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.
From a backtest point of view, the Schoenberg-Corwin argument is impeccable, since we know what p is for the historical period. You might argue, however, that financial markets is not quite stationary, and in my example, if the historical value of p was less than 1/4, it is quite possible that the future value can be more than 1/4. This is why I never make too much effort to optimize parameters in general, and I can sympathize with traders who insist on averaging-in even in the face of this solid piece of research!
36 comments:
I don't have an opinion favoring averaging-in but I feel unsatisfied with the argument as presented.
The main objection I have is that the outcome (that the futures price will go back to 3) is assumed in this argument.
The outcome is obviously not known and averaging-in could minimize losses if the futures price does not go back up. At the same time one does not want to let the opportunity go and so scaling in may be prudent on a risk-adjusted basis?
I'm not married to my argument but I suspect its an argument that the traders would make in support of averaging-in. Whats the argument against this?
JB,
You are correct about the effect of averaging in on losses. Our studies showed that averaging in reduced both the max draw down and the average draw down. Unfortunately, the averaging in schemes that were most effective in controlling the draw downs were the same schemes that were the least profitable.
Al Corwin
Trading Desk Strategies
I read the paper and was quickly able to find a sentence summarising my gut reaction to your blog post:
"Re-running the simulation[...] without averaging-in[...], the profit is greater even though the percent winning trades is smaller"
Averaging-in is great... if you want to reduce your odds of losing trades - but that will come at the expense of overall profitability.
As Paul Tudor Jones said: "Losers average losers" ;-)
Hi Ernie,
As you mentioned you are using RediPlus, who is your provider of choice for automated trading on that platform (OrcSoftware, Flextrade, Trading Screen, etc.)?
Thanks.
Ernie,
I believe you did come up with a scenario that we did not anticipate. I never thought of a price change that doubled the number of units you could buy. That does change the equation considerably, but I would think that would be an extremely rare situation.
Note that because of that, you don’t make your second bet until the first bet is a fifty percent loss. With less than a fifty percent drop in the value of your first investment before you make the second one, you don’t have enough money left to buy two units since you spent exactly half on the first bet.
The Connors TPS approach is just looking to grab another two or three percent. It would find zero opportunities to average in if it ever had to wait for a fifty percent decline. Therefore, it would never make the second bet.
We found that we hit the second target less than fifty percent of the time when the second averaging in signal was consistently less than three percent away from the first. That means that when you get the second bet, it makes only three percent more than the first. In that case, I have to hit the second target sixteen out seventeen times to be a winner. I have to be making at least ten percent more on the second bet if it is coming up fifty percent of the time. The problem is that putting the second signal far enough away to allow a ten percent additional profit means that you now have a signal that will come up far less than ten percent of the time.
Thanks,
Al Corwin
Trading Desk Strategies
Anonymous,
I automate trading on Rediplus myself.
Ernie
The ratio of the expected values of the closing price to each of the opening prices of the first and second signals determines when the trader decides to go all-in on the first signal versus all-in on the second signal:
R = E(c/p1) / E(c/p2)
where c is the closing price, p1 the opening price of the first signal, and p2 the opening price of the second signal.
If the probability if the second signal occurring is greater than R, then all-in on the second signal is the most profitable, otherwise all-in on the first signal is most profitable.
A change in probability of a price movement for the second signal only affects the estimate of R which would change when to go all-in on the second signal and would not imply that averaging in would ever be more profitable.
However, as Chan points out, the time series of prices may not be stationary and we would then not be sure we had good estimates of either R or the probability of a second signal. This uncertainty might give an averaging-in method an advantage over an uncertain or poorly estimated all-in method.
On the other hand, for the trading strategies that Al and I have looked at there is no ambiguity. The estimates of R are near or greater than one, and the probability of a second signal never exceeds .5, and thus all-in on a second signal would never be considered. Nor is it likely that any non-stationarity would alter that conclusion.
It would behoove any trader to do some research with historical data and estimate R and the probability of second signals before considering an averaging-in strategy. The result might be unambiguous even in the face of some non-stationarity.
The only thing a change in probability will do is to change the decision point at which one goes all-in on the second signal versus the first signal by changing the expectation of the closing price to the opening price of the second signal.
Hi Ernie,
TickData (you mentioned them) is offering historical tick data. I am wondering how many records in the database of US Equities historical ticks, all symbols, for one year, let's say 2009?
I am reading your book right now and I am trying to find good and cheap intraday data. btw, we have a Bloomberg feed at my university, do you know if I can get all-adjusted, bias-free historical data from them?
Thank you!
Hi Korben,
If you buy the entire US historical tick database for 1 year from Tickdata, it is likely to be very expensive ... though you can of course ask them for a quote.
Bloomberg does provide div/split adjusted historical data. To get survivorship-bias-free data from them is more ticky though.
Ernie
Ernie,
Do you know how many records are in the US Equities historical tick database in one year, e.g. 2009?
Thank you.
Hey,
I am no quant expert but I think there is a basic flaw in the math.
