The marginal predictor density is
$$
x_i\sim N(\mu,A),
\qquad A=\tau^2+\sigma_x^2.
$$
<Gaussian conditional expectation> gives
$$
\xi_i\mid x_i\sim N\!\left(
\mu+\kappa(x_i-mu),v\right),
\qquad
\kappa=\frac{\tau^2}{A},
\qquad
v=\frac{\tau^2\sigma_x^2}{A}.
$$
Integrating this conditional distribution through the response model yields
$$
y_i\mid x_i\sim N\!\left(
\alpha+\beta[\mu+\kappa(x_i-mu)],
s^2+\beta^2v\right).
$$

When $\tau\gg\sigma_x$, $\kappa\to1$, $v\to\sigma_x^2$, and $A\sim\tau^2$. Hence
$$
L\simeq L_1(\alpha,\beta,\sigma^2)L_2(\mu,\tau^2),
$$
where
$$
L_1=\prod_i\varphi\!\left(y_i;\alpha+\beta x_i,
\sigma^2+\sigma_y^2+\beta^2\sigma_x^2\right),
\qquad
L_2=\prod_i\varphi(x_i;\mu,\tau^2).
$$
The approximate maximum-likelihood estimators are
$$
\widehat\mu=\bar x,
\qquad
\widehat{\tau^2}=\frac1N\sum_i(x_i-\bar x)^2.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 219 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-219/2/d/solution