Why is the normal distribution not normal? - page 3
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It's all right, you've got a nice curve!
Nishchak.
(Big banner in the 5th year dorm: ALL NORMAL!)
You don't need any multiplication for this method, though. That's right.
I calculate the reference function using this formula :
So with x in say 50 the absolute value simply can't be a few thousand like in histogram, so you still have to fit,
But for correctness of fitting it is necessary to apply it to all curve members, then the curve appearance does not change (especially on sliding scale).
Still, for normality estimation it is not necessary to multiply anything. But perhaps I don't fully understand your question.
Colleagues, what are you doing?
A researcher hypothesises that a random process under investigation is NORMAL and models its probability or probability density curve on the basis of the NORMAL hypothesis.
The hypothesis is not confirmed. The graphs did not match.
That's it.
Well that's the first step. Yes, abnormal. Next you can speculate how it differs from HP maximally approximating experimental data. >> talk purely :)
To check for normality you don't need to draw any histograms and argue about how to scale them. It is enough to derive M and sigma. The fact that M is around zero is obvious, and so it remains to find out whether sigma is about 3.
There is also the option of drawing a histogram on a logarithmic scale. For a normal distribution we get a parabola.
You don't need to draw any histograms and argue about how to scale them to check for normality. It is enough to display M and sigma. You can see that M is around zero, so all you need to do is to see if sigma is around 3.
Does the shape of the distribution not play a role?
Does the form of distribution not matter?
The shape of the distribution is determined by two parameters: gamma asymmetry and kurtosis and epsilon. It is desirable to deduce the gamma as well, but for now you can estimate it by eye.
There is also the option of drawing a histogram on a logarithmic scale. For a normal distribution we will get a parabola.
As I understand it, the problem of optimal approximation of normal distribution cannot be solved analytically. But there is no need for that. If we plot the series of the first difference for the price VR, we will get a distribution with zero MO and given that the absolute value of the distribution amplitude is not important for us, we will only have one definable parameter - the distribution width.
Here, for example, in the top of the figure, a series of minutiae is shown at the top and its first difference is to the right. At the bottom left is the density of the probability distribution, on the right is the same probability distribution on a logarithmic scale. If the distribution were normal, we would have a parabola here, which it is not, because of the "fat" tails. Basically, we need to fit a least-squares Gaussian here, and then everything will fall into place. I need to throw in a formula for the optimal fit...
Well, here comes Neutron and puts everything in its place. By the way, marketeer also has a point about the kurtosis and asymmetry.
The corresponding Gaussian curve can be plotted however you like, but here it is easiest to simply calculate the sample variance and plot a Gaussian curve with the parameters 0 and sigma. That's when you can see the difference between a real histogram and such a Gaussian curve.
By the way, this Gaussian approximation should be significantly lower than the real histogram at the centre of the curve (at the zero point).
Urain, how much did you multiply the s.c.o. of the samples by?
On the other hand, the c.c.o. estimate for a strongly fat-tailed distribution depends on the sample size, so it's not so simple here.