Q: How do you derive the rayleigh distribution?

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The mode of the Pareto distribution is its lowest value.

The Gaussian distribution is the same as the normal distribution. Sometimes, "Gaussian" is used as in "Gaussian noise" and "Gaussian process." See related links, Interesting that Gauss did not first derive this distribution. That honor goes to de Moivre in 1773.

Easy. The mean deviation about the mean, for any distribution, MUST be 0.

I derive that this question needs to be moved.

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If X and Y are independent Gaussian random variables with mean 0 and standard deviation sigma, then sqrt(X^2 + Y^2) has a Rayleigh distribution with parameter sigma.

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The mode of the Pareto distribution is its lowest value.

To derive the mean of generalized Pareto distribution you must be good with numbers. You must be good in Calculus, Algebra and Statistics.

The total deviation from the mean for ANY distribution is always zero.

Rayleigh Rockets was created in 1949.

Rayleigh Windmill was created in 1809.

Given the fact that the browsers which we are forced to use cannot even handle superscripts, it is virtually impossible to give a sensible answer here. URLs are prohibited so I cannot give you a link but I recommend you search Wikipedia for Rayleigh Distribution.

rayleigh -8 letters :-)

lord rayleigh discovery

var(X) = (xm/a - 1)2 a/a-2 . If a < or equal to 2, the variance does not exist.

The Gaussian distribution is the same as the normal distribution. Sometimes, "Gaussian" is used as in "Gaussian noise" and "Gaussian process." See related links, Interesting that Gauss did not first derive this distribution. That honor goes to de Moivre in 1773.