Combining multiple systems

Quote from mind:

my point is that if you have an effect on a stock that
as forecasting quality, be it from the stock itself or
from exogenous data, you have a tradeable edge. but
out of randomness comes only randomness. systems
without edge cannot be combined to produce edge.
unless (which is a big, big unless) they have certain
features, which i mentioned before.

the monty hall case gets additional data into the equation
in form of the moderator's metaknowledge supplied
for free.

but, well, i did not answer your question. my guess
is 50%. there is no additional information coming
from the fact that it did NOT go DOWN a dollar. the
market is NO moderator.
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Are you sure??? :D Did you try the Bayes formula? Where do you think is the trick? Can you capitalize on it?
 
all prior and conditional likelihoods are 50%, so the
final conditional is 50% as well. it is like coin tossing.
there is no probability different from random. there
is no sequence ... it is repetition of identical trials. or
i am really missing something here ... but i have no
math background, so what do know? enlighten me.

i would say:
P(A) = 50%
P(B) = 50%
P(B/A) = 50%

so P(A/B) is 50% as well. where is my flaw?
 
and: NO way of capitalizing on it. random in random out.
no moderator know how entering on any level, no shift
of any odds.
 
Quote from Thunderdog:

I just don't see it. If two systems are losing individually, I don't see how they can win collectively unless you can time them properly. And if you can time them properly, then they can be profitable individually and, therefore, they are not losing systems.
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Exactly, when two losing systems are simultaneously in a drawdown - which they will be certainly at some point, since they're losing systems. That said, I imagine trading the equity curve can be a very profitable strategy (I've only briefly looked into the subject).
 
Quote from d08:

Exactly, when two losing systems are simultaneously in a drawdown - which they will be certainly at some point, since they're losing systems. That said, I imagine trading the equity curve can be a very profitable strategy (I've only briefly looked into the subject).
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well, if your losing systems have streaks, meaning
they show significant positive auto correlation, then
you might be able to build a positive strategy out of
two of which both are not be able to make it on their
own as a stand alone. the point is that the P()s deviate
from 50% once you have positive auto corr ...
 
Quote from mind:

all prior and conditional likelihoods are 50%, so the
final conditional is 50% as well. it is like coin tossing.
there is no probability different from random. there
is no sequence ... it is repetition of identical trials. or
i am really missing something here ... but i have no
math background, so what do know? enlighten me.

i would say:
P(A) = 50%
P(B) = 50%
P(B/A) = 50%



so P(A/B) is 50% as well. where is my flaw?
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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. Similar to the "boy/girl problem" in the "twins" paradox

http://en.wikipedia.org/wiki/Boy_or_Girl
 
"... If you knew that this stock has in fact gained $1 a share (moderator affect) what is the probability that this stock will gain another $1 a share?"

that was your initial question. i stick to my answer of
1/2, since it is repetition independent trials. i see no
way how this could go to 2/3. your assumption that the
stock makes +1 in t1 does not cancel out just one path
(-1 -1) but two (-1 -1 AND -1 +1). and if you use the
bayesian formula you cannot treat the two paths with the
end result of zero as one. otherwise the conditional
P gets it wrong, which must assume that in t1 a +1
happened.
 
and the comparison to the twins does not hold water,
since you cancel out not only Girl Girl but Girl Boy
either. your assumption is that the first child is a Boy.
and your question is: is the next child a boy. and the
clear answer (here order matters) is: 1/2.

IMHO.
 
Quote from MAESTRO:

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. Similar to the "boy/girl problem" in the "twins" paradox

http://en.wikipedia.org/wiki/Boy_or_Girl
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i do not see how the result 0 can have 1/3 probability
with two paths from four leading there. in my eyes
0 has 50% chance, the others 25%. you have four
outcomes, not three. since it is about conditional probs
you cannot aggregate 2 and 3.
 
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