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p(\boldsymbol{\sigma})

is the product of the single-variable distributions, \...

May 5, 2024

Solution by Steps

step 1

To prove that maximization over

p(\boldsymbol{\sigma})

yields

\lambda_i = h_i

, we will use the method of Lagrange multipliers

step 2

We start by setting the derivative of the entropy

S[p]

with respect to

p(\boldsymbol{\sigma})

equal to zero, which is the condition for a maximum

step 3

Taking the derivative of

S[p]

with respect to

p(\boldsymbol{\sigma})

gives us:

-\log p(\boldsymbol{\sigma}) - 1 + \lambda + \sum_{i=1}^{\mathrm{L}} \lambda_i \sigma_i

step 4

Setting this derivative equal to zero and solving for

p(\boldsymbol{\sigma})

yields:

p(\boldsymbol{\sigma}) = e^{\lambda - 1} e^{\sum_{i=1}^{\mathrm{L}} \lambda_i \sigma_i}

step 5

Since

p(\boldsymbol{\sigma})

is a product of single-variable distributions

p_i(\sigma_i)

, we can write:

p(\boldsymbol{\sigma}) = \prod_{i=1}^{\mathrm{L}} p_i(\sigma_i)

step 6

Substituting the expression for

p_i(\sigma_i)

from the given information, we have:

p(\boldsymbol{\sigma}) = \prod_{i=1}^{\mathrm{L}} \frac{e^{h_i(\sigma_i)}}{\sum_{\sigma=0}^{q-1} e^{h_i(\sigma)}}

step 7

Comparing the expressions from step 4 and step 6, we see that for the products to be equal, the exponents must match, which implies

\lambda_i = h_i

Answer

Maximization over

p(\boldsymbol{\sigma})

indeed produces the result

\lambda_i = h_i

Key Concept

Lagrange multipliers and entropy maximization

Explanation

The method of Lagrange multipliers is used to find the maximum of a function subject to constraints. In this case, the maximization of the entropy function subject to normalization and expected value constraints leads to the result that the Lagrange multipliers

\lambda_i

are equal to the functions

h_i