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Consider the discrete case:

Shannon's entropy is $H(x)=-\sum\limits_i^n p(x_i) log p(x_i)$.

Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$.

What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.

Consider the discrete case:

Shannon's entropy is $H(x)=-\sum\limits_i^n p(x_i) log\space p(x_i)$.

Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$.

What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.

Consider the discrete case:

Shannon's entropy is $H(x)=\sum\limits_i^n p(x_i) x_i$.

Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$.

What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.

Consider the discrete case:

Shannon's entropy is $H(x)=-\sum\limits_i^n p(x_i) log p(x_i)$.

Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$.

What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.

Just study the discrete case:

Shannon's entropy is $H(x)=\sum_i^n p(x_i) x_i$

Probability based on Prefix-free Kolmogorov's complexity maybe is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov's complexity of $x_i$

Now what is the relation between $R(x_i)$ and $p(x_i)$, Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.

Consider the discrete case:

Shannon's entropy is $H(x)=\sum\limits_i^n p(x_i) x_i$.

Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$.

What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal?

I remember vaguely that a book has discuss on the relation without any precise result.