Hedging with Kelly Criterion on a single event Let's suppose I'm betting on the flip of a coin. Each outcome has equal probability $p$. Fractional odds per single unit stake on each outcome are:
$heads=h=6/5$
$tails=t=7/5$
If I was to bet on a single outcome, the optimal portion $x$ of my bankroll should be wagered on $tails$ given that it has the greatest positive long term expectation of the two choices. According to the Kelly Criterion in this case: 
$$x=\frac{pt-(1-p)}{t}$$
However, the following conditions are met for hedging:


*

*All outcomes belong to a single event.

*All outcomes are disjoint (none can't occur simultaneously along with any other).

*All have positive long term expectation.

*The probabilities of each respective outcome add up to 1 when taken all together.


Consequently, I can wager my entire bankroll on both outcomes (the entire event) to profit from either one of these ocurring. What should $x_t$ and $x_h$ be according to the Kelly Criterion?
In order to avoid daunting mathematics, I simplified the problem utilizing Kelly's derivation in the following manner with a single unknown variable $x$:
$$f(x):=p\ln(1+hx-(1-x))+(1-p)\ln(1+t(1-x)-x)$$
$$=\ln((hx+x)^p(1+t-tx-x)^{(1-p)})$$
Obtain the derivative for $f(x)$, then equal to $0$:
$$\frac{df(x)}{dx}=0$$
Finally, solve for $x$:
$$x=p$$
Implying $x$ should be apportioned for each outcome in strict correspondence with its probability (i.e. if outcome $h$ has probability $p$ with positive expectation, apportion $p$ of your bankroll on that outcome; and so on, succesively, with the rest of the outcomes).
Is there a rigorous demonstration of the same principle being always optimal on more than 2 outcomes that meet the 4 stated conditions above?
 A: Yes, at least if you accept the principle of logarithmic utility. What you do is that you express the expected utility as a function of your bets:
$$u(x) = \sum_j p_j \ln \left(1+(1+b_j)x_j - \sum_k x_k\right)$$
and then you differentiate it:
$$\partial_l u(x) 
= \sum_j {p_j ( (1+b_j)\delta_{jl} - 1) \over \left(1+(1+b_j)x_j - \sum_k x_k\right)} 
\\= {p_l (1+b_l) \over \left(1+(1+b_l)x_l - \sum_k x_k\right)} 
- \sum_j {p_j \over \left(1+(1+b_j)x_j - \sum_k x_k\right)} $$
Now of course you realize that the gradient is never zero inside the region $\sum x_j\le 1$, as you always can expect more utility if you bet more in some way - the only thing that limits that is your bank roll. 
So we're looking for a optimum on the edge $\sum x_j = 1$. To find that we have to project the gradient on the tangent plane and it has to vanish. This means that the gradient of $u$ must be parallell to $(1,1,\cdots,1)$ that is all partial derivates has to be the same.
Since only the first term formally depends on $l$ we reach that:  
$${p_l (1+b_l) \over \left(1+(1+b_l)x_l - \sum_k x_k\right)} = {p_l\over x_l}$$
Should not depend on $l$ and the only way that can happen is if $x_l\propto p_l$ and as $\sum x_j = \sum p_j = 1$ that would mean $x_l = p_l$. 
