Proof that Charting could not be Mathematically consistent?

[euroazn]

Hey guys, what do you think of this primitive proof that charting (technical analysis) could not be mathematically consistent? It's not particularly rigorous and I'm sure there's some errors, but is the idea worth looking into?

I kind of glossed over some details, tell me if you need any clarifications.

Thanks!

 

[RussianMike]

Does it make money though?

 

[euroazn]

Haha, no, there are no arbitrage ideas
Two views, but only one comment... and even that one was said in jest.

Come on guys, if it's fatally erroneous, tell me!
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9 Views and still nothing :[

 

[Eugene Krel]

i opened the thread, i see the word document, i leave the thread.

 

[euroazn]

Word is inconvenient?

Why didn't you say so?

The existence of valid charting would suggest the following: there is some function P that takes in the chart of a security as a parameter and returns the probability that its price will increase in the next time-step. Charts themselves can be represented by a series of some constant N time-steps. For any chart that would require a smaller time frame, the function could simply give zero weight to the first few time-steps.

Charting concerns itself only with the shape of the price graph, so of relevance here is percentage changes rather than nominal changes. For this reason, we will assume all securities start at an equivalent price at the first time-step; this is a trivial distinction as it designates that a fractional amount of the security was bought. In addition, since charting does not concern itself with any fundamentals, we can assume that charting can be applied equally well to any security or any portfolio of securities.

There are so many securities available in the market that we can rather easily assume that there exist at least N+1 (decently) uncorrelated securities.¹

Now suppose we take N of these securities. Because we are only concerned with N time-steps, we can construct a portfolio of these securities such that the chart of this portfolio (that is, the last N time-steps) is identical to the chart of the leftover security. This is a consequence of basic linear algebra.

Supposing that charting works, the probability P that the security/portfolio will go up in the next time-step should be identical because the charts are identical. However, the portfolio’s P value is also equal to the weighted sum of the P-values of its constituents, and so, by the transitive property, the security’s P value must equal to the weighted sum of the P-values of the portfolio’s constituents.

This leaves us with a system of equations, with N unknowns (namely, the weighting of the securities in the portfolio²) but N+1 equations! (Specifically: the first N equations dictate that the price of the portfolio is equivalent to the price of the security at each time-step, while the last equation stipulates that the weighted sum of the probabilities of increase equals the security’s probability of increase.) For there to be a valid, unique solution, one of these equations must be redundant.
But by our definition of “(decently) uncorrelated” securities, none of the first N equations (those equations dealing with time-step values) can be redundant.
Then the last equation must be redundant. But redundant due to which other equation? The answer is obvious when we repeat the above procedure with each security. The prices at the first time step are all equal (as per our definition) and therefore the first equation always states that the sum of the weights is one. If the last equation is redundant due to the first, then clearly the P values of all securities must be equal! So therefore, if a function based on a chart could predict the probability of a security increasing in value, it is merely a constant function – a predictor of a random walk! QED
¹I do not mean uncorrelated in the traditional statistical sense, but rather as defined by the following: a security is (decently) uncorrelated with a group of securities if there exists no subset of that group, save the entire group, such that a portfolio, containing only investments from that group, can be constructed to have the exact same historical performance (in this case we are considering the last N time-steps) as that security.
²Note that the P’s are not unknowns; they must be determinable by for charting to be an effective tool.

 

[euroazn]

Can somebody please offer their opinion?

 

[Andy Nguyen]

You don't seem to be a very patient person. Many people here aren't shoot-from-the-hip type. Besides, TA isn't exactly the most favorite topic among the quantitative finance community.
Elitetrader.com may be a more appropriate audience for chart reading discussion.

 

[euroazn]

You don't seem to be a very patient person. Many people here aren't shoot-from-the-hip type. Besides, TA isn't exactly the most favorite topic among the quantitative finance community.
Elitetrader.com may be a more appropriate audience for chart reading discussion.

lol the first part is probably true. It's just that people actually took the time to download the file, and therefore (presumably) read it over...
This is why it is surprising that nobody commented.

Besides, this proof attempts to disprove technical analysis... Elitetrader.com would not be a good place for that xD

 

[Miket]

I'm not sure if I'm understanding your points correctly, so let me know if I'm wrong

You're saying a function P takes a vector (time series) of percentage changes, lets denote them S1, S2, S3..
P(S1) gives some probability that security S1 will increase in the next time step.
So say you have vectors of 3 time steps for 4 (decently) uncorrelated securities S1,S2,S3 and S4
You can solve for a,b,c where aS1 + bS2 + cS3 = S4
And then your 4th equation is P(aS1+bS2+cS3) = P(S4)? I'm a little confused where your argument goes next, it seems that this equation is unnecessary as it is true by default if aS1+bS2+cS3=S4

 

[Anthony DeAngelis]

Every major bank has chartists and people who subscribe to tech analysis. If you want a job and to sell something it is a convenient story to tell consumers.

 

[euroazn]

I'm not sure if I'm understanding your points correctly, so let me know if I'm wrong

You're saying a function P takes a vector (time series) of percentage changes, lets denote them S1, S2, S3..
P(S1) gives some probability that security S1 will increase in the next time step.
So say you have vectors of 3 time steps for 4 (decently) uncorrelated securities S1,S2,S3 and S4
You can solve for a,b,c where aS1 + bS2 + cS3 = S4
And then your 4th equation is P(aS1+bS2+cS3) = P(S4)? I'm a little confused where your argument goes next, it seems that this equation is unnecessary as it is true by default if aS1+bS2+cS3=S4

I believe that it should also be aP(S1)+bP(S2)+c(S3) = P(S4) but I have a feeling I didn't denote P() correctly ;-)
I think it is a relatively minor fix though, if someone could help me out :P

Thanks Miket, and PS: yes, I suppose the proper notation is time series (again, I'm a high school noob that knows nothing ;-).)

