Continuing with Exercise $6.3 .14$, besides estimation there is also a nice geometric interpretation for testing. For the model $(6.3 .26)$, consider the hypotheses
$$
H_{0}: \theta=\theta_{0} \text { versus } H_{1}: \theta \neq \theta_{0}
$$
where $\theta_{0}$ is specified. Given a norm $\|\cdot\|$ on $R^{n}$, denote by $d(\mathbf{X}, V)$ the distance between $\mathbf{X}$ and the subspace $V$; i.e., $d(\mathbf{X}, V)=\|\mathbf{X}-\widehat{\boldsymbol{\mu}}\|$, where $\hat{\boldsymbol{\mu}}$ is defined in equation $(6.3 .27) .$ If $H_{0}$ is true, then $\hat{\mu}$ should be close to $\mu=\theta_{0} 1$ and, hence, $\left\|\mathbf{X}-\theta_{0} \mathbf{1}\right\|$ should be close to $d(\mathbf{X}, V) .$ Denote the difference by
$$
R D=\left\|\mathbf{X}-\theta_{0} \mathbf{1}\right\|-\|\mathbf{X}-\widehat{\boldsymbol{\mu}}\|
$$
Small values of $R D$ indicate that the null hypothesis is true, while large values indicate $H_{1}$. So our rejection rule when using $R D$ is
Reject $H_{0}$ in favor of $H_{1}$ if $R D>c$.
(a) If the error pdf is the Laplace, $(6.1 .6)$, show that expression $(6.3 .31)$ is equivalent to the likelihood ratio test when the norm is given by $(6.3 .28)$.
(b) If the error pdf is the $N(0,1)$, show that expression (6.3.31) is equivalent to the likelihood ratio test when the norm is given by the square of the $l_{2}$ norm, $(6.3 .29) .$