Variance proportional to a regressor squared: divide it out

Question

You have strong reason to believe the error variance grows with the square of a positive regressor: $\operatorname{Var}(\varepsilon_i \mid x_i) = \sigma^2 x_i^2$ .

Show how a simple transformation turns this into a constant-variance regression, and explain what estimator that yields and why it is preferable to OLS with robust standard errors.

Accepted Answer

Divide the whole equation by the standard deviation. The error standard deviation for row i is \sigma xi. Start from yi = \beta0 + \beta1 xi + \varepsiloni, and divide every term by xi: \frac{yi}{xi} = \beta0 \frac{1}{xi} + \beta1 + \frac{\varepsiloni}{xi}. Check the new error. The transformed error is \varepsiloni / xi, whose variance is \operatorname{Var}\!\left(\frac{\varepsiloni}{xi}\right) = \frac{\operatorname{Var}(\varepsiloni)}{xi^2} = \frac{\sigma^2 xi^2}{xi^2} = \sigma^2, constant. So OLS on the transformed variables (regressing yi/xi on 1/xi and a constant, note the intercept and s…

Variance proportional to a regressor squared: divide it out

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