Let's say you are betting on the outcome of some random variable. You can invest any amount $X$, and each outcome multiples that amount $X$ by a certain amount $ \ge 0$. You must choose $X$ less than how much capital you have. A Kelly Bet is one in which you bet an amount that maximizes the logarithm of your capital. The Kelly Bet will allow your capital to grow larger than any other betting scheme in the long term (i.e., as the number of bets approaches infinity).
What is the proof that previous statement?
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$\begingroup$The statement is false.
For example consider the bet where you double your money with probability $2/3$ and lose it otherwise. Now consider the strategy "bet \$1 more than the Kelly bet on the first turn, and then use the Kelly bet after that". If this wins on the first turn it will forever have more money than the Kelly strategy, and if it loses on the first turn it will forever have less money than the Kelly strategy. So no matter how large the number of bets becomes, it has a $2/3$ chance of having a larger amount than the Kelly strategy.
$\endgroup$ $\begingroup$Say in total I have 1 dollar to start with and I bet $X$ amount of my current net worth each time. Say if I win a bet (with probability $p$), I will get $X b$ back (net odd of $b$). Then after the bet with probability $p$, I got back
$1+Xb$
and with probability $1-b$, I will have
$1-X$
left.
Now, consider betting for $N$ times, then there will be about $Np$ win and $N(1-p)$ loss. So, the final amount is about
$(1+Xb)^{Np} (1-X)^{N(1-p)}=((1+Xb)^{p} (1-X)^{1-p})^N.$
To maximize this gain, we just need to maximize $(1+Xb)^{p} (1-X)^{1-p}$ with respect to $X$. Or we can maximize $f(X)=\ln [(1+Xb)^{p} (1-X)^{1-p}]$ instead. Setting $\frac{d f}{d X}=0$, we have $\frac{pb}{1+Xb}-\frac{1-p}{1-X}=0\Rightarrow X= \frac{bp-(1-p)}{b}.$
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