# Shannon Information | Understanding from a Microstate Perspective

So Shannon's information is a way to quantify "distinct knowledge" by means of combination of microstates. So say 1 bit of information in binary system conveys 2 sets of information due to two possible microstates

$$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

2 bit of information in binary conveys $$2^2=4$$ set of information due to 4 possible microstates.

$$\begin{pmatrix} 0\ 0 \\ 0\ 1 \\ 1\ 0 \\ 1\ 1 \end{pmatrix}$$

Now how to understand this concept using probabilities? If a particular outcome of a random variable say a biased coin flip has a probability of 0.3 for heads then what does it really mean when we say that it conveys $$-\log_2(0.5)=1.73$$ bits of information? How does the outcome of the coin has 1.73 microstates?

There is a difference between entropy and number of microstates when dealing with a random process that is not equally probable. In the example of a single coin flip there are only two microstates regardless of coin bias, the coin can come up heads or tail but the entropy will be different for the cases of even coins or biased coins. For an even coin the entropy can be calculated the usual way,

$$H(X) =- \sum_{i \in h,t}p_i \log_2 p_i =- (0.5 \log_2 0.5 + 0.5 \log_2 0.5) = 1\;\mathrm{bits}$$

or because each microstate is equally probable $$H(X) = \log_2 2 = 1 \;\mathrm{bits}$$

For the biased coin where heads has a probability $$p_h = 0.3$$ the entropy is,

$$H(X) = - ( 0.3 \log_2 0.3 + 0.7 \log_2 0.7) = 0.88\;\mathrm{bits}$$

The entropy for the biased case is lower because we are less uncertain about the outcome of a coin flip (our intuition tells us that tails is more likely to occur). Another simple example is if we have a random process where we take two coins and flip them then there are four possible microstates $$X =\{hh,ht,th,tt\}$$

For even coins where each microstate is equally probably the entropy $$H(X) = \log_2 4 = 2 \;\mathrm{bits}$$

and for two biased coins the entropy is $$H(X) = 1.76 \;\mathrm{bits}$$ .

Again the entropy of the biased coins is less than the equiprobable case because we know that the coins are weighted towards tails.

Entropy is really tedious to understand because it has use in chemistry, statistical mechanics, and information theory. In my opinion the best and most clear understanding of entropy is "Where We Do Stand on Maximum Entropy"[pg. 12-27] by E.T. Jaynes

• The entropy of a dice is not 6 bits, it's 2.585 bits, you need the logarithm. The probability distribution tells us about the entropy, and the probability of a microstate. If the entropy is less than the equally probable distribution entropy then a microstate must be more probable and a microstate must be less probable (probabilities must sum to 1). The logarithm in the entropy formula does not mitigate the issue of bias. To me "the heart of understanding entropy" is expanded in "Where Do We Stand On Maximum Entropy". Apr 6, 2020 at 16:39
• +1 and kudos for referencing Jaynes. @GENIVI-LEARNER the logarithmic relationship between microstate probabilities and (thermodynamic) entropy was already part of Boltzmann's formulation in the 1800s. Jaynes demonstrated the fundamental connection between Shannon entropy and the entropy that had been developed earlier in thermodynamics and statistical mechanics. See this answer and the links from it for further details.
– EdM
Apr 6, 2020 at 18:38
• @GENIVI-LEARNER Boltzmann wanted to find the most probable distribution among states, subject to fixed energy and number of molecules. Maximizing the log of (something proportional to a) probability was more tractable. See page 15 of the Jaynes paper linked in this answer for how the Stirling approximation for factorials leads to a form similar to Shannon entropy. If all microstates are not equally probable then you must generalize to Gibbs entropy. Changing the log base only introduces a constant scale factor.
– EdM
Apr 6, 2020 at 21:40
• @GENIVI-LEARNER I'm convinced that you have an understanding of what microstates are. A microstate is a unique and specific configuration of the system you are studying. For a coin it's simple, the microstates are H or T. For a box of gas with indistinguishable particles, in Boltzmann's case, it is every single possible arrangement of atoms which is N!. N! is both the permutations of atoms and each permutations is a specific microstate. Logarithm does not "measure the permutations a microstate could take" because there is only one permutation for a microstate, the microstate itself. Apr 6, 2020 at 21:56
• I'm not sure what you mean when you say "so is each distinct microstate can be considered distinct information?" but the existence of a specific microstate is important for information theory but to quantify information/entropy you need a probability distribution. Apr 6, 2020 at 22:30