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
\[\sup_\theta R(\theta, \phi_c) = \max\{\sup_{\theta\leq 100}P(X > c), \sup_{\theta > 100}P(X \leq c)\}\]

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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.

The Bayesian Admissiblity

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.

現在市面上的統計計算工具多的令人眼花繚亂,有令何參不知如何下著之感。
之前從internet上發現了R這個工具後,因為其免費、開放(有問題的話,可以看程式碼,有時對理解一些統計計算的細節有很大幫助)的特性,便一直用下去。
現在則在學SAS 。 SAS是大公司的產品,成熟穏定,當然,價格也不低,以本校為例,校園授權的SAS一年要$200美金,對學生來說,不是一筆小數目,還好還有Lab的PC可以用,先不必煩惱這一類的問題。
當然還有其他很多類似的工具, SPSS, Stata, Minitab, …… 族繁不及備載。
這些工具的共同特點就是它們都是在PC上進行交談式的操作與計算。 這樣一來,很有可能,介面就變得比功能還重要了,如果有強大的計算能力而使用者卻不知如何操作,也是罔然。
如果可以透過手機或其他輕薄短小的通訊工具進行計算呢?
這些工具可能就必需調整成主從式架構以達到這種要求,而且要穩定。
那麼,以後的通訊服務商是否也可能成為計算服務商呢?
再來看看吧。