## Berger (1985) Exercise 5.33 (page 382)

### April 26, 2009

According to Theorem 17, if $\delta_0^*\in\mathscr D$ is Bayes with respect to $\pi_0\in\Theta^*$, and
$R(\theta, \delta_0^*) \leq r(\pi_0, \delta_0^*)$
for all $\theta\in\Theta$, then $\delta_0^*$ is minimax and $\pi_0$ is least favorable.
Now we consider $R(\theta,\phi_c) = K_1P_\theta(X > c)$ if $\theta \leq \theta$; $R(\theta,\phi_c) = K_2P_\theta(X \theta$.
Now, in the case $X\sim\mathscr N(\theta, 100)$, we have the risk function
$R(\theta, a) = \int_{-\infty}^{100}f(x|\theta)dx +3\int_{100}^\infty f(x|\theta)dx$
Now we have a easier approach:
Since
$\sup_\theta R(\theta, \phi_c) = \max\{\sup_{\theta\leq 100}P(X > c), \sup_{\theta > 100}P(X \leq c)\}$

## Berger(1985) exercise 8.29

### April 23, 2009

According to Berger (1985), ex. 8.29 we can not find
$K_1(\theta_1, \theta_2)$ and $K_2(\theta_1, \theta_2)$ such that
$L(\theta_2, a) \leq K_1(\theta_1, \theta_2)L(\theta_1, a) + K_2(\theta_1, \theta_2)$
where $L(\theta, a) = \exp[(\theta – a)^2]$
this problem seems not easy. We may use contradiction to show:
If $K_1$, $K_2$ then $\exp\{(\theta_2 – a)^2\} \leq K_1(\theta_1,\theta_2)\exp\{(\theta_1 – a)^2\} + K_2(\theta_1, \theta_2)$
thus,
$\exp\{\theta_2^2 – \theta_1^2 + 2\theta_1a – 2\theta_2a\} = K_1(\theta_1, \theta_2) + K_2(\theta_1, \theta_2)\exp\{-(\theta_1-a)^2\}$
As $a$ goes infinity, the RHS will goes infinity but LHS will converges to $K_1(\theta_1, \theta_2)$, which bounded. A contradiction.

### April 19, 2009

Consider R-better:
$R(\theta, \delta^*) \leq R(\theta, \delta)$
And since we know that $\delta^\pi$ is inadmissible, then there exists a rule $\delta$ with $R(\theta_i, \delta) \leq R(\theta_i, \delta^\pi)$ for all $i$,
with strict inequality for, say, $\theta_k$. Hence,
$r(\pi,\delta) = \sum_{i=1}^\infty R(\theta_i, \delta)\pi(\theta_i) < \sum_{i=1}^\infty R(\theta_i, \delta^\pi)\pi(\theta_i) = r(\pi,\delta^\pi)$
Theorem: If a Bayes rule is unique, it is admissible.\\
Proof. Let $\delta^\pi$ denote a Bayes rule with respect to $\pi$. If $\delta^\pi$ is inadmissible, then there exists a rule $\delta$ such that
$R(\theta, \delta) \leq R(\theta, \delta^\pi) \forall \theta$
with strict inequality for some $\theta$. Hence,
$r(\pi, \delta) = \int_\Theta R(\theta, \delta)dF^\pi(\theta) \leq \int_\Theta R(\theta, \delta^\pi)dF^\pi(\theta) = r(\pi, \delta^\pi).$

Thus, $\delta$ is Bayes, which contradicts the fact that a Bayes rule is unique.
Therefore, $\delta^\pi$ must be impossible.