Least/Most Favorable Paths when dynamically hedging a short straddle

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In a risk adjusted sense, yes, your expectated terminal profit over the stdev of that profit will be higher. Your expected profit, before transaction costs of the hedging, is price of the option (or the spread in this case) minus the price of the option (spread) priced at your forecast vol.

The thing is, that is the same expectation as not hedging at all. If you sell an option at price a, and it is fairly priced at b, then your expectation is a minus b. So unless you can delta hedge at mid and pay no comms/fees, your expectation is actually higher with no hedge (but your sd is worse, much worse). Also, note the expected profit line in the plots in the posts above is essentially flat, your exact hedging scheme affects sd, but not terminal profit much (ignoring hedge transaction costs)


No, but you do have the lowest terminal standard deviation (highest sharpe). You'll also change the shape of the terminal distribution -- the short strangle no-hedge distro is negative skewed and fat tailed negative also. The hedge will make the terminal distro much closer to normal. Sharpe doesn't account for skew and kurtosis, but Kelly does. A four moment Kelly Fraction function is below, if anyone is interested:

kFf4M <- function(mu,sigma,skew,kurt)
{ mu <- max(mu,0)
  rawSigma <- sqrt(sigma^2 + mu^2)
  rawSkew <- skew * sigma^3
  rawSkew <- rawSkew + 3 * mu * rawSigma^2 - 2 * mu^3
  rawKurt <- kurt * sigma^4
  rawKurt <- rawKurt + 4 * mu * rawSkew - 6 * rawSigma^2 * mu^2 + 3 * mu^4
  f = mu * (rawSigma^4 - mu * rawSkew) /
  (rawSigma^6 + rawKurt * mu^2 - 2 * mu * rawSkew * rawSigma^2)
  return(f)
}

This kelly bet size uses only the first 4 moments of the distribution of the returns? How did you derive this? I though to get the 4th order taylor polynomial inside the integral
integral ln(1+x*r)dP(r)
where r is the returns and x is the betsize but that couldn't be accurate b/c the losses for when the arg of the log becomes 0 are the essential to the bet size and wouldn't be captured by any finite order taylor polynomial. Sorry if it's common knowledge but I've never seen it before. I always use the sample distribution, ie, the uniform distribution over all samples:

def kellyReward(x):
        reward = 0
        for i in range(len(returns)):
            reward += np.log(1+returns[i]*x)
        return -reward
    lBound = -1 
    rBound = -1/min(returns)-.01 # this is the betsize such that the largest loss equals a 100% drawdown
    res = minimize_scalar(kellyReward, bounds = (lBound,rBound), method = 'Bounded')

That said though I always calculate kelly, I never use it because the tails are necessarily fatter than observed.