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April 26-May 1, 2005
Hedging 101: A group of Specs discusses options and stops

The Gardiner principle: Be small when wrong, and big when right, by James Sogi

Stops are a mechanical fixed means of decision avoidance akin to wearing a raincoat and carrying an umbrella all week long after hearing the forecast for rain on Monday even though by Tuesday it is sunny already. A proper forecast will change as time goes by and should be the basis for a dynamic decision based trading model. A driver would not consider looking ahead for 500 feet, and seeing the road clear, set the steering and accelerator and look away. The apparent rationale for a stop is that the trader will be incapacitated in the midst of a draw down and unable to make a decision or to make an accurate forecast as he did upon entry. Certainly decision making under pressure is hard, but not impossible, and should be an area of greater attention and study, as it seems to be quite a confused area and fails to get the attention that entries and even exits receive.

Francis Diebold, in the Elements of Forecasting, 3d Ed, raises the issues of the decision environment and loss function. He also gives a very clear summary and explanation of many of Chair's favorite points and is very useful as a study guide for Practical Speculation. Diebold examines the loss structure as being symmetric or asymmetric. The idea is to alter the forecast as time passes to decrease the risk, rather than increase the possibility of larger losses. The forecast is either up, or down. Forecast horizons may differ, so under the Parsimony Principle, the forecast methods should be and the best are simple and robust.

Application of the Gardiner principle, "Be small when wrong, and big when right", depends on whether one is a trendist or contrarian. The trendist will pyramid, insuring the above. The contrarian will phase in in the opposite direction as the market stretches away from its mean, and phase out as it reaches the top, insuring that he will buy the bottom tick at maximum size, and be phased out as the market peaks, insuring maximum return and minimum risk. This will be true on all time frames. The amount and timing of the phase would reflect the changing forecast on the appropriate time horizon, with an eye to reducing risk and increasing return. The phasing reduces the magnitude of any one mistake or adverse tail. Proper phasing would balance leverage against the expected or maximum variance and execution risk.

Musical reference: Forecast Calls for Pain, from Strong Persuader by Robert Cray

Philip J. McDonnell comments:

"But then, that's really the trick isn't it?" - Han Solo

Star Wars quips aside, what is the effect of money management techniques such as pyramiding and lightening up as the market moves in our direction. Does it change the average amount we make or alter the probability of making money?

To begin, let's assume a random walk with no drift and ignore stops and profit targets which can complicate things. Then I would assert that pyramiding will reduce the probability of a win. An example should illustrate. Suppose we buy 1 unit at 50, then a second unit at 60. Now the average price is 55. Given that we are now at 60 it is much more likely that we will fall below 55 than below our previous breakeven point of 50. We have increased the probability of a loss and thus reduced our chance of a profitable trade.

Because we are now long 2 units we have clearly doubled our risk as well. However the expectation or average return remains unchanged. The reason the expected return is still zero (by our original assumption) becomes clear if we break the trade into two pieces. The first trade at 50 has an expectation of zero. Let's pretend we close that trade out at 60 on paper, not in reality. We know that the trade with a profit target of 60 will still have an expectation of zero from our previous discussion on stops and targets. Now we can replace it with two new positions at 60. Clearly by our random walk assumption the two new positions have a zero expectation also. So the expected return of original single position and the double position is simple zero plus zero. Essentially using pyramiding will result in a more skewed distribution with a higher probability of loss and increased risk but unchanged average return.

As for a contrarian "lightening up" strategy something nearly the opposite occurs. Suppose we go in at a full 2 units at 50 and selling one unit when 55 is reached and sell the last unit at 60. This will result in a distribution with a higher probability of success because it is effectively just two different positions with a simple profit target. However it will be negatively skewed in the sense that if both positions fall the losses will be double. Again the expected return is unchanged at zero plus zero.

As with stops and profit targets, money management schemes cannot normally change the basic expected return of any system. In this sense there is no free lunch. However all of these techniques can change the probability of success and alter the skew characteristics of return. As a generality, increasing the probability of success will create negative skew, whereas reducing the probability will increase the skew. If used to manage the probability and skew characteristics these techniques can be useful.

