Why is the normal distribution not normal? - page 2
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This is the pattern I'm getting. EURUSD, M15, 20,000 bars
The strong suspicion is that Urain took similar characteristics of the resulting series as input parameters of expectation and variance. But this may not be the case.
Unlikely. Then the middle part of the graph would be close to a histogram.
Unlikely. Then the middle part of the graph would be close to the histogram.
Yes, then the areas under the red line and the histogram would be the same too.
I was talking about Erlang, but that's not the issue here. The normal distribution has 2 parameters - MO and variance. In this case MO = 0, but the variance is not zero and in order to draw a graph we need to set its value. So I'm asking, how did Urain choose the variance value?
In general, in order to compare graphs, they must somehow be reduced to a common basis. Depending on the choice of this base, there can be completely different patterns.
If we take the variance as a common basis, the graph will be narrower, but it will have thick tails.
For the reference function, the variance and MO are taken from a series of quotes (also calculated there) and set to the same value, but the only manipulation is with the absolute values of the benchmark, here we have to add each term to the coefficient in order to combine vertices.
For a benchmark function, the variance and MO are taken from a series of quotes (also calculated there) and set to the same value, but the only manipulation is with the absolute values of the benchmark, here we have to multiply each term by a coefficient to match the vertices.
This is not very correct, in the sense of multiplying by the coefficient.
For the benchmark function, the variance and MO are taken from the series of quotes (also calculated there) and set to the same value, only manipulations are performed with absolute values of the benchmark, here we have to add each term to the coefficient in order to combine the vertices.
Probably, variance for non-stationary series is not quite correct as it may not exist :). It is more correct to choose one so that analytical distribution coincides with experimental one at most. I.e. approximate it. imha.
Probably the variance for a non-stationary series is not quite right as it may not exist :). It is more correct to find one so that the analytical distribution coincides at most with the experimental one. imha
This is not very correct, in the sense of multiplying by a coefficient.
Colleagues, what are you doing?
A researcher puts forward a HYPOTHESIS about the NORMALITY of the random process under study and models its probability curve or probability density based on the NORMAL HYPOTHESIS.
The hypothesis was not confirmed. The graphs didn't match.
That's all.
This is not very correct, in the sense of multiplying by a factor
I calculate the reference function using this formula:
so with x in say 50 the absolute value just can't be a few thousand like in the histogram so you still have to adjust,
For the fitting to be correct, it has to be applied to all terms of the curve, so the curve looks the same (especially on sliding scale).
It's okay, you've got a nice curve!
Pisser.
(Big banner in the dormitory of the 5th year of uni: EVERYTHING is OK!)
Colleagues, what are you doing?
A researcher hypothesises that a random process under investigation is NORMAL and models its probability curve or probability density on the basis of the NORMAL hypothesis.
The hypothesis is not confirmed. The graphs did not match.
That's all.
Why? This is one crude way of checking for stationarity, and it should be noted that it is not the worst. Let me specify just in case. Expectation and variance are measured for the analyzed time series. A random sequence is formed (created by some "normal" generator with exactly the same input characteristics as the original one). Further, one distribution is subtracted from the other. Obtained errors, I do not remember exactly, must in turn obey something, their characteristics are evaluated and the final conclusion is drawn. Everything is normal, I mean the method is normal :o)