Why you should add to winners and never add to losers

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In its simplest form, the trader starts out with 2 contracts, exits the first at a small profit (the “peel off”), and then holds the second as a “runner,” going for big profits. In some cases when the first contract is exited, the second contract is then modified to have a breakeven stop.

Like almost all of Kevin’s work it’s a very interesting piece based upon supporting analysis. Kevin actually back tests the exit rules upon three different strategies but the results are inclusive

Results of the out-of-sample analysis are mixed. The ES is the same, the GC strategy is better with the “peel off” method, and the JY strategy is better with the baseline (always trading 2 contracts).

While Kevin’s blog was looking at exit scenario’s, Ernie’s blog was looking at entry rules. It was also more theoretical in nature, but his conclusion was more decisive.

Ron Schoenberg and Al Corwin recently did some interesting research on the trading technique of "averaging-in". For e.g.: Let's say you have $4 to invest. If a future's price recently drops to $2, though you expect it to eventually revert to $3. Should you
A) buy 1 contract at $2, and wait for the price to possibly drop to $1 and then buy 2 more contracts (i.e. averaging-in); or
B) buy 2 contracts at $2 each; or
C) wait to possibly buy 4 contracts at $1 each?
Let's assume that the probability of the price dropping to $1 once you have reached $2 is p. It is easy to see that the average profits of the 3 options are the following:
A) p*(1*$1+2*$2) + (1-p)*(1*$1)=1+4p;
B) 2; and
C) p4*$2=8p.
Profit A is lower than C when p > 1/4, and profit A is lower than profit C when p > 1/4. Hence, whatever p is, either option B or C is more profitable than averaging in, and thus averaging-in can never be optimal.

As I started thinking about Chan’s analysis I didn’t like the way he considered the one situation “that the probability of the price dropping to $1 once you have reached $2 is p” and excluded other scenario’s. In a moment of boredom I decided to expand Chan’s analysis to a binomial tree where we have a probability p of going down $1, but also a probability (1-p) of going up $1.

Hence at time T+1 we have probability p of trading $1, and a probability (1-p) of trading $3. Then at time T+2 we have probability p*p of trading $0, a probability 2p(1-p) of trading $2 and a probability of (1-p)*(1-p) of trading $4.

For those of you were not aware this is what we would call a binomial tree which is used to derive the Binomial Distribution (Wikipedia, Wolfram Mathworld) and there are many option and derivative pricing models based upon binomial trees and even more based upon other type of trees.

If we consider as an example scenario A) where we buy 1 contract at $2, and wait for the price to possibly drop to $1 and then buy 2 more contracts.

In case i) where the market goes up at T+1, and T+2 to $4 we never buy the additional contracts, hence we buy a single contract at $2, that has a $2 profit. Since this has a probability of (1-p)*(1-p) the expected payout of case i) is 2*(1-p)*(1-p)

Case ii) the market goes up at T+1, and then down at T+2 to $2 we never buy the additional contracts, and we have no profit. Hence the expected payout of case ii) is 0.

Case iii) the market drops at T+1, and then rallies back to $2 at T+2. In that case our initial contract purchased at $2 is breakeven, but when prices dropped to $1 at T+1 we added two more contracts. These two contracts now have a total profit of $2. Hence the expected payout of case iii) is 2*p*(1-p).

Finally in case iv) the prices drops at T+1 and T+2 and ends at $0. In this case we lose $2 on our initial one contract purchase and lose an additional $2 on our two extra purchases at $1. Hence we have a total loss of $4 and an expected payout of -4*p*p

Adding this all together our expected payout for scenario A is
2*(1-p)*(1-p) + 0 + 2*p*(1-p) - 4*p*p = -4p^2 -2p +2
It follows that the other expected payouts are B (-8p +4) and C (-8p^2 +4p)

If we plot these functions we can see that while A) and C) might outperform B) (well actually lose less) when p>0.5 (ie market is more likely to go down) but they underperform even more when p<0.5 and hence the expected returns of A) and C) are both negatively skewed in relation to B).

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While this doesn’t completely agree with Chan’s analysis that’s it’s never better to scale in, it does support his analysis in that from a risk reward perspective it’s never better to scale down into a position.
I then added a fourth scenario D) where we Buy 1 at $2, and if prices go up we buy 2/3rds more at $3. I know fractions make it messy but it’s the only way to keep it an equal investment size.
The expected payout of D) is +1.33p^2 - 6p +2.67

This is the complete opposite of A) and C). It underperforms 2) when p>0.5 but outperforms by a greater margin when p<0.5, hence it has a positive expectancy skew!

