From theory to practice - page 617

[Igor Makanu]

Evgeniy Chumakov:

Ran on the pound

Erm, sorry to cut in on the leisurely discussion.... Actually, the question is:

Where? "4trades in total."

Are you serious? ))))

[Alexander_K2]

Evgeniy Chumakov:

Ran on the pound

Aaaaaaaaaaaaaa!!!!

Trade signal come on, Eugene! Let me drink from the Grail - I've been looking for it and suffering for so long...

[Renat Akhtyamov]

Evgeniy Chumakov:

Run it through the pound.

i.e. the pound jumped 3 times during the month and a maximum of 50 pips of the 4 digits?

m`yeah.

systemo...

And clearly with a wait-and-see attitude.

[Oleg Papkov]

[Violetta Novak]

Can anyone help, looking for generation of random variables with Laplace distribution in Excel, found for exponential -LN(SLCHIS())/lamda, but for Laplace I can not find. There is something: mean(mu)+LN(SLCHIS())/lamda, but there is something wrong with this formula, who knows, throw me a link, thanks

[Vladimir]

Novaja:

Can anyone help, looking for generation of random variables with Laplace distribution in Excel, found for exponential -LN(SLCHIS())/lamda, but for Laplace I can not find. There is something: mean(mu)+LN(SLCHIS())/lamda, but there is something wrong with this formula, who knows, throw me a link, thanks

It seems to be found, haven't tried it myself http://forum.sources.ru/index.php?showtopic=210488:

"The Laplace distribution is a two-sided exponential distribution (divided in half)
For a distribution centered at zero
p[lap](x) = lambda/2 * exp(-lambda * |x|)
p[exp](x) = lambda * Exp(-lambda * x) -
i.e. having an exponential distribution, it is easy to pass to Laplace

while the exponential distribution can be obtained by inversion from the uniformly distributed by 0..1 values U
1/lambda * Ln(U) "

[Violetta Novak]

Vladimir:

Seems to be there, haven't tried it myself http://forum.sources.ru/index.php?showtopic=210488:

"The Laplace distribution is a two-sided exponential distribution (divided in half)
For a distribution centered at zero
p[lap](x) = lambda/2 * exp(-lambda * |x|)
p[exp](x) = lambda * Exp(-lambda * x) -
i.e. having an exponential distribution, it is easy to pass to Laplace

while the exponential distribution can be obtained by inversion from the uniformly distributed by 0..1 values U
1/lambda * Ln(U) "

Thank you very much. I specifically need Laplace on the same principle as here"the exponential distribution can be obtained by inversion from uniformly distributed at 0...1 values of U

1/lambda * Ln(U)" and the other side of the distribution would be: -1/lambda*Ln(U), for Laplace we need to connect these two sides.

I found in Wadzinski as I wrote, mean(mu)+LN(SLCHIS())/lambda, but I am doing it wrong there LN is not from(U)uniformly distributed value is considered, but from the ratio of these random variables, what in this case the entry should be, I do not know.

[Алексей Тарабанов]

Alexander_K2:

I think the dough will turn up, too.

I just feel extremely sorry for the time wasted on junk like ACF and Hearst. They don't give you anything... And on the forum Prival confused everyone with this lousy ACF, and he forgot to show his state :)))

One thing I can say: trading in the channel is the only sensible solution. It's debatable whether to go with the trend or against it... I personally am a supporter of counter-trend trading.

The main thing is to see the "tails". And the quantile before the sigma must be dynamic. But how can we define the type of the current distribution? It is difficult and resource consuming using standard methods. And within the framework of anomalous diffusion this question is solved by itself - there is no such a notion as "quantile" and the support/resistance lines are determined as if by themselves. Self-tuning, so to speak.

Alrighty then...

[Vladimir]

Novaja:

Thank you very much. I specifically need Laplace on the same principle as here"the exponential distribution can be obtained by inversion from uniformly distributed at 0...1 values of U

1/lambda * Ln(U)" and the other side of the distribution would be: -1/lambda*Ln(U), for Laplace we need to connect these two sides.

In Wadzinski I found as I wrote, mean(mu)+LN(SLCHIS())/lambda, but I am doing it wrong there LN is not from(U)uniformly distributed value is considered, but from ratio of these random variables, what in this case the notation should be, I do not know.

Based on http://sernam.ru/book_dm.php?id=6 formula (1.5) made exponential and Laplace, looks similar, but no consistency checks:

MS Excel file attached. It (and the picture) is incomplete, in cell J3 it should read "y = 2x-1".

[Alexander_K2]

So that no one suffers in the future, I would like to report that abnormal diffusion coefficients also do not work well in the market. The problem is the same - it is impossible to find and theoretically explain why one chooses the proportion not the "root of T", but the "cube root of T". It doesn't work well. Tested on a deposit.