How to Show that AdaBoost Weighted Error is Exactly 1/2 I am trying to prove that in an AdaBoost model $Y \rightarrow [-1,1]$
$err_t'= \frac{\sum_{i=1}^{N}w'_i1\{h_t(x^{(i)})\neq t^{(i)}\}}{\sum_{i=1}^{N}w'_i} = \frac{1}{2}$
here, $w_i' = w_i exp(-\alpha t^{(i)}h_t(x^{(i)})$ is the reweighted weight at step $t+1$, but $h_t$ is the classifier at step $t$. So in other words, trying to show that total error weighted with regards to updated (next step) weights is exactly 1/2. Also, this is assuming AdaBoost is using the usual procedure of setting $\alpha_t= \frac{1}{2}log(\frac{1-err_t}{err_t})$.
Here is what I have so far
break into two scenarios, error and not error
$err' = \frac{\sum_{i:h_t(x^{(i)})\neq t^{(i)}}w_i'*1}{\sum_{i=1}^{N}w'_i} + \frac{\sum_{i:h_t(x^{(i)}) = t^{(i)}}w_i'*0}{\sum_{i=1}^{N}w'_i} =$
second part is multiplied by zero so it goes away
$= \frac{\sum_{i:h_t(x^{(i)})\neq t^{(i)}}w_i'*1}{\sum_{i=1}^{N}w'_i} =$
plug in what we know about the new weight
$= \frac{\sum_{i:h_t(x^{(i)})\neq t^{(i)}}w_i exp(-\alpha_t(-1))}{\sum_{i=1}^{N}w'_i} =$
plug in what we know about $\alpha$
$= \frac{\sum_{i:h_t(x^{(i)})\neq t^{(i)}}w_i exp(\frac{1}{2}log(\frac{1-err_t}{err_t}))}{\sum_{i=1}^{N}w'_i} =$
log and exp() cancel out, and take error out of the sum as it is a constant w.r.t. $i$
$= \frac{\sum_{i:h_t(x^{(i)})\neq t^{(i)}}w_i }{\sum_{i=1}^{N}w'_i}*(\frac{1-err_t}{err_t})^\frac{1}{2} = ??$
I am stuck here.. 
 A: I figured it out. I am sharing this because I was looking for a solution online for many hours and mostly I found people discussing this fact or showing the derivation of $\alpha$, but I could not find a detailed proof. Here we go.
To prove:
$err_t'= \frac{\sum_{i=1}^{N}w'_i1\{h_t(x^{(i)})\neq t^{(i)}\}}{\sum_{i=1}^{N}w'_i} = \frac{1}{2}$
Given:
$w_i' = w_i exp(-\alpha t^{(i)}h_t(x^{(i)})$ 
$\alpha_t= \frac{1}{2}log(\frac{1-err_t}{err_t})$.
Proof:
Break into two scenarios, error and not error. To make things easier to write down define:
$E: h_t(x^{(i)})\neq t^{(i)}; E^C = C: h_t(x^{(i)}) = t^{(i)}; $
$err' = \frac{\sum_{i \in  E}w_i'*1}{\sum_{i=1}^{N}w'_i} + \frac{\sum_{i \in C}w_i'*0}{\sum_{i=1}^{N}w'_i} =$
Second part is multiplied by zero so it goes away:
$= \frac{\sum_{i \in  E}w_i'*1}{\sum_{i=1}^{N}w'_i} =$
Plug in what we know about the new weight:
$= \frac{\sum_{i \in  E}w_i exp(-\alpha_t(-1))}{\sum_{i=1}^{N}w'_i} =$
Break denominator sum into E and C, move $exp(\alpha)$ to denominator, substitute $w'_i$, simplify:
$= \frac{\sum_{i \in  E}w_i}{\sum_{i = 1}^N \frac{w'_i}{exp(\alpha)}} = \frac{\sum_{i \in  E}w_i}{\sum_{i \in E} w_i\frac{exp(\alpha)}{exp(\alpha)}+\sum_{i \in C} w_i\frac{exp(-\alpha)}{exp(\alpha)}} = \frac{\sum_{i \in  E}w_i}{\sum_{i \in E} w_i +exp(-2\alpha)\sum_{i \in C} w_i}$ 
Now we take a break from this logical line and we point out two statements:
1) $\alpha = exp(\frac{1}{2}log(\frac{1-err_t}{err_t}) \Rightarrow e^{2\alpha} = \frac{1-err_t}{err_t}$
2) $\frac{\sum_{i \in E}w_i}{\sum_{i \in E}w_i+\sum_{i \in C}w_i} = err_t \Rightarrow \sum_{i \in C}w_i = \frac{\sum_{i \in E}w_i}{err_t} - \sum_{i \in E}w_i $
Going back to the logical path and plugging in the two statements above:
$\frac{\sum_{i \in E}w_i}{\sum_{i \in E}w_i + e^{-2\alpha}(\frac{\sum_{i \in E}w_i}{err_t} - \sum_{i \in E}w_i)} = \frac{\sum_{i \in E}w_i}{\sum_{i \in E}w_i + \frac{err_t}{1-err_t}(\frac{\sum_{i \in E}w_i}{err_t} - \sum_{i \in E}w_i)} = $
$ = \frac{\sum_{i \in E}w_i}{\sum_{i \in E}w_i + \frac{\sum_{i \in E}w_i}{1 - err_t} -  \frac{err_t}{1-err_t} \sum_{i \in E}w_i} = $
Notice that at this point $\sum_{i \in E}w_i$ simply cancels out:
$ = \frac{1}{1 + \frac{1}{1 - err_t} -  \frac{err_t}{1-err_t} } = \frac{1}{1 + \frac{1-err_t}{1 - err_t}  } = \frac{1}{1 + 1  } = \frac{1}{2}$
