Calculating Bayes decision boundary on a practical example In "The elements of Statistical Learning", Chapter 2, the following example is presented: first generate $10$ means $m_k$ from a bivariate Gaussian distribution $\mathcal{N}((1, 0)^t, I)$ and label this class BLUE. Similarly $10$ more is drawn from $\mathcal{N}((0, 1)^t, I)$ and are labelled ORANGE. Then for each class we generate $100$ observations as follows: for each observation pick $m_k$ at random with probability $1/10$ and then generate a $\mathcal{N}(m_k, I/5)$.
At the end of that chapter there's a question (Exercise 2.2):

Show how to compute the Bayes decision boundary for the simulation example.

In an unofficial solution manual for the book the following answer is present:

For this problem one is supposed to regard the points $p_i$ and $q_i$ below
  as fixed.  If one does not do this, and instead averages over possible
  choices, then since the controlling points are $(1,0)$ and $(0,1)$, and all
  probabilities are otherwise symmetric between these points when one
  integrates out all the variation, the answer must be that the boundary is the
  perpendicular bisector of the interval joining these two points.  The
  simulation draws 10 points $p_1, \dots , p_{10} \in \mathbb{R}^2$ from
  $\mathcal{N}((1, 0)^t, I)$, and $10$ points $q_1, \dots, q_{10} \in \mathbb{R}^2$ from $\mathcal{N}((0, 1)^t, I)$. The formula for the Bayes
  decision boundary is given by equating likelihoods. We get an equation in the
  unknown $z \in \mathbb{R}^2$, giving a curve in the plane:
$$\sum_i \exp \left( -5 ||p_i - z||^2 / 2\right) = \sum_j \exp \left( -5 ||q_j - z||^2 / 2\right).$$
In this solution, the boundary is given as the equation of equality between
  the two probabilities, with the $p_i$ and $q_j$ constant and fixed by
  previously performed sampling. Each time one re-samples the $p_i$ and $q_j$,
  one obtains a different Bayes decision boundary.

I have a few questions about the proposed solution:
1) Why likelihoods are just sums.
2) Why can we keep $p_i$ and $q_j$ fixed (the reasoning isn't clear)? Why the answer must be that the boundary is perpendicular bisector?
3) Shouldn't summation indices range over a set $[1, \dots, 100]$?
 A: 1) I am guessing that the likelihoods are sums because the gaussian means are extracted with equal probability. Actually all members of the sum should have a 1/10 weight but because all them have the same weight it was simply ignored. 
2) The data were generated with the given pi and qj - that is why they are fixed
3) They are only 10 gaussians for each of the two classes - you do not need to sum to 100.
A: 
1) "Why likelihoods are just sums."?

The answer in the unofficial solutions manual has lots of typos in it (and I am not at all sure that it is an answer to the question asked in the  OP's first paragraph).  In particular, I believe that the likelihoods are stated incorrectly with the summation signs are in the wrong place. Most likely, they should be of the form 
$$C\exp\left[-\frac{1}{2\sigma^2}\sum_{k=1}^{10}\lVert p_k -z\rVert^2\right]  = C\exp\left[-\frac{5}{2}\sum_{k=1}^{10}\lVert p_k -z\rVert^2\right]$$
and 
$$C\exp\left[-\frac{1}{2\sigma^2}\sum_{k=1}^{10}\lVert q_k -z\rVert^2\right]  = C\exp\left[-\frac{5}{2}\sum_{k=1}^{10}\lVert q_k -z\rVert^2\right].$$
Note that since the variance is $\frac 15$, the $p_k$'s are clustered around $(1,0)$ while the $q_k$'s are clustered around $(0,1)$.
The maximum likelihood decision regions (which is the same as the Bayesian decision regions for prior probabilities $\pi_0 = \pi_1 = \frac 12$) then are the sets of points in $\mathbb R^2$ for which
\begin{align}
C\exp\left[-\frac{5}{2}\sum_{k=1}^{10}\lVert p_k -z\rVert^2\right]
&\gtrless C\exp\left[-\frac{5}{2}\sum_{k=1}^{10}\lVert q_k -z\rVert^2\right]\\
&\Downarrow\\
\sum_{k=1}^{10}\lVert p_k -z\rVert^2 &\lessgtr \sum_{k=1}^{10}\lVert q_k -z\rVert^2\\
&\Downarrow\\
\sum_{k=1}^{10}\lVert p_k \rVert^2 -2 \langle p_k, z \rangle + \lVert z\rVert^2 &\lessgtr \sum_{k=1}^{10}\lVert q_k \rVert^2 -2 \langle q_k, z \rangle + \lVert z\rVert^2\\
&\Downarrow\\
D &\lessgtr \sum_{k=1}^{10} \langle p_k-q_k, z \rangle 
\end{align}
where $D = \frac 12 \sum_{k=1}^{10}\lVert p_k \rVert^2 - \lVert q_k \rVert^2$
So, the decision region boundary is the set of points where a linear function of $z\in \mathbb R^2$ equals a constant, that is, it is a straight line, and a little thought will show why this straight line is perpendicular to the line through $\bar{p} = \frac{1}{10}\sum_k p_k$ and
$\bar{q} = \frac{1}{10}\sum_k q_k$.
