Quote from mind:
hm. thnx for the patience. i struggle with this:
"A hint: Number 2 and 3 result in "0" loss/gain. So you can reduce the problem to this 3 outcomes: gain, loss, break even with 33% probability each."
i simply think this is wrong. i have four end states, each
of them equally likely:
+1 +1 2
+1 -1 0
-1 +1 0
-1 -1 -2
i have no idea how the outcomes 2, 0, and -2 can have
equal likelihood. i am really stuck. well ... maybe i get it
overnight ...
Quote from MAESTRO:
I am 100% sure you will. And once you get it you will never look at the markets the same way. I promise.
Hey Maestro,
I always seem to find you on the more interesting threads (at least from my perspective). I wanted to chime in on some of the conversation that has unfolded from my perspective.
Firstly, I commented before on Parrondo's Paradox. There was no further discussion. A paper was referenced, from which,
"The key is to know the rules of the games exactly, so one can know which game to play in which situation. This is clearly the reason why one cannot gain money by applying parrondo's paradox to the stock market" was quoted.
The paper:
http://gi.cebitec.uni-bielefeld.de/people/rahmann/parrondo/rahmann-report.pdf
-----------------------------------------------
While I have not personally run an exhaustive study of the feasibility of parrondo's method, there are a few common sense ideas which lead me to believe (along with the academic response I quoted, although for different reasons) that there are a few very difficult issues to make it practical in markets.
1) The basic premise is quite simple and easily studied from a timing diagram. Although they often quote 2 systems being switched, I prefer to think of it as three (2 of which have dependencies depending on the modulo relationship of the current capital)-- A, B1, and B2. Even though system A and B have net expectation of zero (I've run the numbers), the critical requirement IMO is that the third system, or B2 if you like, must have a very high hit rate. You need to have a reliable system that has consistent ~75% hit rate, which is very difficult to do reliably. Even though the combined b system looks low, it is A with it's 50% likelihood that tends to make up for the deficiencies from the B1 low probability. Timing is also critical. It may seem simple to assume a 10%, 50%, and 75% likely-hood, but to have all three consistantly perform in such a manner over time is no simple feat.
They tend to cancel out and replace portions of their time domain windows such that the aggregate expectation with the random switching has a slightly positive expectation.
But what's even more important than getting the probabilities in the time domain stationary in order for the combined system to work properly is that the assumption under parrando's game is that the system's outcomes are binary. This is a huge problem in reality, as you can run many simulations and get the aggregate sequence of the stitched ones and zero windows such that the expected outcome of the three switched systems are profitable, however, once you are no longer dealing with binary outcomes it is VERY DIFFICULT to align a myriad of continuous return possibilities to stitch together profitably in the same manner as a binary system. It is similar to the problem of assuming simple binary outcomes in kelly betting. Under continuous outcomes, the entire timing and cancellation/replacement of sequential window bet magnitudes that Parrondo's paradox depended upon is null and void. I won't elaborate much further on that, but needless to say, if you have figured out a way around this, why not send it to the noble prize committee (heck another guy, R. Taylor, was nominated for claiming he found the secret to market timing based on phases of the moon).
Lastly, to address the conditional probability issue, I pretty much agree with mind's comments. It is a simple manner to derive a truth table of the 4 possible outcomes you posit.
Let each set of two sequence events be P(A), P(B), P(C), and P(D). All are independent and have equal probabilities of 2/4 or 1/2. For example P(A) represents the row comprised of the set of +1,-1 or +1,+1 (again easier to see on truth table).
Assuming P(B) represents -1,+1 or +1,+1 (the vertical column) the intersection of P(A) and P(B) is 1/4 (common element being +1 as second term). Thus the conditional probability
is P(B)/P(A) = P(B intersect A)/P(B) = (1/4)/(1/2) = 1/2 = 50%. Which also shows they are independent since P(B intersect A) = P(A)*P(B).
With regards to sequencing and the probability of outcome of 4 events, we assume NO compounding as the results in both parrondo's as well as the simple truth table outputs are drastically different under compounding.
Here we only assume the simple +1 win, -1 lose fixed +1 bet binary outcome.
F(1,1)= 2
F(1,-1) = 0
F(-1,1) = 0
F(-1,-1) = -2
Thus,
P(1,1) = 25%
P(1,-1) = 25%
P(-1,1) = 25%
P(-1,-1) = 25%
or reduced to three states the respective p()s are 25%, 50%, 25%.
However, this is the answer looking at it from the perspective of sequential events. I do not see another way around that.
There is another way to look at it to get to your 33.3% scenario.
Simply think of systems A (binary digit 1)
and system B(binary digit 2) running in parallel. From that perspective, you can envision three states per event:
1) both bits are equal and high -net outcome = +2
2) both bits are opposite -- net outcome = 0
3) both bits are equal and low -- net outcome = -2
In this case, they are running in parallel and the sequential order is irrelevant,
only the outcome of the 2 bit state is relevant for each event, thus the probabilities are divided equally into three states (hence 1/3 per state).
I still don't see how that relates to parrondo or a successful edge in combining losing systems, but hopefully this adds food to the discussion.
I'm always interested to look at new ideas.