Absolutely!! Questions??


Posted by Richard Reid on July 02, 1997 at 18:24:50: In Reply to: We desperately need a "Kelly" FAQ posted by Steve Heston on July 02, 1997 at 11:38:32:

Steve wrote:
> "Kelly" proportional betting maximizes the expected logarithm of bankroll.
> This is always less than the logarithm of expected bankroll.

Steve:

You're absolutely right when you say that we need a Kelly FAQ. I'm more than willing to save any Kelly stuff that is posted here on the math site and I'm studying hard to get a handle on this stuff so I understand and know what is correct. Let me know if and where I'm wrong in the following:

In a previous post, you stated:
> You want to maximize the _expected_ logarithm. To simulate a 1% advantage try plotting y=.505*ln(1+x)+.495*ln(1-x)

I took this formula for the expected logarithm and generalized as follows:
The expected logarithm may be shown by the following basic formula y= N * (p*ln(1+f)+.q*ln(1-f) ),
where p is the probability of a win,
q is the probability of a loss,
f is the fraction of bankroll that is bet, and
N is the number of bets

Furthermore, it seems to me that the formula for the expected logarithm can be derived from the the basic formula for "exponential growth", Bf = ((1+f)^W)*((1-f)^L) * Bi

For instance, the exponential growth rate (G) is equal to the final bankroll (Bf) divided by the initial bankroll (Bi). In other words,

G = Bf/Bi = ((1+f)^W)*((1-f)^L),

Now, W is the number of wins and can be rewritten as the probability of winning (p) multiplied by the total number of bets (N), and L is the number of losses and can likewise be rewritten as the probability of losing (q) multiplied by the number of bets (N). In other words,

W = p*N, and
L = q*N

Therefore, G can can be rewritten as,

G = ((1+f)^(p*N))*((1-f)^(q*N)).

So, what am I leading up to? Well, correct me if I'm wrong, but it looks to me that the logarithmic growth rate (y) is nothing other than taking the natural log (or ln ) of the exponential growth rate. Therefore, taking the ln of G, we get,

Ln (G) = ln(Bf/Bi) = ln (((1+f)^(p*N))*((1-f)^(q*N)))

We can simplify ln (((1+f)^(p*N))*((1-f)^(q*N))) to become N * ( p*ln(1+f) + q*ln(1-f) )

Therefore,

[A] y = ln (G) = N * ( p*ln(1+f) + q*ln(1-f) )

This is essentially the formula that you gave me to check on the Function Plotter, with N=1, p=0.505 and q=0.495

Here is the point of confusion for me. I thought the above process determined the logarithm of expected bankroll, but we end up with the formula for the "expected logarithm of bankroll".

"Is y the "expected logarithm of bankroll"? Or is y the "logarithm of expected bankroll"? And is [A] the formula for "logarithmic growth"?

Sincerely,
Richard Reid



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