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I Maximum of entropy and Lagrange multiplier

  1. Oct 30, 2016 #1
    Hello, I have to find the density of probability which gives the maximum of the entropy with the following constraint


    [tex]\bar{x} = \int x\rho(x)dx[/tex]
    [tex]\int \rho(x) dx = 1[/tex]

    the entropy is : [tex] S = -\int \rho(x) ln(\rho(x)) dx[/tex]

    [tex]L = -\int \rho(x) ln(\rho(x)) dx - \lambda_1 ( \int \rho(x) dx -1 ) - \lambda_2 (\int x \rho(x)dx -\bar{x})[/tex]

    [tex]\frac {\partial L } { \partial \rho(x) } = \int ( - ln(\rho(x)) -1 - \lambda_1 - x \lambda_2 ) dx= 0[/tex]

    [tex]\rho(x) = e^{-(1 + \lambda_1 + x \lambda_2)}[/tex]

    Now I use the normalisation

    [tex]\int \rho(x) dx = 1 = e^{-(1 + \lambda_1) } \int e^{-x\lambda_2} dx \Rightarrow e^{-(1 + \lambda_1) } = \frac{1}{\int e^{-x\lambda_2} dx}[/tex]

    [tex]\rho(x) = \frac{e^{-x \lambda_2}}{\int e^{-x\lambda_2} dx} [/tex]

    From there I don't really know what to do. What shall I do to get a better expression of this ?
     
  2. jcsd
  3. Nov 1, 2016 #2
    Does anyone have an idea ? Maybe I can't do better
     
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