From theory to practice. Part 2 - page 62
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))) Once again, Nikolaev wrote that the equity curve on SB has an MO equal to 0.
Not the MO of SB, but the MO of equity.
I'm telling you: you lack understanding, so find out what's what, ask your idol, he should be able to explain it to you.
Well, you will then share sacred knowledge, if it is not secret, because this knowledge is the basis of your sect's doctrine. ;)))Aleksey Nikolayev:
Ох уж эта набившая оскомину двухходовочка
three-way) the last move is always a well-deserved ban
First, it's a two-pronged, tired trick. First, it's about attracting attention through emotion, and then it's about advertising the company that runs the contests.)
That's a lot of stupidity to come up with! Is that all your "arguments"?
Three-way) the last move is always a well-deserved ban
I couldn't help myself, so I put in my five kopecks ;))))))))))
First, it's an attention-grabbing, emotional trick, and then it's an advertisement for a company that runs contests.)
Three-way) the last move is always a well-deserved ban.
He even asked the moderators to remove him from the forum. Then complained to them why they weren't removing him. Came back... It must be hard not to socialize...
He even asked the moderators to remove him from the forum. Then he complained to them why they wouldn't remove him. He came back... it must be hard without communication...
without results.
To be fair, it should be noted that the MO of both white noise and integrated white noise = the initial reference point. In experiments, most often = 0.
However, this does not change the essence of the matter.
A. Nikolaev asserts that when trying to earn on SB, the deposit will remain equal to the initial one, i.e. the expected payoff = 0.
But this assertion is more than dubious. Here I am inclined to trust Wizard.
Alexander, this changes the crux of the matter.
The MO of a classical SB is zero (by definition). Then the sampling OM of such SB is equal to zero (by the property of the sampling mean).
I.e. for any sample of SB(position opened - position closed) the IR is equal to zero.
There is nothing to argue about, unless of course you are inclined to trust mathematics rather than wizards.
You are apparently not making money on a classic SB, but on an instrument with a Hurst of just under 0.5. That's a very different case.
The point is that any oscillator shows roughly the same thing throughout a trend, so theoretically this method seems dubious. Unless you add additional signs.