- #1
Nico045
- 9
- 0
Hello, I have to find the density of probability which gives the maximum of the entropy with the following constraint\bar{x} = \int x\rho(x)dx
\int \rho(x) dx = 1
the entropy is : S = -\int \rho(x) ln(\rho(x)) dx
L = -\int \rho(x) ln(\rho(x)) dx - \lambda_1 ( \int \rho(x) dx -1 ) - \lambda_2 (\int x \rho(x)dx -\bar{x})
\frac {\partial L } { \partial \rho(x) } = \int ( - ln(\rho(x)) -1 - \lambda_1 - x \lambda_2 ) dx= 0
\rho(x) = e^{-(1 + \lambda_1 + x \lambda_2)}
Now I use the normalisation
\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}
\rho(x) = \frac{e^{-x \lambda_2}}{\int e^{-x\lambda_2} dx}
From there I don't really know what to do. What shall I do to get a better expression of this ?
\int \rho(x) dx = 1
the entropy is : S = -\int \rho(x) ln(\rho(x)) dx
L = -\int \rho(x) ln(\rho(x)) dx - \lambda_1 ( \int \rho(x) dx -1 ) - \lambda_2 (\int x \rho(x)dx -\bar{x})
\frac {\partial L } { \partial \rho(x) } = \int ( - ln(\rho(x)) -1 - \lambda_1 - x \lambda_2 ) dx= 0
\rho(x) = e^{-(1 + \lambda_1 + x \lambda_2)}
Now I use the normalisation
\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}
\rho(x) = \frac{e^{-x \lambda_2}}{\int e^{-x\lambda_2} dx}
From there I don't really know what to do. What shall I do to get a better expression of this ?
