Var(X)=∑
(pi*(xi-µ))2 for i =1 to n
variables
The variance is traditionally defined as the square of a
parameter of a random distribution, σ2, where σ is also known as the
standard deviation, SD. The other parameter
of a random distribution, µ, the mean, which is equal to the median of a
normal distribution, is
µ=∑ pi(xi-µ)*xi
for i=1 to n variables.
However it is not
correct to say that the variance, which is the square of the standard deviation,
SD, can be solved by Pythagoras’ formula. This is only true on a flat surface. In fact on a flat surface, rather than a
summation, this should be expressed as a product as
SD(X)=cos-1(∏
cos(pi(xi-µ))) for i=1 to n variables.
On a flat surface, this
is identical to
SD(X)=√( (pi(xi-µ))2
for i =1 to n variables or √( (xi-µ))2/n).
That Bessel’s correction,
n/(n-1), is necessary for the SD is an indication that the surface is not flat.
On a spherical surface,
the formula should be
SD(X)=R*cos-1(∏
cos(pi(xi-µ)/R)) for i=1 to n variables.
Here R is the radius
of the spherical surface. As R
approaches infinity, e.g. the sphere become very large compared to any pi(xi-µ),
this becomes identical to the formula for a flat surface.
On a hyperbolic surface,
the formula should be
SD(X)=cosh-1(∏
cosh(pi(xi-µ))) for i=1 to n
The variance, σ2,
of a random logistics distribution, where the parameters are µ and s,
is s2*π2/3.
Because this is the summation of a square, it is correct to say that σ2=s2*π2/3,
but it is NOT correct to say that because of this s=σ√3/π. This is only true for a flat surface.
If the surface is hyperbolic, then the correct formula is
s=cosh-1(cosh(σ√3/π))
Because cosh-1(x)=ln(x±√(x2-1)),
cosh2(x)-sinh2(x)=1, cosh(x)=½*(ex+e-x),
and sinh(x)= ½*(ex-e-x), this can be expressed as
s=ln(cosh(σ√3/π) ±√( cosh2(σ√3/π)-1))
s=ln(cosh(σ√3/π) ± sinh(σ√3/π))
s=ln(½*(eσ√3/π +e-σ√3/π) ± ½*(eσ√3/π -e-σ√3/π))
s= σ√3/π ± σ√3/π= 2* σ√3/π or 0
This is because the variance, the square of the standard
deviation, is a summation, NOT a single variable.
Only if the surface is flat is it true that s=σ√3/π.
This means that the random normal logistics distribution,
the hyperbolic secant distribution,
PDF = (1/(4s))*sech2((x-µ)/2s);
CDF = ½*tanh((x-µ)/2s)+ ½
can be expressed in terms of the parameters µ and σ as
PDF = (π /(8* σ√3))*sech2(π
*(x-µ)/(2* σ√3));
CDF = ½*tanh( π *(x-µ)/(2* σ√3))+ ½.
When the standard deviation, σ, is π/(4√3)
then this simplifies to
PDF = ½*sech2(x-µ);
CDF = ½*tanh(x-µ)+ ½ which is identical to saying that s=½.
In this case, when x=µ, then the PDF is 50% and the
CDF is also 50%. If π is the absolute, and negative numbers are not
allowed, then, according to Pearson’s Second Coefficient of Skewness, µ=π/2. If s=½, then σ2=(½)2π2/3=0.822467.
If the universe is hyperbolic, as proposed by Mabkhout (Mabkhout, 2012), the absolute is
defined as π, x is an absolute measurement which means that 0 is
an absolute zero, numbers x<0 are not defined, AND the universe is random,
then if the universe is normal, it has both a mean and a variance which can be
expressed in terms of its absolute. That universe also allows choice, s=½,
where there is one choice: the absolute or absolute zero.
If the universe is hyperbolic and random, then random
events do not repeat except on an imaginary plane. Random events may appear
to repeat, be cyclical, but it is not in the real surface where the coefficient
of the imaginary axis is 0. I.e., Mark Twain
was correct. History does NOT repeat,
but it sure does rhyme.
Mabkhout, S. (2012). The infinite distance horizon
and the hyperbolic inflation in the hyperbolic universe. Phys. Essays,
25(1), p.112.