Russell Sears adds:

Pointing out the fundamentals of what I said earlier, and perhaps stating the obvious.

There is nothing wrong with stops or limit orders.... when you have a bias in the direction.

The Chair has given us a textbook case for using these in oil and commodities as he has implied, it is not "IF" but a matter of "when." Further, the long-term performance of stocks versus commodities suggests that eventually you would be glad with a decision to get off the commodity bull.

Conversely, when the bet is on speed, or both speed and direction option purchases/ writing, position adjusting or replicating is better.

As doggerel dogma:

If the trend is your friend you want to be "in."

If you know the beat, dancing with the market can be sweet.

James Sogi: The Dragon's Left Tail

An inquisitive colleague asks:

Q. Why is this the wrong question to ask? Given the market has been down three days in a row and the current day is up, what are the chances that the current change will be above five points? The answer is very surprisingly 4 to 1, but this is the kind of useless query that we encourage you never to make. Always look ahead not at the random probabilities of a previous event occurring.

We thought we'd ask the Chair and the List for more comment.

I opined that the frequency or magnitude of past changes alone is not predictive. The relationship of the mean and standard deviation of the sample to the mean and sd of the population that determines the probabilities as measured by a Z or T score according to the tables. Out of sample testing or randomization helps to eliminate bias of the model. These establish future probabilities with a standard error.

The Bernoulli problem of discrete probability remains: managing risk with issues of full information, independent trials, and the relevance of quantitative valuation. Under the Central Limit Theorem adequate sample size reduces dispersion around the mean, but where each occurrence is independent, Isn't it true that the probabilities of the right tail or the left tail within the expected distribution occurring next are quite close? If the distribution is roughly normal, or only slightly skewed, the probability of the right and left tail would be roughly equal, or in other words, as between them, random. This is not the issue of fat tails, rather the probability of occurrence of the left vs the right tail in a normal distribution. We would always like to catch the right tail, but every distribution has a left tail and most significantly distributed samples still have a standard deviation exceeding the mean. So though you may have a good edge, the Dragon's left tail may lash out and get you. Even with an edge, the risk of ruin remains, turning the problem into a money management problem.

My esteemed correspondent replies: Fat tails mar otherwise picturesque distributions. As for me, I treat them like a cancer doc would, I cut them out. Thus the use of stops.

The first question is whether the expectations of the right or left tail occurring in a normal distribution are roughly the same or random.

The second question is whether stops change the distributions and possibly affect the right tail as well as the left.

Steve Ellison replies:

I thought it was simpler than that. I thought the right way to frame the question would be, "What happens when the market has been down three days in a row?" That is a completely forward-looking question. Conversely, one cannot know in advance whether the current day will be up or down, so odds of 4 to 1 that the advance will exceed five points are not actionable.

Yishen Kuik adds:

I suspect that the effect of stops on a distribution is non trivial.  Say while the probability of stock being down 15% or more in a week is only 10%, this representing the total area under the left-hand tail of the bell curve.    If one were to then place a stop at the down 15% mark to protect a long  position, one would find that the stop is hit far more often than 10% of the time, because the path the stock takes during that week can often bring it into below 15% territory and then back up again.  Because this problem is path-dependent, one needs to consider it as a knockout option on a probability tree. It can be modeled, but I don't think it's trivial. Therefore it is hard to guess how often a stop would be triggered without some modeling work.  

Kim Zussman suggests:

A good course on stops.

Jeremy writes:

Options aren't equivalent to stops, because, as I think most people

recognize, the stop cuts off future paths that might come back into the money. In that sense, stops are different than options; but they're free, so they are not worse in a "fair value" sense. Else, dealers would charge for placing stops and arbitrage them against options!

One important insight not mentioned in the discussion, is that once the stock price drifts down, the average length of time it will stay down is pretty large (statistically). Likewise once it goes up. There are natural "runs" of price-up and price-down that are implicit in the statistical behavior. So less is given up than one might think by stopping out - especially if one finds alternative investments for the money.

