Im trying to proof the non asymptotic part of Condorcet's jury theorem:
"In a homogeneous committee composed of n=2k+1 independent experts with the same expertise 0.5 < , any addition of an even number of specialists with expertise increases the probability of the majority rule being right. "
I tried it by induction on the number of added experts but I get too complicated equations of binomial probability.
http://en.wikipedia.org/wiki/Condorcet%27s_jury_theorem
Here is a proof in wikipedia but I am don't understand it correctness
Thanks
There is nothing new here, I just follow the proof from wiki. Let $X_n \sim \text{Binomial}(n, p)$. Note that $X_{2k+3}$ has the same distribution as $X_{2k+1} + X_2$ if $X_{2k+1}, X_2$ are independent. We want to prove $$ \Pr\{X_{2k+1} \geq k+1\} < \Pr\{X_{2k+3} \geq k+2\} $$ when $\displaystyle p > \frac {1} {2}$, for all $k = 0, 1, 2, \ldots$
Now we list the cases (events) in which only the first two cases the additional $X_2$ will change the majority of $X_{2k+1}$:
So by law of total probability,
$$ \begin{align} &~ \Pr\{X_{2k+3} \geq k + 2\} \\ =&~ \Pr\{X_{2k+1} + X_2 \geq k + 2\} \\ = &~ \sum_{i=1}^6 \Pr\{X_{2k+1} + X_2 \geq k + 2|M_i\}\Pr\{M_i\} \\ = &~ 1 \times \Pr\{X_{2k+1} = k\}\Pr\{X_2 = 2\} \\ &~ + 0 \times \Pr\{X_{2k+1} = k+1\}\Pr\{X_2 = 0\} \\ &~ + 0 \times \Pr\{X_{2k+1} = k\}\Pr\{X_2 \leq 1\} \\ &~ + 1 \times \Pr\{X_{2k+1} = k+1\}\Pr\{X_2 \geq 1\} \\ &~ + 0 \times \Pr\{X_{2k+1} \leq k-1\} \\ &~ + 1 \times \Pr\{X_{2k+1} \geq k+2\} \\ = &~ \binom{2k+1} {k} p^k (1-p)^{k+1} p^2 + \Pr\{X_{2k+1} = k+1\}(1 - (1-p)^2) + \Pr\{X_{2k+1} \geq k+2\} \\ = &~\Pr\{X_{2k+1} \geq k+1\} + \binom{2k+1} {k} p^{k+2} (1-p)^{k+1} - \binom{2k+1} {k} p^{k+1} (1-p)^{k} (1-p)^2 \\ = &~\Pr\{X_{2k+1} \geq k+1\} + \binom{2k+1} {k} p^{k+1} (1-p)^{k+1} (p - (1-p)) \\ > &~\Pr\{X_{2k+1} \geq k+1\} \end{align}$$
So that is just what we want.