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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?

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    $\begingroup$ Hi: If $\sigma^2$ is the variance of the price and $f$ is the rate of return, then the variance of the profit and loss ( i.e, the profit ) = $f^2 \sigma^2$. Essentially, you use $var(ax) = a^2 var(x)$. $\endgroup$ – mark leeds Oct 24 '18 at 13:29
  • $\begingroup$ @David The Kelly Criterion is about maximising the expected value of terminal wealth. Terminal wealth = $G(f) = f\frac{S_t}{S_0} + (1-f)e^{rt}$. The aim is to maximise the expectation of that, but you have your expectation of $S_t$ inside the logarithm, and I don't understand why you have a second logarithm term. $\endgroup$ – Tim Wilding Nov 1 '18 at 18:10
  • $\begingroup$ Okay. I have fixed that. I guess I am just having an issue discretizing the phenomenon. Intuitively, I feel as though terminal wealth is proportional to the fraction of wealth allocated times the percent return. I nonetheless accept your prior response as mathematically correct. $\endgroup$ – David Addison Nov 1 '18 at 18:48
  • $\begingroup$ I think your intuition about terminal wealth is correct, but he Kelly Criterion is about maximising the expected value of the logarithm of the total wealth $\mathbb{E}(ln(G))$. See en.wikipedia.org/wiki/Jensen%27s_inequality for why $\mathbb{E}(ln(G))$ doesn't equal $ln(\mathbb{E}(G)$. $\endgroup$ – Tim Wilding Nov 2 '18 at 9:41
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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.

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  • $\begingroup$ I accept this as correct. But could you briefly explain why the second term of $f'(G)$ is negative? The wiki derivation has this term as positive. (en.wikipedia.org/wiki/…) $\endgroup$ – David Addison Oct 24 '18 at 20:36
  • $\begingroup$ Oops, it was a typo. Should be fixed now. $\endgroup$ – Tim Wilding Oct 25 '18 at 8:58
  • $\begingroup$ There is a typo in the third equation. It should read $dG/G=dB/G + dS/G$ and the rest is correct since by definition $G=B+S$ $\endgroup$ – Ezy Jan 8 at 1:58

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