But one argument presented in these discussions (see link) *does* imply that it's important to use the sigma/drift calculation to select a reasonable floor price to place your stops. Remember, for any future threshold price and holding period, we can calculate very accurately the probability that the stock will close at-or-above that threshold price. Those calculations really do work.

So if you DO set stops, and move them up when the stock moves up, you will at least improve the likelihood of not losing money. It would be interesting to analyze the prob. of making money according to the Random Walk model, at different price thresholds. The difficulty with doing so is NOT modeling the stops, it is to decide when to ENTER! In other words, if you stop out, when do you get back in?

So: the deep problem with some of the analysis presented (see link) is that it is not a differential-arbitrage model like option pricing, which assumes there is no cost to trace the option and stock against a risk-free bond at any moment. The option model doesn't have the problem of deciding when to get back in.

Tony Corso responds:

Regarding this point, one of the questions I am occasionally asked is, "Why can't you duplicate an option by placing buy and sell stop orders at the option strike? 

I.e., what is the difference twixt owning a Microsoft call option struck at 25, and just selling when Microsoft falls below 25, and buying when Microsoft rallies above 25?

Well, let's work it out , . . .

Microsoft ticks up from 24.99 to 25.00, so you buy a share, but, dammit, the next tick isn't to 25.01, it's back down to 24.99 . . . you've made an unfortunate choice trying to duplicate the option.

Similar scenario if Microsoft is at 25.01, and you decide to sell it out when it ticks to 25, and swear at all that is holy when it ticks back up to 25.01, cause you again made an unfortunate choice . . .

Of course, if you just own a call struck at 25, you don't have to make those choices, so you can't make an unfortunate choice.

The value of an option is merely the expected value of the avoided costs associated with those unfortunate choices.

Brian J. Haag writes:

TonyC wrote: > the value of an option is merely the expected value of the avoided costs associated with those unfortunate choices.

I believe Mr. C has flippantly tossed us a nugget. He has shown that an option is just the expected value of its replication. In the real world, that replication is very costly, and those costs are also reflected in options prices. This is an excellent point, and gives an interesting perspective on the decline of equity implied vols. The costs of doing business in the stock market and its derivatives have plummeted in recent times, while execution expertise has improved remarkably. That is to say, THE COSTS OF REPLICATION HAVE GONE DOWN. Only a factor in the decline in vols, but a factor nonetheless.

Personally, I see equity implied volatility going much lower as this trend continues. This is recently reinforced by the exchange merger announcements.

Jeremy responds:

Option pricing assumes the cost of replication is zero.

Tony Corso replies:

It assumes the cost of TRADING is zero: "...markets are frictionless and continuous, market agents both borrow and lend at the riskless interest rate...."

Michael Cohn comments:

Hidden behind all my CFA books is a book on Portfolio Insurance and Option Replication Strategies circa 1987 (plus or minus). Kind of an interesting year. In comparing dynamic replication to static hedging (one off option purchase=static hedging) the winning hand is the replication strategy if it can be executed for an implicit all in cost less than the option implied vol purchased.

I would imagine that a random model with a narrow range of prices is the worst world for dynamic replication as this strategy does best in trending markets which if I could only identify same I would rule the world....That being said, your model may well describe the world in which we live....

Do you get a better test of your hypothesis if you randomly select from a pool of daily actual returns that include more of the non normal world in which we live? How would you describe the distribution of prices from your model in terms of the actual distribution of market prices?

Russell Sears lays it out:

If you are striving for a end result there are only two ways to evaluate your "system":  you over- or under-achieved it.

For example, you are trying to get the upside on a stock, but limit your loss to X. There are generally three ways to "hedge" or limit your loses:

  1. The gold standard: Buy a put option to go with your stock. The results are known. Max (X, Stockend) less put cost less stock purchase price.
  2. The stop loss. With one exit, you raise your chance of being taken out before the stock takes off. With multiple entries and exits, you raise your chance of being churned, and will pay multiple bid-ask spreads.

The mistress loves stop losses. You will find that the market has a way of "knowing." It will periodically test your limits so as to take as much money as possible from you. And don't think your size will help you, because often the multitudes are thinking the same thing. The market knows what the herd is thinking.

