Proof that log-odds minimize binomial deviance How do you prove that minimizing the binomial deviance estimates the log-odds? i.e:
$$
\ln{\left ( \frac{p(x_i)}{1-p(x_i)} \right )} = \underset{f(x_i)}{argmin} \ \mathbb{E} \left [y_i \ln \left ( \frac{e^{f(x_i)}}{1+e^{f(x_i)}} \right ) + (1-y_i) \ln \left ( 1-\frac{e^{f(x_i)}}{1+e^{f(x_i)}} \right ) \right ]
$$
where
$$
y_i \in \{ 0,1 \} \\
p(x_i)=\frac{e^{f(x_i)}}{1+e^{f(x_i)}}
$$

Research:
According to Hasti, Tibshirani, Friedman; Elements of Statistical Learning 2nd Ed,  the minimizer should be $1/2$ of the log odds, just like for exponential loss, however no proof is provided.
A proof for the minimizer of the exponential loss (not the binomial deviance) is derived here by Weatherwax and Epstein, which is straight-forward and is based on the $\{-1,1\}$ notation. However the same principles don't seem to apply (?) to the binomial deviance with the $\{0,1\}$ notation, since one term becomes 0 in the expectancy.
A statement is made at this wikipedia page and some proof instruction is somewhat given, however I find the notation confusing. The coding there is $\{-1,1\}$ too (I think).
Any proof/ intuition/ idea of how to go about this would be much appreciated.
 A: So it seems the solution is much simpler than I though. Hasti, Bishirani and Friedman in the source mentioned above mention the gradient of the (multinomial) deviance, which is:
$$
g_{ik} = -I(y_i=G_k)+p_k(x_i)
$$
This holds for binomial deviance as well, and is fairly easy to show that:
$$
\frac{\partial L(y_i,f(x_i))}{\partial f(x_i)} = \frac{\partial}{\partial f(x_i)} \left ( y_i \ln \left ( \frac{e^{f(x_i)}}{1+e^{f(x_i)}} \right ) + (1-y_i) \ln \left ( 1-\frac{e^{f(x_i)}}{1+e^{f(x_i)}} \right ) \right ) =-y_i+p(x_i)
$$
By Leibniz' integration rule it holds that $\frac{\partial}{\partial f(x_i)} \mathbb{E}[\cdot]=\mathbb{E}[\frac{\partial}{\partial f(x_i)}( \cdot)]$, thus:
$$
\frac{\partial \mathbb{E}[L(y_i,f(x_i))]}{\partial f(x_i)} = 
\mathbb{E} \left [ \frac{\partial L(y_i,f(x_i))}{\partial f(x_i)} \right ] \overset{!}{=}0 \\
\Leftrightarrow  \mathbb{E}[-y_i+p(x_i)] \overset{!}{=} 0 \\
\Leftrightarrow  (-y_i+p(x_i)) P(y_i=1)+(-y_i+p(x_i)) P(y_i=0) \overset{!}{=} 0 \\
\Leftrightarrow  (-1+p(x_i)) P(y_i=1)+(-0+p(x_i)) P(y_i=0) \overset{!}{=} 0 \\
\Leftrightarrow  (-1+p(x_i)) P(y_i=1)+(-0+p(x_i)) P(y_i=0) \overset{!}{=} 0 \\
\Leftrightarrow  p(x_i)=\frac{P(y_i=1)}{P(y_i=0)} \\
\Leftrightarrow  \underbrace{\frac{e^{f(x_i)}}{1+e^{f(x_i)}}}_{=(1+e^{f(x_i)})^{-1}} =\frac{P(y_i=1)}{P(y_i=0)} \\
\Leftrightarrow f(x_i)=\ln \left ( \underbrace{\frac{P(y_i=1)}{P(y_i=0)}}_{=log-odds} \right )
$$
