Minimum error rate classification rule for deciding between two values of a parameter I have a qualifying exam later this week based on material similar to what is covered in Casella and Berger, and am studying past exams. It appears that past exams used other texts besides Casella and Berger (which I'm familiar with), and for the most part, I've been able to figure out the concepts and solve the problems in these situations. However, I've been stuck on this particular problem.

Below is a table giving $\theta = 0$ and $\theta = 1$ pmfs $f(x \mid
 \theta)$ for a discrete random variable $X$.
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  & x = 1 & x = 2 & x = 3 & x
 = 4 & x = 5 & x = 6  & x = 7 \\ \hline \theta = 1 & .10 & .20 & .15 & .10 & .05 & .30 & .10 \\ \theta = 0 & .20 & .10 & .20 & .05 & .10 &
.15 & .20 \\ \hline \end{array}$$
Suppose that a priori there is probability .6 that $\theta = 1$.
  Identify a minimum error rate (minimum 0-1 loss Bayes risk)
  classification rule for deciding between $\theta = 0$ and $\theta = 1$
  based on $X$. (Give values of a decision rule $d(x)$ for $x = 1,
 \dots, 7$.)

Here's what I know:


*

*We've assigned a prior distribution to $\theta$, where $\pi(\theta) = \begin{cases}
.6, & \theta = 1 \\
.4, & \theta = 0\text{.}
\end{cases}$

*Zero-one loss is a loss function which is $1$ if you classify a point incorrectly, $0$ if you don't.

*To find an estimator which minimizes the Bayes risk - say $T = h(X)$, such an estimator is such that at each fixed data point $x$ of the random variable $X$ that $\mathbb{E}_{\theta \mid X}[L(h(x), \theta)]$, with $L$ being the loss function, is minimized. 

*The likelihood ratio $\Lambda(X) = \dfrac{f(X \mid 1)}{f(X \mid 0)}$ is a minimum sufficient statistic for $\theta$.


I do have the solution to this problem: it starts off by saying 
$$d(x) = \begin{cases}
1 & .6\cdot f(x\mid 1) > .4\cdot f(x \mid 0) \\
0 & .6\cdot f(x\mid 1) < .4\cdot f(x \mid 0) 
\end{cases}$$
but I have no idea how the condition $.6\cdot f(x\mid 1) > .4\cdot f(x \mid 0)$ is derived. This appears to just be saying that $f_{\theta \mid X}(1 \mid x) > f_{\theta \mid X}(0 \mid x)$ where $f_{\theta \mid X}$ is the posterior pmf of $\theta$, but I don't see how this makes sense.
 A: Any decision rule for the likelihood matrix
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  & x = 1 & x = 2 & x = 3 & x
 = 4 & x = 5 & x = 6  & x = 7 \\ \hline H_1 & .10 & .20 & .15 & .10 & .05 & .30 & .10 \\ H_0 & .20 & .10 & .20 & .05 & .10 &
.15 & .20 \\ \hline \end{array}$$
can be defined by marking (e.g. by making it boldface) one entry in each column of the likelihood matrix, say like
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  & x = 1 & x = 2 & x = 3 & x
 = 4 & x = 5 & x = 6  & x = 7 \\ \hline H_1 & \mathbf{.10} & \mathbf{.20} & .15 & \mathbf{.10} & .05 & .30 & .10 \\ H_0 & .20 & .10 & \mathbf{.20} & .05 & \mathbf{.10} &
\mathbf{.15} & \mathbf{.20} \\ \hline \end{array}$$
meaning that when $x$ equals $1, 2$, or $4$, we decide that $H_1$ is true while if $x$ equals $3,5,6$, or $7$, we decide that $H_0$ is true. The false-alarm probability $P_{FA}$ (a.k.a. probability of Type I error), the probability of choosing $H_1$ when in fact $H_0$ is true is the sum $(0.35)$ of the unbolded entries on the $H_0$ row, while the missed-detection probability $P_{MD}$ (a.k.a  probability of Type II error) is the sum $(0.6)$ of the unshaded entries on the $H_1$ row. (Hey, I never said that it was a good decision rule!).  
The average error probability of this decision rule is 
$$P_e = P_{FA}P(H_0) + P_{MD}P(H_1) = 0.35\times 0.4 + 0.6\times 0.6
= 0.50$$
and an easy way of visualizing this is to convert the likelihood matrix into the joint probability matrix by multiplying the entries in each row by the probability of the hypothesis, while retaining the boldfaces.
This gives us 
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  & x = 1 & x = 2 & x = 3 & x
 = 4 & x = 5 & x = 6  & x = 7 \\ \hline H_1 & \mathbf{.06} & \mathbf{.12} & .09 & \mathbf{.06} & .03 & .18 & .06 \\ H_0 & .08 & .04 & \mathbf{.08} & .02 & \mathbf{.04} &
\mathbf{.06} & \mathbf{.08} \\ \hline \end{array}$$
and $P_e$ is just the sum of all the unbolded entries in the
joint probability matrix.
So, which decision rule minimizes $P_e$? Well, we have to mark 
one entry in each column of the joint probability matrix and whichever one we don't mark contributes to $P_e$, and the answer is obvious:

mark the larger of the two entries in each column of the joint probability matrix!

The minimum error probability of error rule is thus
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  & x = 1 & x = 2 & x = 3 & x
 = 4 & x = 5 & x = 6  & x = 7 \\ \hline H_1 & .06 & \mathbf{.12} & \mathbf{.09} & \mathbf{.06} & .03 & \mathbf{.18} & .06 \\ H_0 & \mathbf{.08} & .04 & .08 & .02 & \mathbf{.04} &
.06 & \mathbf{.08} \\ \hline \end{array}$$
and it achieves a $P_e$ of $0.35$.

We can gussy up this simple notion by saying that when we observe $x$, we decide in favor of $H_1$ precisely when $P(x,H_1)>P(x,H_0)$, that is, when $P(H_1\mid x)P(x) > P(H_0\mid x)P(x)$, which is the same as 
$P(H_1\mid x) > P(H_0\mid x)$: 

MAP (maximum a posteriori probability) decision rule: Choose the hypothesis with the larger a posteriori probability

and arrive at the usual claim that the MAP decision rule minimizes the error probability, but why it does so is more intuitively obvious to me via the development above; ymmv.
A: As you have pointed out, minimum error rate means that you choose $\theta=1  $ if for a given $x$: $$P(\theta =1 | X) > P(\theta =0 | X) $$
or stated another way, you choose the state of nature $\theta$ that maximise your a posteriori probability function.
The intuition behinds this is that for a given $x$, usually called evidence, you choose the $\theta$ this evidence is more likely to have come from.
after a little algebra with the bayes formula you get the same decision rule but stated now with the a priori distribution and the likelihood
$$P(X | \theta =0)P(\theta =0) > P(X | \theta =0) P(\theta =1)$$
and you have all this probabilities in the table.
