Question about quadratic form of f* in the Continuous Kelly Criterion

Question

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)

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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 1

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.