0
$\begingroup$

The Kelly criterion for the discrete case is: $$\tag{1}f=\frac{pW-(1-p)L}{LW}$$ where $f$ is the fraction of wealth to invest, $p$ the probability of winning, $(1-p)$ the probability of losing, $W$ the return if win, $-L$ the return if loss.

The continuous case is: $$\tag{2}f=\frac{E[r]}{V(r)}$$ where $E[r]$ is the expected return, and $V(r)$ is the variance of returns.

I am trying to derive the discrete case from the continuous case. So, the expected return $E[r]$ when the distribution of returns is discrete (it takes two values $W$ and $-L$) is actually equal to $pW-(1-p)L$, which is the numerator in the discrete case. So far so good. However, $V(r)$ should be equal to $pq(W-L)^2$, which is quite different from $LW$ in formula 1. Why is it so?

$\endgroup$

1 Answer 1

0
$\begingroup$

So imagine you specified your discrete outcomes by virtue of a classic "stop-loss" and "target" price level. You buy/sell and hold, until either critical level is met, creating your discrete set of outcomes.

Assuming: p is the probability of hitting the upside barrier before the downside equivalent U is the upside payoff D is the downside payoff L is your leverage, ie your Kelly bet.

Your expected log wealth (ELW) becomes ELW = p * ln(1+LU) - (1-p) * ln(1+LD) I want to maximise DPW with respect to L.

The derivative of ELW wrt L = pU/(1+LU) - (1-p)D/(1+LD) Equals zero at a maximum, so the two parts above must equate

Do the algebra, and you will end up with: L = (p*U - (1-p)*D) / (U * D)

Which is "edge over odds" in the traditional formulation. Why U*D represents "odds" is maybe not intuitive. Just divide everything above by D so U is an x:1 odds bet. Which gives you the classic Kelly discrete formula.

hope this explains to your intuition and satisfaction, DEM

$\endgroup$
1
  • $\begingroup$ Thanks, but why the formula for the continuous case is different? $\endgroup$ Commented Feb 23, 2022 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.