1
$\begingroup$

In Faraways Linear Models (2ed.) page 77, he mention: "when non-constant variance is seen in the plot of $\hat{\epsilon}$ against $\hat{y}$", a transformation of the response $y$ to $h(y)$ where $h()$(a comment on what h() means here is appreciated) can be chosen so that $\operatorname{var}(h(y))$ is constant."

Then he gives an example on how to consider $h$

$$h(y) = h(Ey) + (y-Ey)h'(Ey)+ \cdots \\ \operatorname{var}(h(y)) = 0+h'(Ey)2\operatorname{var}(y)+ \cdots$$ We ignore higher order terms. For $var(h(y))$to be constant we need

$$h'(Ey) \propto (\operatorname{var}(y))^{-\frac{1}{2}}$$ which suggests: $$h(y) = \int\frac{dy}{\operatorname{var}(y)} == \int\frac{dy}{\text{SD}y}$$ (a comment on what the '$==$' means here is appreciated).

By an example with R, how exactly would this transformation plan out on data with non-constant variance?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The very next sentence in Faraway gives you two examples:

For example if $\operatorname{var} y=\operatorname{var}\varepsilon\propto(Ey)^2,$ then $h(y)=\log y$ is suggested while if $\operatorname{var}\varepsilon\propto(Ey),$ then $h(y)=\sqrt{y}.$

Furthermore, on page 219 you can see an example of the logarithm transformation done in R:

lmodu <- lm(log(longevity) ~ activity, fruitfly)

I imagine the $==$ just means "is equivalent to", similar to $\equiv$ in logic.

Improve this answer
$\endgroup$
5
  • $\begingroup$ How do we test for $var(y) = var(\epsilon)$, he uses the example of var.test before this paragraph, but he's testing the variances between residuals. Do we suppose something like var.test(data$response, residuals(lmod)) where lmod is a fitted linear model? $\endgroup$
    – Working dollar
    Jun 9, 2022 at 20:45
  • 1
    $\begingroup$ Faraway prefers, and I think I agree with him, graphical methods to check for constant variance: plot(fitted(lmod), residuals(lmod)) or plot(<predictor>,residuals(lmod)). You're looking for patterns. Does the spread tend to stay the same from left to right? If you see no discernable patterns, that's a good sign for that particular model. $\endgroup$
    – Adrian Keister
    Jun 9, 2022 at 20:50
  • $\begingroup$ oh I see, I agree also however I am working on a function that performs the calculations for these test to give some output for decisions prior to plotting. I might have got slightly confused with the notation and misguided myself with the dependent variable and the predictor. Thanks for showing me the example! $\endgroup$
    – Working dollar
    Jun 9, 2022 at 20:54
  • $\begingroup$ I have an additional question, what does it mean to plot $\hat{\epsilon}$ against $x_i$, I understand that $\hat{\epsilon}$ is the residuals, and $x_i$ are the independent variables? Therefore, plotting the residuals against the variables used in the model, something like plot(data$predictor, residuals(lmod))? But because it's $x_i$ we plot against all the predictors used in the model? <- What is the importance behind this, or perhaps I should open a new post? $\endgroup$
    – Working dollar
    Jun 9, 2022 at 21:46
  • $\begingroup$ You've got it exactly: plot(data$predictor, residuals(lmod)), and you're looking for the same thing as when you do plot(fitted(lmod), residuals(lmod)): patterns. If you see random behavior, that's generally good. $\endgroup$
    – Adrian Keister
    Jun 10, 2022 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.