Traders Joking

A frozen Lake Michigan by Stacey Anne Leeson

Where the city meets the ocean. Gold Coast – Australia by Glen Anderson

Rain storms over Ragged Top Peak in the Silverbell Mountains by Lori Grace Bailey

Mates
- photo by Emmett Sparling (Cape Town, South Africa)

Photo by Rodica Racu

WHERE IS THE CAT IN THIS PICTURE?

"My cat's name is Norah, although in my family no one agrees on whether there should be an 'h' on the end," Hinds told me in an email. "We adopted her a little over ten years ago from a local rescue group."

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Kamikaze Trading - script for MetaTrader 5

This script opens order with maximal amount of volume available with active chart symbol.

Take Profit and Order Type is specified by user.

There is no Stop-Loss. Trade will be open until account has been blown up.

Elephant in the room

Two traders are talking (taken from this mql5 thread):

Consider a linear regression model xi = a + b * i + ei in time i = 1, 2, ..., n, where the errors ei are white noise with the Laplace distribution. The error density then has the form p (x, c) = 0.5 * c * exp (-c * | x |), log (p (x, c)) = log (0.5) + log (c) -c * | x |

The likelihood function for the noise will have the form L = p (d1, c) * p (d2, c) * ... * p (dn, c), where di = xi-ab * i are the residuals of the model. Logarithm of the likelihood function LL = n * log (0.5) + n * log (c) -c * S, where S = | d1 | + | d2 | + ... + | dn |. S does not depend on the parameter c, therefore the problem of maximizing LL is solved in two stages

and it is the reply:

This is all true. The question is what exactly to take for the sliding between the two rows. For example, there is a traditional opinion that the length of the perpendicular to the regression line. But it seems to me that this is not quite the right way. For it gives a separation not relative to the previous values, but relative to a certain midpoint of them. Such a substance as the "asymmetry" of the opening is lost, and I would like to feel it.

And what do you think?

• minimization of S (since c> 0) with respect to a and b ?
  or
• maximization of LL with respect to the parameter c, with the found value of S ?

Originally Posted by mql5

Two traders are talking (taken from this mql5 thread):

Consider a linear regression model xi = a + b * i + ei in time i = 1, 2, ..., n, where the errors ei are white noise with the Laplace distribution. The error density then has the form p (x, c) = 0.5 * c * exp (-c * | x |), log (p (x, c)) = log (0.5) + log (c) -c * | x |

The likelihood function for the noise will have the form L = p (d1, c) * p (d2, c) * ... * p (dn, c), where di = xi-ab * i are the residuals of the model. Logarithm of the likelihood function LL = n * log (0.5) + n * log (c) -c * S, where S = | d1 | + | d2 | + ... + | dn |. S does not depend on the parameter c, therefore the problem of maximizing LL is solved in two stages

and it is the reply:

This is all true. The question is what exactly to take for the sliding between the two rows. For example, there is a traditional opinion that the length of the perpendicular to the regression line. But it seems to me that this is not quite the right way. For it gives a separation not relative to the previous values, but relative to a certain midpoint of them. Such a substance as the "asymmetry" of the opening is lost, and I would like to feel it.

And what do you think?

• minimization of S (since c> 0) with respect to a and b ?
  or
• maximization of LL with respect to the parameter c, with the found value of S ?

I am impressed with your knowledge. So, according to your explanation - it is possible to make money on Forex, right?