First flaw: A is better than B at many values of p (Eg. p=1/4). Note that all are giving output profit as 2 but A has possibly less investment than others and this should not be ignored.
Expected investment in first case: (1-p)*2 + p(4) = (2+2p)$
Expected investment in the second case: 4$
Expected investment in the third case:
4*p$
Second more fatal flaw:
You have assumed that the price would go to $3. If you knew that this would happen, then there is no uncertainty.
Lets do the calculation again with 1-q as the probability that the cost does not move once it has hit its bottom (Crude I know! But I am demonstrating that ur assumptions are wrong).
Case A: Buy 1 contract at $2, with prob p the price drops to $1. Now, with prob q, the price moves to $3 and in the other case it stays at its lowest.
So, pq(1*$1+2*$2) + p(1-q)(1*-1$) + (1-p)q(1*1) = q+5pq-p
Case B: p(1-q)(2*-1$) + p*q(2$) + (1-p)*q(2$) = 2(q+pq-p)
Case C: pq(4*2$) = 8pq
Now comparing them is not that direct and should be used for decision.
I hope I am correct!!
-
Pratik (Math Enthusiast)
http://pratikpoddarcse.blogspot.com
Pratik,
I think you are right (I didn't check your math).
Just to add to your point about the 2nd fatal flaw, if we knew apriori the outcome is 3 with certainity, forget averaging-in! We should go in with everything and leverage up as much as we can muster!
The description from Ernie above is not the correct dollar cost average approach.
The idea of a diversified portfolio is to minimize idiosyncratic risk of a particular company. The correct dollar cost average approach is a way to minimize the risk of a single entry timing. If we accept the idea of a diversified portfolio, we should also accept the idea of a diversified entry timing.
Right.. Thanx a ton..
Pratik
http://pratikpoddarcse.blogspot.com
I do not when it is not averaged
Korben,
I have no idea how many records there are in the US historical tick database. However, you can certainly ask Tickdata.com about that.
Ernie
Pratik,
An overall point in my example is to illustrate whether averaging-in generates more profit (when mean-reversion does happen) with the same account equity, not whether the returns on investment (i.e. gross market value) is higher. Whether or not you actually invest in the second position, you still need a fixed account equity to prepare for that event. Therefore returns on equity is a more useful measure than returns on investment, especially for futures trading example that I constructed.
If you assume there is no mean-reversion, you can of course construct a different example to illustrate whether the loss will be greater or less with averaging-in.
Note that I constructed an example for illustration, not a proof, so the example would not encompass all possibilities of future price movements. For a proof, you should refer to Ron and Al's original paper.
Ernie
Hi Ernie,
A correct comparision must be risk adjusted. When method A is handicap with up to 50% of capital not put into play, the risk profile of methods A, B & C are very different. The conclusion draw from this is misleading.
Also, I think it is fair to assume mean reversion will happen. However, it isn't fair to assume mean reversion will happen before investor capitulation. This is the important missing piece in the probability model. After all, if you are certain of mean reversion and certain you'll survive to profit from it, the right play should be to go all in with maximum leverage.
By the way, your description of average in is based on price. Why this approach and not the approach of average in base on time? For example, buy a fraction (say 1/10 of the account equity) every week for 10 weeks regardless of price movement. What are your thoughts of this method of average in to minimize a major timing mistake?
-Crazy Hog
Hi Crazy Hog,
You are right in saying that the best strategy is the one that maximizes Sharpe ratio (a point which I made repeatedly in my book). However, this is not the point of my example. My example is not to tell you the ideal strategy, it is merely to compare the profitability of averaging-in vs all-in (at different thresholds).
Also, I disagree with your assertion that just because mean-reversion is assured, you should go "all-in", because it begs the question: all-in at what price? In the face of uncertainty, you would not know whether to go all-in at $2 or at $1, since you may or may not reach $1.
Finally, there is a theory that dollar-cost averaging is an optimal investment strategy, though it is more relevant to a buy-and-hold portfolio than a trading strategy. For details, see http://epchan.blogspot.com/2007/01/universal-portfolios.html.
Ernie
Ernie
Hi Ernie,
On your point of all-in at what price, the answer should be don't wait and go all in at $2 and with maximum leverage immediately. Waiting is just risking not to profit at all. Don't be greedy by waiting for better price. Be greedy by leveraging up.
-Crazy Hog
Hi Crazy Hog,
Whether your suggestion is optimal under realistic market conditions is not what I was discussing in this example.
Under the assumptions of my example, if p is close to 1, you would have forgone close to half of your profits when you go all-in at $2, regardless of what leverage you deploy. On the other hand, if p is close to 0, then indeed we should follow your suggestion.
Ernie
Hi Ernie,
Here is the question I don't know the answer to.
Can p be determined reliably? What I mean is this. There are various techniques (mean reversion, correlation, fundamental analysis, etc.) that allow us to predict current price at A should be at B. On the other hand, I honestly don't know any reliable way to determine the probability p (and the magnitude) of further price drop before reverting back. Is there a way to determine p?