In short, your end result is Max (X, Stockend) less sum of buys (price w/transaction cost less X) less sum of sales (X less price w/transaction cost) less the initial purchase price of the stock.

The third way is portfolio insurance. I agree that it can be thought of in a much higher theoretical way. But in the end it is a simple equation: If the stock closes below X, you are simply left with the average price you sold at since you are no longer in the stock. If it is higher than X, you are holding the stock but you must subtract out average price you bought at.

This reduces portfolio insurance down to the fundamental counting line of Clint Eastwood: "You are asking yourself, was that five or six shots?"...."Are you feeling lucky today?"

This is of course ignoring interest and dividends which you could tag on without too much effort.

Professor Pennington offers:

It's been suggested here that a put option can be replicated by a system in which one always exits the stock when it's below the strike price and goes long when the stock is above the strike price. TonyC said that such a put option has value because it lets us avoid the trading costs and slippage that would be involved in such a system. (Caution though that I may have misunderstood.) Brian Haag suggests that some but not all of the option's value can be ascribed to this effect.

Using a numerical simulation, I'll try to demonstrate here that, for a random walker, the stop-loss system doesn't have any value, EVEN IF trading costs, such as the "bid-ask spread" are zero. An option, by contrast, does have value.

My stock starts at time t=0 with a price of 100. At each time increment I'll add +1 or -1 to the price, with the sign determined by a random number. (I choose a random number between 0 and 1. If it's bigger than 1/2, then I take +1; otherwise -1.) I'll look at 100 time steps. So typically the stock will end up somewhere between 90 and 110. Occasionally it will end up higher than 120 or lower than 80.

Here are three portfolios/strategies:

1) Buy and Hold--The portfolio owns one share of stock and holds it throughout.

2) Stop-Loss system--The portfolio starts with one share of stock. If/when the stock price ticks from 95 at time n to 94 on time n+1, then the portfolio will SELL at time n+2, so that it's value won't change during the interval from n+2 to n+3. Whenever the stock ticks from 94 (at time m) to 95 (at m+1), the portfolio will re-enter at time m+2.

3) Buy and Hold, and hold one put option (a gift from a friendly stranger) at strike price 95 and expiration at t=100.

I ran this simulation 1000 times, using Excel along with a most excellent Monte Carlo macro written by Mr. Tom Downing that kept me from having to write down 3000 numbers by hand.

Here are the statistics for the ending values of these three portfolios:

1 (Buy and Hold): Avg final value 100.73, std dev 10.3, std err 0.33, N=1000

2 (Stop Loss system): Avg final value 100.78, std dev 9.3, std err 0.30, N=1000

3 (Buy and Hold + put option with strike at 95) Avg final value 102.63, std dev 7.8, std err 0.24, N=1000

4 (Portfolio 2 minus Portfolio 1) Avg final value 0.04, std dev 4.3, std err 0.13, N=1000

5 (Portfolio 3 minus Portfolio 1) Avg final value 1.90, std dev 4.2, std err 0.13, N=1000

Notice that portfolio 2 and portfolio 1 have the same final value--the stop loss didn't help or hurt.

Portfolio 3 is more valuable than Portfolio 1 by about $1.90. That represents the value of the call option that Porfolio 3 held. That's a pretty reasonable value. (I plugged the numbers for this simulation into the Black-Scholes formula and got $1.22; I'm not sure at the moment about the source of this discrepancy, which looks bigger than what would be expected statistically.)

The point? Well, even if you could execute stop-losses without any trading costs, the stop-loss system wouldn't add any value. It's not worth any money. Even if there were no trading costs, there's no reason to do it.

The option, by contrast, is worth money, about $1.90 in this case.

Steve Wisdom steps it up a notch:

Dr. Pennington wrote: > TonyC said that such a put option has value because it lets us avoid > the trading costs and slippage that would be involved in such a system. > (Caution though that I may have misunderstood.) Brian Haag suggests > that some but not all of the option's value can be ascribed to this effect.

