Brain-training tasks related to trading in one way or another. Theorist, game theory, etc. - page 9
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I have proved the inequality, viz:
p(AA) + p(BB) >= p(AB) + p(BA)
no matter what the value of p(A) is, i.e. greater than 0.5, less than or equal to this very 0.5.
It's a hot summer, the grass is good.
But you're right:
If the outcomes of the events are independent and
0 <= p(a) <= 1,
0 <= p(b) <= 1,
p(A) + p(B) = 1.
Then
p(AA) + p(BB) >= p(AB) + p(BA)
It's a hot summer, the grass is good.
But you're right:
If the outcomes of events are independent and
0 <= p(a) <= 1,
0 <= p(b) <= 1,
p(A) + p(B) = 1.
then
p(AA) + p(BB) >= p(AB) + p(BA)
Actually, it is strange that this "kindergarten" ( p(AA) + p(BB) >= p(AB) + p(BA)) has caused such controversy and confusion in the brain...
nevertheless, the formula is correct.
Yes, x^2 + (1-x)^2 >= 2x(1-x)
Proof: transfer the right part to the left part and count: x^2 + 1 - 2x + x^2 - 2x + 2x + 2x^2 = 4x^2 - 4x + 1 = (2x-1 )^2 >=0
P.S. By the way, PapaYozh, it is not necessary that the sum of probabilities equals 1. A more general inequality is also true:
x^2 + y^2 >= 2xu
Of course true, just like 2 x 2 = 4, just like any other "kindergarten"... The question was about what follows from it. And nothing follows.
Theoretically, you could, with a cheeky face, continue to insist that nothing follows from this, but:
p(AA) + p(BB) >= p(AB) + p(BA)
corresponds:
p(AA) + p(BB) - p(AB) - p(BA) >= 0
If we play a game of foxes and make unit bets on series AA and BB, therefore, we win in the size of the bet if these very series will fall out, or lose in the size of the same unit bet if series AB or BA will fall out
Therefore, the above inequality is the expected payoff for our betting system:
MO = 1 * (p(AA) + p(BB)) - 1* (p(AB) + p(BA)) = p(AA) + p(BB) - p(AB) - p(BA) >= 0
For some pseudo-scientific commentators, maturity is nothing, a brazen twisting of an opponent is everything.Hence, the above inequality is the expected payoff for our betting system:
MO = 1 * (p(AA) + p(BB)) - 1* (p(AB) + p(BA)) = p(AA) + p(BB) - p(AB) - p(BA) >= 0
P.S. By the way, PapaYozh, it is not at all necessary that the sum of probabilities equals 1. A more general inequality is also true:
x^2 + y^2 >= 2xu
Yes, of course.
But in the groups of outcomes considered by Reshetov it is also important that one group has probability >= 0.5 . This requires the condition: p(A) + p(B) = 1.0
Yes, x^2 + (1-x)^2 >= 2x(1-x)
Proof: transfer the right part to the left part and count: x^2 + 1 - 2x + x^2 - 2x + 2x + 2x^2 = 4x^2 - 4x + 1 = (2x-1 )^2 >=0
P.S. By the way, PapaYozh, it is not necessary that the sum of probabilities equals 1. A more general inequality is also true:
x^2 + y^2 >= 2xu
Alexei, is this p(AA) how to read correctly ? the probability of two tails ( notionally ) in a row ?
Provided there is a constant trend, a bent coin that is more likely to go heads than tails. Naturally, the expectation of playing with such a coin will be greater than zero.
Once again, for the particularly gifted near-scientific commentators:
- Your comments are a special case in point. This is a blatant exaggeration of your opponent. I don't consider special cases in my problem. Even a drunken hedgehog understands without your malacholy comments that if a coin rolls eagle more often, and the player knows it, he will bet on the side of a coin with statistical advantage.
- The above inequality is true whether the coin goes heads or tails more often, or vice versa: heads or tails more often, or neither side has a winning position. That is, it is a general case for a game of tails with any coin: crooked, slanted, perfectly even, or even cheating, i.e. with heads on both sides or with tails on both sides.