I am trying to understand how to approximate the kelly criterion when I have multiple winning events.
The Taylor approximation says that $\log(1+fr) = fr + \frac{(fr)^2}{2}$
If I have 3 winning events, wherein one I have $1/216$ chances to win $30$ times my bet, then $15/216$ to win $2$ times my bet, and $75/216$ to win my bet back, and $125/216$ to lose my bet.
Using the Taylor approximation, would this become:
$E(\log(1+fr)) = f30 + \frac{(f30)^2}{2} + f2 + \frac{(f2)^2}{2} + f + \frac{(f)^2}{2}$
$E'= 33+905f = 0 $
$f = -33/905 = -0.036$
This cant be the case since f must be positive.
Let:
$ p_1, p_2, ... , p_n $ the probabilities of the events $ e_1,e_2, ..., e_n $
$ r_1, r_2, ..., r_n $ the returns of events $ e_1,e_2, ..., e_n $
$ e_1,e_2, ... e_n $ events are mutually exclusive
$ p_1 + p_2 + ... + p_n = 1 $
$ p_1r_1 + p_2r_2 + ... + p_nr_n > 0 $ (positive expected value)
Then:
$ G(f)= p_1\log(1 + fr_1) + p_2\log(1 + fr_2) + ... + p_n\log(1 + fr_n) $
with $ G(f) $ denoting logarithmic wealth growth.
as it was given that $ \log(1+x)\approx x-\frac{x^2}{2} $ then
$ G(f) \approx p_1r_1f - \frac{1}{2}{p_1r^2_1f^2} + p_2r_2f - \frac{1}{2}{p_2r^2_2f^2} + ... + p_nr_nf - \frac{1}{2}{p_nr^2_nf^2} $
To find the value of $ f $ for which the growth rate is maximized, denoted as $ f^* $, we differentiate the above expression and set this equal to zero.
$ p_1r_1 - p_1r^2_1f + p_2r_2 - p_2r^2_2f + ... + p_nr_n - p_nr^2_nf = 0 $
$ \Rightarrow $
$ p_1r_1 + p_2r_2 + ... +p_nr_n = p_1r^2_1f + p_2r^2_2f + ... + p_nr^2_nf $
$ \Rightarrow $
$ f.(p_1r^2_1 + p_2r^2_2 + ... + p_nr^2_n) = p_1r_1 + p_2r_2 + ... + p_nr_n $
$ \Rightarrow $
$ f^* \approx \frac{p_1r_1 + p_2r_2 + ... + p_nr_n}{p_1r^2_1 + p_2r^2_2 + ... + p_nr^2_n } $
For the specific case of the question raised, it is not possible to calculate Kelly, as the expected value is negative. But let's assume a similar situation where:
$ p_1=\frac{1}{216} $ , $ r_1=30 $,
$ p_2=\frac{15}{216} $ , $ r_2=10 $,
$ p_3=\frac{75}{216} $ , $ r_3=0 $,
$ p_4=\frac{125}{216} $ , $ r_4=-1 $
In that case we have:
$f^* \approx \frac{1/216*30 + 15/216*10 + 75/216*0 + 125/216*(-1) }{1/216*30^2 + 15/216*10^2 + 75/216*0^2 + 125/216*(-1)^2} \approx $ 2.18%