I think p is usually is not knowable in practice. Therefore, if given a guranteed profitable trade, a bird in hand is better than waiting for the 2 in the bush. Especially since good opportunities are usually fleeing.
-Crazy Hog
p.s. I don't think technical analysis can be used to reliably determine p.
Hi Crazy Hog,
p can certainly be estimated with many statistical techniques. In fact, in a crude way, you can simply backtest your strategy with different assumptions of p, and the value that gives you the best Sharpe ratio is the estimated value of p. (This is similar to a Maximum Likelihood Estimation.)
As I have said in my article, price series is generally not stationary. So if p has a borderline value, decisions based on past estimation of p do not guarantee optimal future performance. However, in other cases, the value of p is stable and far from borderline. In such cases, backtest and past performance does predict the future. In this second scenario, there is no rational reason to follow your suggestion, and I would agree with Ron and Al's analysis.
Ernie
The explosion in the market on Friday was a good example of how averaging in can work, because the future is unknowable. P is make believe. I've found that by averaging down - I'm able to lower my basis enough, so that on normal volatility, I'm able to exit a trade at close to B/E, if I want out. It has worked for me and against me, but I would never rule it out based on a mathematical equation
Yes, MPConvert, I agree. Indeed, the future is unkownable, which is why what's optimal in backtest may not be optimal in the future.
Ernie
Great discussion here. It echoes what I have found in my systems development: that averaging-in is too often just a repair for habitually bad entries.
My studies have shown that with most GOOD systems, a larger initial entry size beats a number of smaller entries averaged-in. This is because when a trade moves in the intended direction soon after initial entry, planning to "average-in" later on leaves you with a smaller position as the trade takes off.
In essence, find a good system in which a favorable move occurs sooner rather than later - and "load the boat" early on.
The original question doesn't state (1)where the capital is coming from or (2) whether there is any cost of using the capital (which is indirectly addressed by the comment regarding risk).
Question: Let's say that if one is running a fully invested portfolio, to have capital for a "new" position, one must sell the old position. It seems intuitive that as the expected return of the new-example position increases (i.e. as the price declines), it makes sense to sell increasing amounts of the old open position. This argues for the averaging in approach, doesn't it?
Anon,
My example stated that we have $4 to invest, which is the maximum buying power. Since we anticipate the possibility of buying up to 2 contracts, we would not have invested all the buying power into the first contract.
Best,
Ernie
main problem is that the best strategy assume the price to go back to 3$ somehow, which is less probable from 1$ than from 2$. so the bigger pnl is balanced by the probability of never seeing 3$ (or even 2$ for that matter) again... The other issue is that the existence that there's a better strategy por p<1/4 and ANOTHER better strategy for p>1/4 doesn't make the averaging down a suboptimal strategy.
Thanks, Marc, for the insightful comment. Yes, I agree with your first observation. But not exactly sure about the second.
Ernie
What about the following scenario:
We trade based on a known market phenomenon. We observe something, following this, we usually see price pull back before going up. BUT it does not always do this. This is something that happens usually within 2 bars on a chart, depending on time period we trade this can happen within 5 minutes, 15 minutes, 1 hour etc. we are not talking about huge swings and averaging down, we are talking about averaging IN.
Our total risk for a trade will be 3%. Our ideal entry is in the pullback area. If price pulls back there immediately that is where we enter at full size with stop level N. However sometimes price does not pull back and continues up, we are expecting it to pull back but sometimes it does just blow away. In this instance we put on 1% (1/3) position at stop level N. If it then blows away we are still in but if price pulls back afterwards to the level we were looking for (50%) we enter in at double size for another 1%, if price falls to 25% we enter in again at quadruple size for another 1% with stop still at level N. If we are wrong we still only lose our 3% assuming no gapping, slippage etc. If we are right, based on the phenomenon we are trading on price will explode upwards and we are in for much larger size than we would have otherwise been in for. Of course we could just miss out on the trade entirely and wait for the next one that's 'textbook' but that is not such an easy thing to do when day trading\scalping for example.
Hi Anon,
Thanks for your contribution. I am assuming you are thinking of a momentum (trending) strategy, am I right?
The averaging-in technique applies only to mean-reverting strategies. I have not thought about how we might go about averaging in if we are trading momentum.
Ernie
Hi,
Yes.
The technique is momentum based, being specific it is based on post momentum analysis and how larger players will often try to get in at a better price and will hunt stops to do so.
I realise this is an old thread. I just looked through the original post, and saw that irrespective of p, the average down strategy is always the second best (whether p>0.25 or p<0.25). Which sort of makes sense. Averaging down avoids extreme outcomes, which means you don't maximise the return, but in a relative sense you minimise the variance of the outcome.
Hi manuka,
That's a good observation. Yes, option A (averaging-in) is always in the middle of profit ranking.
Ernie
Hi Ernie,
The link to the original paper is broken. Can we get a new link or a copy via email?
Many Thanks.
Hi SeanP,
I unfortunately have not saved their article. But I have summarized some important points in my book Algorithmic Trading. I don't believe the authors are involved in publishing their research any longer.
Ernie
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