​The maximum entropy constraints are as follows:

  • Over the interval [0,infinity]
  •  sum{kappa=0}{N}{P(x_i)}=1       …. sum over all probabities must = 1
  •  sum{kappa=0}{N}{P(x_i){x_i}}=mu     …. given an average value AKA "mean"

The langrangian is formed as follows:

 L=sum{kappa=0}{N}{{-P(x_i)}{log_2 P(x_i)}}+lambda_0(1-sum{kappa=1}{N}{P(x_i)} )+lambda_1(mu-sum{kappa=1}{N}{{P(x_i)}{x_i}})  

 {partial L} / {partial P_i}= {-log_2 P(x_i)}-1-lambda_0-lambda_1{x_i}=0    ….setting equal to zero to find the extrema point

 {log_2 P(x_i)}=-1-lambda_0-lambda_1{x_i}

Allowing  {lambda_1} to take up the slack to turn base 2 log into natural log:

 {P(x_i)}=e^{-1-lambda_0} e^{-lambda_1{x_i}}

Using the sum of probabilities =1 criteria

 sum{kappa=0}{N}{P(x_i)}=e^{-1-lambda_0} {1/{1-e^{-lambda_1}}}=1  

 sum{kappa=0}{N}{x_i}{P(x_i)}={e^{-lambda_1{x_i}}/(1-e^{-lambda_1{x_i}})^2}={mu}    ( See below for derivation)

 

Derivation of mean value infinite sum:

                 sum{kappa=0}{N}{x_i}{e^{-lambda_1{x_i}}}=0+1*e^{-lambda_1}+2*e^{-2{lambda_1}}+3*e^{-3{lambda_1}}cdots  

           e^{-lambda_1}{sum{kappa=0}{N}{x_i}{e^{-lambda_1{x_i}}}=..................1*e^{-2{lambda_1}}+2*e^{-3{lambda_1}}+3*e^{-4{lambda_1}}}cdots  

Subtracting we get the same old geometric series that we all know

(1-e^{-lambda_1{x_i}}){sum{kappa=0}{N}{x_i}{e^{-lambda_1{x_i}}}}=0+1*e^{-lambda_1}+1*e^{-2{lambda_1}}+1*e^{-3{lambda_1}}cdots

Rearranging terms:

(1-e^{-lambda_1{x_i}}){sum{kappa=0}{N}{x_i}{e^{-lambda_1{x_i}}}}={e^{-lambda_1{x_i}}/(1-e^{-lambda_1{x_i}})}

{sum{kappa=0}{N}{x_i}{e^{-lambda_1{x_i}}}}={e^{-lambda_1{x_i}}/(1-e^{-lambda_1{x_i}})^2}

Another way of looking at the series:

Infinite Series Multiplication Table – The product of the 2 series is the sum of all the product entries ad infinitum
e^{-3{lambda_1{x_i}}} e^{-3{lambda_1{x_i}}}
e^{-2{lambda_1{x_i}}} e^{-2{lambda_1{x_i}}} e^{-3{lambda_1{x_i}}}
e^{-lambda_1{x_i}} e^{-lambda_1{x_i}} e^{-2{lambda_1{x_i}}} e^{-3{lambda_1{x_i}}}
1 1 e^{-lambda_1{x_i}} e^{-2{lambda_1{x_i}}} e^{-3{lambda_1{x_i}}}
  1 e^{-lambda_1{x_i}} e^{-2{lambda_1{x_i}}} e^{-3{lambda_1{x_i}}}

The table uses 2 exponential series each starting with 1.  In order to get the same series as the solution in the derivation above multiple the result by e^{-lambda_1{x_i}}

It forms a sort of number wedge or number cone. I wonder if it extends to 3 dimensions?

Observations  ( Need to complete this ) 

  • ….delay like Z transform
  • continuous form correspondence with discrete form

Research Links


2 Comments

Prof Von NoStrand MonsterID Icon Prof Von NoStrand · June 29, 2014 at 2:11 pm

I wanna see graphs and pictures and stuff!

Freemon SandleWould MonsterID Icon Freemon SandleWould · June 29, 2014 at 2:13 pm

You want math porn!

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