# Antilog of Shannon entropy (exponential entropy)

In the source article at the bottom, the antilog of entropy is introduced as the exponential of Shannon entropy:

$$H(N) = -\sum_N w_i \ln w_i$$

$$e^{H(N)} = \prod_N w_i ^{w_i}$$

1. Could someone show the steps in between $$e^{H(N)}$$ and how to get to $$\prod_N w_i ^{w_i}$$?
2. and can the last measure also be applied to point-wise probabilities of a random variable like how Shannon entropy normally is used for, i.e. $$p(x)$$ instead of $$w_i$$ so that, for random variable $$X$$,

$$e^{H(X)} = \prod_N p(x) ^{p(x)}?$$

Straathof, S. M. (2007). Shannon's entropy as an index of product variety. Economics Letters, 94(2), 297-303.

• By the laws of indices/logarithms, $$\exp H(N)= \exp\Big(-\sum_{N} w_i\log w_i\Big) = \prod_{N} \exp(-w_i\log w_i) = \prod_{N} \exp(\log(w_i^{-w_i})) = \prod_N {w_i}^{-w_i},$$ since $\exp$ and $\log$ are inverses of each other. Commented Nov 10, 2020 at 22:36
• thanks, could you change the comment to an answer Commented Nov 10, 2020 at 22:37
• @deverlarist I did it as a comment cause I didn't really answer your second point. Commented Nov 10, 2020 at 22:40

By the laws of indices/logarithms, $$\exp H(N)= \exp\Big(-\sum_{N} w_i\log w_i\Big) = \prod_{N} \exp(-w_i\log w_i) = \prod_{N} \exp(\log(w_i^{-w_i})) = \prod_N {w_i}^{-w_i},$$ since $$\exp$$ and $$\log$$ are inverses of each other.
• can the laws of indices/logarithms also be used to produce the same result if Shannon entropy were replaced with differential entropy: $$h(X) = \int f(x) \log f(x) dx ?$$ Commented Nov 10, 2020 at 23:15
• @develarist Not easily, no. There is no equivalent to $\exp\big(\sum{}\cdot{}\big) = \prod\exp({}\cdot{})$ for integrals. Commented Nov 10, 2020 at 23:16