Underlying all this is a "foundations of mathematics" question. With any luck, one of the full-academics or major quants on the List will offer a more rigorous overview, but in a nutshell, one thinks aloud: Shouldn't options be worth "zero"? After all, in the idealized model you can costlessly "replicate" by delta-hedging in real time. And since you're always exactly delta-hedged, you never lose any money. So you've replicated an option for "zero".

More generally, "how can a lot of zeros add up to more than zero" has been kicked around since the early days of calculus. What's the area under the normal curve? Well, the definite integral from -Inf to +Inf, with some trickery, works out to sqr(2*pi) == 2.5. But shouldn't integration, ie taking an infinite number of "paper-thin slices" and adding them together, result in either Inf*nothing == 0, or Inf*something == Inf, depending on whether the slices are zero or non-zero? So how can 2.5 be the answer?

Chair's friend Mitchell Jones, an autodidact, belletrist ("Dogs of Capitalism", inter alia) and amateur scientist in the spirit of Francis Galton, for years argued that calculus isn't "real". He doesn't believe in "instantaneous", or "limits", or "Riemann sums". He believes in numbers you can count on your fingers. And it's non-trivial to show him he's wrong (feel free to try at the next Spec party!)

Philip J. McDonnell augments:

An alternate way to create a risk free put is the put call parity relationship. The idea is to short the stock and buy a protective call at strike K. The trader is in a position to sell a put at strike K which is hedged by the short stock/call position. Arbitrage pricing theory says this risk free trade should yield no more than the risk free return.

Reversing everything, buy the stock, buy the put now we can sell a call without risk. Again APT says we are entitled to no more than a riskless rate. These relationships give the following Put Pricing Arbitrage model:

put price = call price + K - stk price + dividends - carrying cost

The above relationship can also be used to create synthetic long or short stock positions typically with margin requirements of 20% (lower than NYSE requirements).

Synthetic Long Stock - Buy 1 call, Sell 1 put at strike K

Synthetic Short Stock - Buy Put, Sell Call at strike K

The nice thing about the above positions is that they are linear. Usually the payoff functions of options are curves. You need to worry about the higher order derivatives of the curves in up to 6 dimensions. But the put and call at the same strike are mirror image functions and thus neatly cancel out all the non-linear issues. This also has the effect of eliminating dynamic hedging. Assuming adequate margin, once you make the trade you can just hold it to expiration and collect the return.

Dr. Zussman queries:

The above relationship can also be used to create synthetic long or short > stock positions typically with margin requirements of 20% (lower than NYSE > requirements). > > Synthetic Long Stock - Buy 1 call, Sell 1 put at strike K > > Synthetic Short Stock - Buy Put, Sell Call at strike K

Why would one want to create synthetic stock positions which cash-out at expiry? Even as an anti-emotional ploy to stick to a specific holding period, morons like the undersigned could still sell the long option and/or buy the short during intervening panics.

Big Al wants the bottom line:

Okay, enough of this fancy-shmancy derivatives stuff. Who can understand these synthetically-delta-hedged-short/long-option-replicating-double-mocha-half-decaf things anyway? What I want to know -- and I'd appreciate it if we could just get to the point here, please -- is this: What are the Open, High, Low and Close for the S&P going to be...*tomorrow*? Thanks in advance.

Philip J. McDonnell applauds:

Finally Big Al has asked the right question. Here is my super secret formula:

Tomorrow's S&P.X = tomorrows Call.X - tomorrow's Put.X

where you can substitute Open, High, Low or Close for ".X" and of course Calls and Puts are on the S&P's.


George Zachar queries:

I am a mere price taker, not price maker, but this discussion about risk-free rates prompts me to ask this question:

What lessons can we draw from Japanese derivatives markets, where the risk free rate has long been zero nominally, and negative in real terms?

Philip J. McDonnell responds:

Although I barely qualify as a pseudo minor quant poseur I shall attempt an answer.

The answer is contained in the derivation to the original Black-Scholes model. Essentially they set up an arbitrage where one buys 100 shares of stock and sells an "appropriate" number of calls against it so as to continuously adjust the position to delta neutrality. The resulting arbitrage is riskless assuming all of the usual academic simplifying assumptions. Given that it is riskless the return to the arbitrageur should be the riskless rate.