This would imply that not only should you not “average down” but that adding to winning positions has a better risk reward.

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While D has the best risk/reward it does have a lower reward than the base scenario B). Hence to achieve the same desired results, you can/should scale the positions appropriately. This is actually a great way to visualize what we are discussing as you can see below. You will see that while 3 of the 4 have similar positive profiles, scenario D has the least worse profile when the market goes against the position.

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I then expanded the analysis significantly, going out to T+5, and including additional scale in scenarios. The conclusions were the same though. Averaging down has a worse risk reward profile than buying your entire position at entry, while adding to winners has a better risk reward profile.

Disclaimers/Notes/Flaws/Weaknesses:
A couple of the obvious flaws of this analysis are
- it uses a fixed price movement of $1 as opposed to a percentage move - while this is actually easy to fix for the purposes of this example it would add undue complication since +1-1=-1+2 but +1%-1%<>-1%+1%
- it assumes a binomial distribution when in reality a log normal distribution would be a better fit.
- Price (changes) in reality are not log normally distributed. As we all know tails are a lot lot fatter.
- Stops could potentially completely change the analysis.

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Thank you for sharing this analysis.
How do you add to winners? A separate entry rule? The market is going in the right direction so add?
I have been struggling with each of these options off and on. I learned not to do A) the hard (expensive) way. B) Yields the best results for me. I have been testing D) and it has been exhausting.

Great analysis! I'm new to the futures.io (formerly BMT) forum and already a fan of yours.

While I absolutely agree with your model mathematically, I can not help to think about it from an another perspective. Given the natural law of demand and supply, where a lower price will attract more demand, which in theory will eventually bring the price back up once demand overwhelms the supply. A mean-reversion trader would live on such law and buy low sell high on the expectation that probabilistically prices have lower chances of going lower when it is already low or going higher when it is already high. This is also why people average in the first place in my understanding.

So when it comes to p (probability of price dropping) in your model, if in T+1, p happens, then when it comes to T+2, the probability of price dropping again should be in theory, less than p, say theta*p, where theta<1 and the same goes for T+3,T+4.....T+t. I tried to model the exact same thing but with such tweak, the curve on the averaging in scenario improved as you might have imagined. The magnitude of improvement varies based on the input of theta and investment time horizon t.
My opinion is that averaging losers might not be as toxic given the right strategy, right trading system,right trading time frame and right mind.

I see that you are a CL spread trader just like myself, what is your opinion when it comes to averaging in spreads trading?

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What you are describing is similar to what is called a mean reverting process, and I would agree that this analysis would not hold for a mean reverting price process. The most common way to model this is with a Ornstein–Uhlenbeck process which behaves similarly to the way you describe, but your 'theta' is a function of how far away you are from the mean, rather than what the previous days movement was. When price changes are dependent upon prior price changes we call this serial or auto correlation. Doug Huggins and Christian Schaller's book Fixed Income Relative Value Analysis has an excellent chapter on mean reversion. Unfortunately some of the math is over my head and much of the rest of the book is outside my sphere of understanding.

Yes I trade a lot of CL spreads and butterflies, that I hold anywhere from minutes to weeks. While CL spreads are often correlated to absolute prices, I believe CL Butterflies are mean-reverting once you are past the front of the curve. As such I am comfortable averaging into (and out of) many of my positions. I do not make 'primarily directional trading positions' though, so unsure how appropriate it would be for that style of trading.

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I suspect we are talking about different things...

To put this in perspective below are two slides from a presentation I gave a few months back.
CL01 refers to the prompt month for outrights, and to the CL01-CL02 spread for spreads and to the CL01-CL02-CL03 Butterfly for Butterflys.
Other than that I think they are self explanatory.
Remember that vol scales proportionally to the square root of time. So if the 1 day standard deviation of CL01 price change is $1.37 we would expect the 1 month/20 Day standard deviation to be $1.37 * SQRT(20) = $6.12. On the other hand we would expect the 1 Month standard deviation of the CL10-11-12 Butterfly to be $0.05!

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Yes, in a potentially big way. What's the assumption of this analysis when it comes to stops? Are we assuming independent stop levels for each independent trade or are we assuming a common stop? I don't know many people who scale in with independent stop levels as it's not really "scaling in" at that point. There's some parts of this analysis that while mathematically may indicate one thing, intuitively it doesn't totally add up for me.