Theory testing - beyond binary case Given $n$ hypothesis $\{H_1, \dots, H_n\}$ which are mutually exclusive and exhaustive:
$$\forall_{i \neq j }P(H_iH_j) = 0\text{ and }\sum_{i=1}^nP(H_i)=1$$
Also we have acquired $m$ data points - $\{D_1,\dots,D_m\}$ 
It is common to assume logical independence of the data points given $H_i$ :
$$P(D_1\dots D_m\mid H_i) = \prod_{j=1}^mP(D_j\mid H_i)$$
Show that if we assume that:
$$P(D_1 \dots D_m\mid\overline{H_i}) = \prod_{j=1}^mP(D_j\mid\overline{H_i})$$
and $n>2$ then at most one of the factors $\frac{P(D_j\mid H_i)}{P(D_j\mid\overline{H_i})}$ is not $1$ for fixed hypothesis $H_i$.
Solution so far:
Using the Bayesian rule we can compute that: 
$$P(D \mid \overline{H_i}) = \frac{\sum_{k \neq i} P(D \mid H_k)P(H_k)}{\sum_{k \neq i} P(H_k)} = \frac{P(D) - P(D \mid H_i)P(H_i)}{1 - P(H_i)}$$
Substituting $D=D_1D_2 \dots D_m$ and using the first equation we get that:
$$P(D_1D_2 \dots D_m \mid \overline{H_i}) = \frac{P(D_1D_2 \dots D_m) - P(D_1D_2 \dots D_m \mid H_i)P(H_i)}{1 - P(H_i)} = \frac{\prod_{j=1}^mP(D_j) - P(H_i)\prod_{j=1}^mP(Dj \mid H_i)}{1 - P(H_i)}$$
Now substituting $d=D_j$ we can get that:
$$P(D_j \mid \overline{H_i}) = \frac{P(D_j) - P(D_j \mid H_i)P(H_i)}{1 - P(H_i)}$$
From this follows that
$$\prod_{j=1}^mP(D_j\mid\overline{H_i}) = \frac{\prod_{j=1}^m(P(D_j) - P(D_j \mid H_i)P(H_i))}{(1 - P(H_i))^m}$$
The ratios from the problem are
$$\frac{P(D_j\mid H_i)}{P(D_j\mid\overline{H_i})} = \frac{P(D_j\mid H_i)(1-P(H_i))}{P(D_j) - P(D_j \mid H_i)P(H_i)} = \frac{P(D_j\mid H_i)-P(D_j\mid H_i)P(H_i)}{P(D_j) - P(D_j \mid H_i)P(H_i)}$$
From which follows that such ration to be one is equivalent to 
$$P(D_j \mid H_i) = P(D_j)$$
Even so I can not seem to see how does the number of hypothesis play a role?
 A: $n=3$ case: 
So let's arrange in a table.
     H1    H2    H3
D1   a     b     c
D2   d     e     f

where for instance $a = P(D_1 | H_1)$.
Note that by Bayes' rule, to compute for instance
$P(D_1 | H_2+H_3)$, it's a weighted average of $P(D_1 | H_2)$ and $P(D_1 | H_3)$.
Then note that with $h_2 = P(H_2)$ and $h_3 = P(H_3)$,
$P(D_1 | H_2+H_3) = \frac{h_2}{h_2+h_3} b + \frac{h_3}{h_2+h_3} c$ and $P(D_2 | H_2+H_3) = \frac{h_2}{h_2+h_3} e + \frac{h_3}{h_2+h_3} f$
Both independence assumptions yield 
$(h_2+h_3)(h_2 be + h_3 cf) = (h_2 b + h_3 c)(h_2 e + h_3 f)$.
Simplifying yields that $cf+be = bf + ce$.
Note the assumption that $h_2$ and $h_3$ are strictly positive is needed here.
Now, studying this one equation yields that letting $t = e/(e+f)$,
$tb + (1-t)c = tc + (1-t)b$.
One can conclude that either $b=c$ or $e=f$ from this. This is clearly not complete, but wanted to mention that the motivation was to figure out what exactly the second independence assumption adds to the requirements.
Edit: (From chat)
Putting together everything:


*

*$b=c$ or $e=f$,  

*$c=a$ or $d=f$,

*$a=b$ or $d=e$.


All three of these are true, and the conclusion in this case is that either $a=b=c$ or $d=e=f$ (in which case, the ratio is $1$ for whichever row, as one is the average of the other two). Suppose $a=b$, so $d\neq e$ by the third condition. In the first condition, suppose that $b\neq c$ (otherwise $a=b=c$ already), so that $e=f$. Now in the second condition, $a \neq c$, so it must be that $d=f$, and $d=e=f$.
The full problem needs some more work, with more hypotheses.

Alright, here is an outline of the full situation:
Let $a_i = P(D_1 \mid H_i)$ and $b_i = P(D_2 \mid H_i)$
Also let $h_i = P(H_i)$.
Generalizing the 3 hypothesis case, we end up with equations of the form
$0 = \sum_{i<j,\ i\neq k,\ j\neq k} h_i h_j ( a_i b_j + a_j b_i - a_i b_i - a_j b_j)$
or
$0 = \sum_{i\neq k,\  j \neq k} h_i h_j ( a_i b_j - a_i b_i )$  for fixed $k$
Now, to generalize the goal of the 3 hypothesis case, we note the goal of showing that
$0 = (P(D_1 \mid H_1) - P(D_1 \mid \overline{H_1}))(P(D_2 \mid H_1) - P(D_2 \overline{H_1}))$
which can be expanded to showing that
$0 = \left( \sum_{j\neq 1} h_j (a_1 - a_j)\right)\left(\sum_{j \neq 1} h_j (b_1 - b_j)\right)
 = \sum_{i \neq 1,j \neq 1} h_i h_j (a_1 - a_i)(b_1 - b_j)$
Now if you expand, you study the summand $a_1b_1 - a_1 b_j - a_i b_1 - a_i b_j$, you will start to see very similar pieces to the other equations. Now you should have enough to finish. What will happen is that if you study the equations for $H_2,\ldots,H_n$, you will see $a_1 b_1 - a_j b_1 - a_1 b_j \sim -a_j b_j$ (summing). The equation for $H_1$ finishes the job.
