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Consider a simple linear regression model with response $y$, an intercept and an explanatory variable $x$,i.e. $$y_i=\beta_0+\beta_1x_i+\epsilon_i, \;\;\;\;\;\;\;\; i=1,...,n.$$ Assume further that the residual variance proportional to the square of $x_i$: $V(\epsilon_i)=\sigma^2x^2_i$. Consider now the following data transformation: $$\tilde{y}_i:=\frac{y_i}{x_i} \text{ and } \tilde{x_i}:=\frac{1}{x_i}, \;\;\;\;\;\;\;\; i=1,...,n.$$ (1) If we use weighted least squares with weights $w_i =\frac { 1}{x^2_i}$.Is this equivalent to the above considered transformation? (2)Assume $\beta_0 \in \Bbb R $ was known. What other variance stabilizing transformation can be used in this case?

ANSWER (1):

Yes, using weighted least squares with weights $$w_i =\frac { 1}{x^2_i}$$ is equivalent to applying the given transformation $$\tilde{y}_i:=\frac{y_i}{x_i}$$ and then performing ordinary least squares.

This is because the weights $w_i$ effectively adjust $$Var (\tilde\epsilon_i):=\epsilon_i\tilde x_i$$ to be constant, regardless of the value of $x_i$. This weighting scheme implies that observations with higher values of $x_i$(lower variance, since weights are inversely proportional to variance) have more influence on the estimation of the model parameters. The variance of the residuals becomes constant when using these weights, which aligns with the objective of variance stabilization.

Can anyone do (2) by using Box-Cox transformation. and please check 1 ,am i doing it correctly?

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1 Answer 1

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Using WLS with weights $w_i=1/x_i^2$ is clearly equivalent to that transformation.

When $\beta_0\in\mathbb R$ is known, a proper transformation can be obtained from $$ \frac{y_i-\beta_0}{x_i}=\beta_1+\frac{\epsilon_i}{x_i}\implies \tilde y_i=\frac{y_i-\beta_0}{x_i},\tilde\epsilon_i=\frac{\epsilon_i}{x_i}. $$

This transformation satisfies $\operatorname{Var}(\tilde \epsilon_i)=\operatorname{Var}(\epsilon_i)/x_i^2=\sigma^2$, which will lead to an ordinary least square (applied to $\tilde y_i=\beta_1+\tilde\epsilon_i$).

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