3.3 Overbetting and the shape of $g$

Where growth turns negative#

Proof

With $b = 1$, $f^* = p - q =: e$ and $p = (1+e)/2$, $q = (1-e)/2$. For $|f| < 1$,

$$g(f) = p\ln(1+f) + q\ln(1-f) = \sum_{k\ge1}\frac{f^k}{k}\big[(-1)^{k+1}p - q\big] = \sum_{k\text{ odd}}\frac{e\,f^k}{k} - \sum_{k\text{ even}}\frac{f^k}{k},$$

because for odd $k$ the bracket is $p - q = e$ and for even $k$ it is $-p - q = -1$. At $f = 2e < 1$, pair the odd term $k$ with the even term $k+1$:

$$\frac{e(2e)^k}{k} - \frac{(2e)^{k+1}}{k+1} = (2e)^{k+1}\Big[\frac{1}{2k} - \frac{1}{k+1}\Big] = (2e)^{k+1}\frac{1-k}{2k(k+1)}.$$

For $k = 1$ the pair is 0; for $k = 3, 5, \dots$ it is negative. The series converges absolutely, so $g(2e) < 0$. Since $g(0) = 0$, $g$ is strictly concave and $g(f^*) > 0$, $g$ has exactly one root $f_0$ in $(f^*, 1)$, and $g(2f^*) < 0$ places it below $2f^*$. Concavity then gives $g < 0$ on $[2f^*, 1)$.∎

Remark 3.5 (the rule is not universal) For long-shot bets the root can lie above $2f^*$. With $p = 0.02$, $b = 100$ (a 1-in-50 chance at 100:1), $f^* = 0.02 - 0.98/100 = 0.0102$ and $g(2f^*) = 0.02\ln(3.04) + 0.98\ln(0.9796) = 0.02224 - 0.02020 = +0.0020 > 0$. Positive skew moves the zero of $g$ to the right. The "$2f^*$" rule is exact in the continuous model (3.4) and true for even-money bets; for skewed payoffs compute $g$ directly.

The growth parabola#

Second-order picture Write $m = E[X]$ and $v = E[X^2]$. For small $f$, $g(f) = fm - \frac12 f^2v + O(f^3)$. This parabola peaks at $f^* \approx m/v$ with $g(f^*) \approx m^2/(2v)$, crosses zero at $2m/v \approx 2f^*$, and satisfies

Betting half Kelly ($c = \frac12$) keeps 75% of the growth; double Kelly ($c = 2$) keeps none. A fractional Kelly bettor gives up little growth for each unit of variance removed.

$c$ (multiple of Kelly)	0.25	0.5	0.75	1	1.5	2
Growth as share of max, $2c - c^2$	44%	75%	94%	100%	75%	0%