The noise has the product <Laplace distribution> density
$$
g(y)=\prod_{j=1}^ke^{-2|y_j|}=e^{-2\|y\|_1},
$$
which is normalized because $\int_{\mathbb R}e^{-2|t|}\,dt=1$. By translation invariance of <Lebesgue measure>, the conditional law of $A(u)+N$ has density
$$
p(y\mid u)=g(y-A(u))
=\exp[-2\|y-A(u)\|_1].
$$
Hence an associated <likelihood function> is
$$
\boxed{L(u;y)=\exp[-2\|y-A(u)\|_1]}.
$$
The map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is <continuous function>[continuous]. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 3 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-326/3/2/a/solution