I wish I had more option experience. So far it been limited to covered calls and long calls.
I agree that protective puts will reduce returns but only if they’re never needed. The popular analogy is insurance … like for a home or a large wager. One doesn’t buy it with the desire to actually use it - and it is a drag on your overall return - but if the consequences of loss are too great, it becomes a requirement. Tail-risk protection should be cheap now with volatility low. I’m going to look into it for my IRA.
So, I have been making claims without really looking at the numbers. I thought I would focus on one small aspect of the big picture. I thought I would actually look at the volatility drag of one of my ports and see how much hedging/options might help with this.
My understanding is that the geometric or compounded return = arithmetic mean - 0.5 * Variance. Where 0.5 * Variance is the volatility drag.
So for one of my ports I calculated: 100 * 0.5 (Monthly Variance of Returns)/(Monthly Arithmetic Mean). This should give me the reduction in my returns due to the variance (in percentage).
I get 7.41%
This tells me that even perfect hedging that reduces my volatility to zero will reduce my overall returns if it costs more than 7.41% of my average monthly returns (per month). I will spare you (and me) the calculus on this for a closer look.
This is, of course, based on backtested volatility which may not hold up in the future. And it is over a long period (Max). And it is my port, not yours. So this may not apply to you if you have a shorter time horizon, think volatility may be different in the future or if you use a different port (which I hope is the case).
And there are issues other than the volatility drag. Including the most important consideration: What will it take to prevent me from getting scared out of the market at the bottom? I do not mean to diminish the importance of the other considerations.
Jim, I may be missing something but it seems to me that a perfect hedge that costs 7.41% of average monthly returns would reduce the returns by the same amount, rather than insure the average monthly returns stay the same. Unless the perfect hedge also compensates for the 7.41% cost. Just trying to understand properly.
Bob,
Thanks for the comment. Volatility drag is really hard to understand. And I think it is easy to find errors about this on the Web. For example, some of the Monte Carlo simulations do not get this right. Also, I am probably making some errors and would appreciate any corrections.
My assumptions are pretty simplified. But the most we could expect from any hedge (even a dynamic one) is that we would reduce the volatility to zero. This would mean no drawdowns whatsoever.
The only other assumption is that there is some cost to the hedge. For some there may not be a cost. Professional may be shorting stocks and hope to make a profit (not a loss) on the shorted stocks. But if you have a constant hedge or continue to buy puts in all types of markets then there will be a cost in a market that grows over a long period.
P123, being as great is it is uses annualized returns. This means they are using the geometric mean of the returns (or log returns). There is no volatility drag with geometric returns. Or more accurately, the geometric returns include the information about the volatility drag.
When you use average returns then what you actually get as far as returns is: Arithmetic mean -0.5 (Variance). Google volatility drag and leveraged ETF for over 78,000 hits on this. It is a real issue.
But the equation shows you that if you can reduce variance you will increase your annualized returns. But to be beneficial to your bottom line it has to cost less than the benefit you receive.
In very smart hands using hedging will improve returns. Especially if these investors use leverage which increases the volatility.
I have calculated a limit to the benefit of hedging for me. In my case and in my port, any hedging method that costs more than 7.41% of my average returns will cause me to lose money—given certain assumptions.
I can use this to see if options or constant hedging would help my overall returns.
Sorry, probably about as clear as mud so far. I am happy to try again or answer any specific questions to the best of my ability.
I fully understand the difference between arithmetic and geometric, or compound, returns and your desire to reduce or eliminate volatility. What I don’t get is the calculation of a maximum cost related to the pre-hedge variance or standard deviation of arithmetic returns. If it is true that a perfect hedge will eliminate all variation, then it must also offset it’s own cost. If that holds, then it would follow that the cost of the hedge is independent of the non-hedged volatility. In essence, it would not matter how much the hedge cost as long as it covered its own cost plus the desired variance offset.
Yes. A matter of different definitions or poor wording on my part, I think.
Let me start over. If I found a hedge that reduced variance to zero. How much would that hedge cost me?
In my search for a hedge, with my limited abilities, it would cost me. Trust me on this. For example, putting a constant short on SPY would cost in the long run. I could try to time the hedge. But with my limited market timing abilities it would end up costing me. There is no reason to think I will ever come up with a good, low-priced hedge in my lifetime.
And a zero cost hedge? Forgetaboutit! Not me anyway
So there is no perfect hedge. Not for me.
But if I got very lucky and had a period with no variance, then even under these great circumstance I would lose money if the cost of shorting SPY was more than 7.41% of the average return of my port. This is the best case. In practice I would lose even with a lower cost hedge because I would never reach zero volatility: not with my limitations.
And I agree if you put the port and the SPY short into a book the number might be a little different than 7.41% but it would be close.
All this does not apply to those who short specific stocks and make money on the shorts. Hmm… Maybe I should try that: see above regarding my limitations.
But Walter says puts can be low cost. So maybe there is hope for me. I am not holding my breath but maybe. It would, I think, have to cost less than 7.41% of my average returns for those puts is all I am saying.
So maybe. And maybe I do it anyway just because I do not want to take a big loss—even if it does cost me a little bit. Walter is probably on to something. I just want to do the math before I do it.
An equivalent way to express most of this in very few words is: I think I am way below optimal Kelly in my non-leveraged ports.
I will just add that the I used to be perplexed at volatility drag, but it turns out that volatility only exists due to the percent metrics we use, because expected returns of random normally distributed stock price follow a Weiner Process.
E*(S_T) = S_t + mt + sZ*sqrt(t)
so:
E*(Ln(S_T/S_t)) = Zs(T−t)^.5+m(T-t)
therefore:
E*(S_T/S_t) = e^(Zs(T−t)^.5+m(T-t))
which should be familiar because:
(S_T-S_t)/S_t = S_T/S_t - 1 = “percent change”
where:
E* is the expected value;
S_T is the terminal price
S_t is the initial price;
Z is a random Gaussian variable;
s is the variance;
T-t is time elpased; and,
m is the logarithmic drift (i.e., instantaneous rate of return) equal to the mean - .5*variance.
Another way to think of this is that you cannot add percents together because the denominator changes. When you are measuring instantaneous change, the denominator changes instantaneously. However, you can add logs of the same base together because math.
So, volatility drag only exists because we measure things in terms of percent returns. If we were to look at logarithmic returns, volatility drag would cease to exist. Although I acknowledge that it is impossible to actually measure instantaneous rate, this is what the “root time” volatility model are built upon. So, it’s a model…