$ and an independent variable $X$, and the model is: $log(100/Y_i) \sim N(\beta_0 + \beta_1X_i, \sigma^2), i = 1, \ldots, n$.

Suppose that we now fit a new model $log(150/Y_i) \sim N(\beta_0^{'} + \beta_1^{'}X, \sigma^2)$, what would the relationship between the regression coefficients $(\beta_0, \beta_1)$ and $(\beta_0^{'}, \beta_1^{'})$ be?

While it is easy to derive this relationship in the case of a simple linear regression, i.e., if $Y_i \sim N(\beta_0 + \beta_1X_i, \sigma^2)$, then $Y_i + 100 \sim N(100 + \beta_0 + \beta_1X_i, \sigma^2)$, and $2*Y_i \sim N(2 * \beta_0 + 2 * \beta_1X_i, 4\sigma^2)$, is there a way to derive the relationships between $(\beta_0, \beta_1)$ and $(\beta_0^{'}, \beta_1^{'})$ in a similar way?

• Hint: $\log(150/Y) = \log(150/100) + \log(100/Y)$. Take it from there.
– whuber
Aug 23 '17 at 18:11
• Hmm.. so the intercepts would differ by log(1.5) while the slopes would remain the same... wonder if this would be a same in case it is a mixed effects model with X as the fixed effect and Z as a simple scalar random effect. Aug 23 '17 at 18:35

You can open the log term.

log(100/Y) -> log(100) - log(Y)

log(150/Y) -> log(150) - log(Y)

Then the difference between both will be a constant term.