Setting the return for the foregoing arbitrage to the riskless rate sets up a differential equation which has the Black Scholes formula as the closed form solution. The main point is that the formula explicitly includes the riskless rate because it is the reward to the risk free arbitrage.

The other equally important point to take away is that this particular solution out of all possible solutions has the serendipitous outcome that the value of the calls happens to exactly coincide with the expected value of the call option under the log normal distribution.

The bottom line is that calls are fairly priced and risk free arbs are paid the riskless rate and there are no free lunches. Mathematics is saved!

Brian J. Haag comments:

Exactly my point.

More and more, any given "position taker" will take advantage of the plummeting cost and skyrocketing effectiveness of execution rather than actually assume curvilinear exposure.

On a related note, I find the recent exchange merger announcements as massively bullish. Doesn't this warrant the "risk premium" coming down? More transparent, faster, more efficient markets, and all that.

All speak to less volatility. Although I do agree (as Mr. E has been hinting at), all these convergence and "normality" bets at some point are going to cause some serious pain.

A distinguished Spec writes:

You would consider paying more for options in a discontinuous world than in a smooth, continuous world of infinitesimal price changes nicely behaved in all respects. So, in the later case where the return is the risk free rate, option values==0 interval to interval; any other case replication has a real cost and options cost money. Sorry I can't make this fancy. They simply cost money to replicate and as such have a value.

One conundrum is that you can only systematically make money in options by selling them so somewhere above I am wrong. Fear has to be systematically overpriced relative to actual results. You could see this in so many markets with implodes trading over actuals for long periods.

This suggests the strategy of selling exchange traded options and delta hedging in the real world with the actuals.....and so went my youth in the 1980s and 1990s...

Russell Sears: Modeling a Stop Loss Is Not Easy

The stop loss guy says, "I will make my average transaction always be at X, by only buying and selling at X." He is betting he knows the direction it is going at X. Buys as goes up past X, sells as goes below X. That he is making a "direction" bet can be seen in building a model.

In a model you are left buy at X + Z above or selling at X - Z Because you can not determine if it crossed X only 1 time or 1 million times between modeled periods. Prof. P's $1.90 would be the sum of those Zs. Though he used 3,000 movements, how many of those were "crossings" of 95. I suspect the answer is few thus explaining the difference between $1.90 and $1.22.

The replicator is shooting at a moving target. thinking the cost will end up within X + or - options expected cost. with a profit (in implied vols) for taking vol risk. He is betting he can predict how fast the target is moving. If he shots at bull's-eye X enough times, throughout the options life he expects to miss that bulls-eye, on average, a given distance due to speed. This distance is $1.22 in Prof. P's example.

Stop loss traders bet directional knowledge, cost is speed of market around X. The Prof showed that this can be quite volatile, in addition to costly.

Delta hedgers bet speed only, not direction. Cost is actual speed whatever path the market takes. Thus given 3,000 tries should, with interest, come close to the $1.22 cost.

James Sogi concludes:

Thank you very much to everyone who added to the wonderful, informative and enjoyable discussion on hedging and stops. What I learned:

Fixed stops are a fixed system to overcome personal indecision and therein is their weakness that affect the probabilities and distributions, and are not the sure cure often touted. Options theoretically are an alternative, but actual cost may present practical barriers to realization of their theoretical benefit as a hedge. Conversely option mispricing presents good speculative 'overlays.'

As time in trade passes cycles change, and the probabilities and situation changes rendering the old fixed stops obsolete. An alternative management solution would be to enter in appropriate size according to probabilities expected, then at a later time, reanalyze the situation, adjust size at a later time, each time with a view to keeping size below the risk of ruin. With modern decreases in vig, this cost might be less than the price of an option, offer flexibility. If probability increases, add size, if probability decreases, reduce size or exit all with a mind to proper execution. Peter Gardiner stated it well. "Position sizing is cheaper, more effective .... the only real way to achieve downside protection is to be small enough when you are really wrong; and to achieve great profits, to be big when you are right."

Peter Gardiner's Final Word:

Problem is, I am most expert in doing the reverse.