Evolution in a nutshell

an alternative outline on evoution

and some consequences concerning valuations

by

Gregor Kjellström

            References          

Theorem 6.2.4. A normal distribution has the highest average information  as compared to other distributions having the same moment matrix M.

Proof: Because it is required that ò x² N(x) dx = s²  and that  ò N(x) dx = 1, we introduce the Lagrangian function

     F(N(x), x) = ò { -N(x) log[N(x) ] + a x² N(x)  + b N(x) } dx =   

          extremum,

with a and b as Lagrange multipliers. This is a problem within the calculus of variations, and we could use the Euler-Lagrange differential equation. But in this case F is independent of  ¶N(x)/¶x and it is therefore sufficient to differentiate with respect to N  and put the differential equal to zero. We have

     dF = (¶F/¶N) dN  = ò { -log[N(x) ] – 1 + a x² + b } dN  dx = 0.

Now assuming that the constants a and b have the values

     a = - 1/(2s²)   and

     b = log[ (2p)^-1/2s ^-1 ] + 1.

Then we have

     dF = ò{-log[N(x)]–1+x²/(2s²)+log[(2p)^-1/2s ^-1]+1}dN dx = 0.

In order to guarantee that the integral will always vanish for all x and all variations dN  we must have

     log[N(x) ] = log[ (2p)^-1/2s ^-1 ] - x²/(2s²),

and thus  N(x)  must be normal, i. e.

     N(x) = (2p)^-1/2s ^-1 exp(- x²/2s² ).

The reason for showing an interest in theorem 6.2.4 is that in a technical application the moment matrix, M, is estimated and therefore it is of interest to find the distribution having the greatest average information assuming M given. But in the natural case this argument is not equally plausible.