 

[koupparis]

I think you're running into trouble with (decently) uncorrelated (written as ~ for brevity). If I understand you correctly, a set that satisfies ~ means that every security satisfies ~ with respect to the remaining securities. I.e. every security can be written as a linear combination (lc)of the rest (all with non-zero coefficients). This in turn allows you to prove that all your securities are just constant multiples of each other, which is clearly too good to be true.

For example, x, y and z satisfy ~. Hence x =ay + cz, y = dx + ez. Insert y into first equation and you get x as a multiple of z, repeat.

 

[euroazn]

Every security in a group of n+1 securities can have n of its timesteps represented by a combination of securities. I don't see how what you claim follows...

Edit: hmmm... let me think about that! Good point.

Edit 2: I believe cancellation to a redundancy (i.e 0=0) occurs.

---------- Post added at 05:16 PM ---------- Previous post was at 05:04 PM ----------

An by the way, thanks for chiming in

 

[koupparis]

The only redundancy is either 1) All securities are multiples of each other or 2) the securities are not (decently) uncorrelated.

 

[esahione]

The induction hypothesis doesn't work for 2 securities, i.e, if you have one security and one price for it, by your 'proof', you could predict the probability of another security going up at time t=2. But this is clearly absurd, since they are uncorrelated.

Here's a little proof that doesn't work in the same way as yours:

Suppose that we have a set of five horses. We wish to prove that they are all the same color. Suppose that we had a proof that all sets of four horses were the same color. If that were true, we could prove that all five horses are the same color by removing a horse to leave a group of four horses. Do this in two ways, and we have two different groups of four horses. By our supposed existing proof, since these are groups of four, all horses in them must be the same color. For example, the first, second, third and fourth horses constitute a group of four, and thus must all be the same color; and the second, third, fourth and fifth horses also constitute a group of four and thus must also all be the same color. For this to occur, all five horses in the group of five must be the same color.
But how are we to get a proof that all sets of four horses are the same color? We apply the same logic again. By the same process, a group of four horses could be broken down into groups of three, and then a group of three horses could be broken down into groups of two, and so on. Eventually we will reach a group size of one, and it is obvious that all horses in a group of one horse must be the same color.
By the same logic we can also increase the group size. A group of five horses can be increased to a group of six, and so on upwards, so that all finite sized groups of horses must be the same color.

So not only it isn't 'mathematically consistent' (whatever this means, since if something is mathematical, it must be consistent), but the proof itself is flawed.

 

[euroazn]

The induction hypothesis doesn't work for 2 securities, i.e, if you have one security and one price for it, by your 'proof', you could predict the probability of another security going up at time t=2. But this is clearly absurd, since they are uncorrelated (your hypothesis, not mine).

This is a proof by contradiction... I'm not sure what you mean. Can you please elaborate?

EDIT : Thanks for your elaboration. What I am saying, is that if such a function (that returned the probability of rising based on a time series) existed, it would return a constant value; this essentially disproves technical analysis by contradiction, since the function would predict a random walk. I am still not sure where exactly you see the error.

---------- Post added at 05:46 PM ---------- Previous post was at 05:45 PM ----------

The only redundancy is either 1) All securities are multiples of each other or 2) the securities are not (decently) uncorrelated.

Again, if you do the math, everything will just cancel out. Try it for a low number of securities.

 

[esahione]

This is a proof by contradiction... I'm not sure what you mean. Can you please elaborate?

This is a proof by induction. Your contradiction depends on an induction, and the induction is flawed, hence, the argument is flawed. Read my post above, I've added a story that illustrates the error.
To make my point clear: by your argument, if you have only two securities, a chartist could predict the price increase on the second one my just looking at the first. As I said, this is absurd, since nobody claims to do this.

Edit:
I know your argument, and I'm telling you that it is flawed because you're not using induction correctly. Whenever you take N objects out of N+1, you're proving something by induction. And in your case, you didn't prove the base case: for N=1.

I'm not saying that chartists can predict the future, I'm saying that your argument isn't valid.

 

[euroazn]

My proof does not make use of induction...
And you understand that I am trying to disprove charting, right?

EDIT:
I disagree. When you're taking N objects out of N+1 objects, it means you're taking N objects... No induction here.

But I digress. What does my proof say if we are only looking at N=1. Then note that we are only considering one historical time-step.
Then that chart (i.e one timestep) can be modeled by a portfolio of a fractional share of the other security... And the proof goes on just fine afterwards. I do not see any issue. We get that if technical analysis is possible on one time step (which is ludicrous at face), then securities must be correlated. Well yeah, that actually makes sense.

N=1 is a really stupid case for this anyway, you can't have a chart with one price, lol

 

[esahione]

There are so many securities available in the market that we can rather easily assume that there exist at least N+1 (decently) uncorrelated securities.¹
Now suppose we take N of these securities.
...
This leaves us with a system of equations, with N unknowns (namely, the weighting of the securities in the portfolio²) but N+1 equations! For there to be a valid, unique solution, one of these equations must be redundant.
...
Then the last equation must be redundant. But redundant due to which other equation? The answer is obvious when we repeat the above procedure with each security. The prices at the first time step are all equal (as per our definition) and therefore the first equation always states that the sum of the weights is one. If the last equation is redundant due to the first, then clearly the P values of all securities must be equal!

This is an argument by induction, and it is as obvious as it can be.

 

[euroazn]

This is an argument by induction, and it is as obvious as it can be.

Ahhh! I didn't make myself clear. Each of the original N+1 securities.