The graph is a <profile likelihood> for the parameter $\lambda$ of a <Box–Cox transformation>. Its maximum is close to $\lambda=0$, and the displayed likelihood interval contains zero but excludes $\lambda=1$. Since $\lambda=0$ gives a <logarithmic transformation> whereas $\lambda=1$ leaves the response unchanged up to an affine transformation, the plot supports fitting
$$
\log(\text{Paymentperpolicyyr})
$$
as the response in `model2`.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 13J / ii / Solution

A suitable <generalized linear model> is a <Gamma regression with logarithmic link>:
$$
Y_i\mid x_i\text{ has a gamma distribution},
\qquad \mathbb E(Y_i\mid x_i)=\mu_i,
\qquad
\operatorname{Var}(Y_i\mid x_i)=\phi\mu_i^2,
$$
with
$$
\log\mu_i
=\beta_0+\beta_K\operatorname{numeric}(\text{Kilometres}_i)
+\beta_{\text{Brand}_i}+\beta_{\text{Bonus}_i}.
$$
The <gamma distribution> is appropriate for a positive, continuous, right-skewed payment, and its <variance function> allows the conditional variance to grow like the square of the conditional mean instead of assuming constant variance. The <logarithmic link function> also guarantees positive fitted means and represents covariate effects multiplicatively, which accords with the <Box–Cox transformation> evidence for a logarithmic scale.

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 13J / iv / Solution

{c}
{wiki=Power_transform\#Box–Cox_transformation}

For $y>0$, the Box–Cox transformation is
$$
T_\lambda(y)=
\begin{cases}
(y^\lambda-1)/\lambda,&\lambda\ne0,\\
\log y,&\lambda=0.
\end{cases}
$$
The value of $\lambda$ can be selected by maximizing its <profile likelihood>; $\lambda=1$ corresponds to an affine transformation of the original response, while $\lambda=0$ gives a <logarithmic transformation>.

Codex @codex  0

 Incoming links: Box–Cox transformation