Deriving the Continuous Kelly Growth Objective with Itô’s Lemma
Summary
The document derives the continuous Kelly objective for a portfolio split between a risky asset following geometric Brownian motion and a continuously compounded risk-free asset. With a fraction of wealth invested in the risky asset, portfolio drift is the weighted combination of the two assets’ returns, while its volatility scales with the risky allocation. Applying Itô’s lemma to the logarithm of portfolio wealth introduces a negative half-variance term.
Taking expectations yields a growth rate that is linear in the allocation’s expected returns and quadratic in its volatility exposure. This explains why maximizing expected log wealth can produce an interior allocation rather than simply favoring maximum leverage. The response also clarifies why taking a logarithm of a weighted sum of expected terminal asset values is not the same calculation: Kelly optimizes expected log portfolio wealth along the stochastic process. The derivation assumes the stated continuous-time model and does not discuss parameter estimation, trading frictions, or constraints on leverage and short positions.
Key ideas
- The portfolio combines risky and risk-free assets in fixed proportions.
- Itô’s lemma applied to log wealth produces a variance penalty in the growth rate.
- The Kelly objective is quadratic in risky allocation because volatility exposure is squared.
- Expected log portfolio growth differs from the logarithm of a weighted sum of expected asset values.
- The derivation omits estimation error, transaction costs, and allocation constraints.
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# Question about quadratic form of f* in the Continuous Kelly Criterion
# Question about quadratic form of f* in the Continuous Kelly Criterion
I am trying to follow the Optimal Kelly derivation on Wikipedia for two continuous assets: one risky and one risk-free.
The derivation begins by assuming that the risky assets follows a GBM (a special type of exponential semi-martingale). It is assumed that we already know the solution for the expected value of $S_t$ (convexity adjustment and all).
The derivation then says that our expected rate of return from investing $f$ in the risky asset and $(1-f)$ in the risk-free asset is:
(1) $${\displaystyle G(f)=f\mu -{\frac {(f\sigma )^{2}}{2}}+(1-f)\ r}$$
How did $G(f)$ become a quadratic function of $f$? Intuitively, the quadratic form makes more sense because otherwise optimizing for a linear $f^*$ would prescribe maximum leverage and thereby assure Gambler's ruin. My sense is that it follows from Ito's lemma (or some analogue of it).
I follow along with the remainder of the derivation in terms of finding $f^*$.
(Correct answer awarded for proof by Ito’s Lemma)
### extra credit
But why is $G(f)$ not the following?
(2) $$G(f)= \ln\left(f\,e^{(\mu -{\frac {\sigma^{2}}{2}})}\ + (1-f)\,e^r\right) $$
Because: $${\mathbb{E}[S_t]} = S_0\,e^{(\mu -{\frac {\sigma^{2}}{2}})t}$$
I.e., how the fractional weights for the risky and riskless asset make their ways into the exponent?
## Answer by Tim Wilding (score 5, accepted)
https://quant.stackexchange.com/a/42362
The Kelly Criterion aims to maximise the expected value of the logarithm of terminal wealth. The derivation starts off by assuming that there is a risky asset that is following a Geometric Brownian Motion:
$$ \frac{\,dS}{S} = \mu \,dt + \sigma \,dZ_t $$
This is combined with a riskless asset that is continuously compounding:
$$ \frac{dB}{B} = r \,dt $$
Hence, if we build a portfolio of the two with proportions $f$ in the risky asset and $1-f$ in the riskless asset, its value, $G$, will follow the equation:
$$ \frac{\,dG}{G} = \frac{(1-f)\,dB + f\,dS}{G} = (1-f) r \,dt + f\mu \,dt + f\sigma \,dZ_t $$
In other words, its value will follow a Geometric Brownian Motion with drift, $(1-f)r + f\mu$, and variance, $f\sigma$.
We can use Ito’s lemma to find the time-dependent process for $\log(G)$. Using Ito’s Lemma with $f(G) = \log(G)$ gives:
$$ \begin{align} \,d f(G) & = f'(G)\,dG + \frac{1}{2}f''(G) (Gf\sigma)^2 \,dt \\[3pt] &= \left((1-f) r \,dt + f\mu \,dt + f\sigma \,dZ_t\right) -\frac{1}{2}(f\sigma)^2\,dt \end{align}$$
In other words, our expected value for $\log(G)$ after 1 period is $(1-f)r + f\mu -\frac{1}{2}(f\sigma)^2$. It is easy to derive $f^*$ from this formula.
I am sure you can tie this in to Shannon’s Demon if you want, but I haven’t done that here.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.