I have been reading "Entropy Optimization Principles with Applications" by Kapur and Kesavan. In the book they derive some distribution of statistical mechanics and then show how they can be applied to everyday scenarios not involving particles and energy.

Below is a quick over-view of their derivation of BE distribution, an application relating to marketing, and my question.

The question is not so much of statistical mechanics although one of its distributions is used, but in when to know to use BE versus other structures of constraints in relation to what information may be known (marketing, urban planning, etc.)

3.2.2 The Bose-Einstein (BE) Distribution

In deriving the Maxwell-Boltzmann distribution, we considered a system with $n$ possible states with energies $\epsilon_1,\epsilon_2, \dots, \epsilon_n$. There we assumed only one moment constraint arising from the knowledge of the mean energy of the particles, $\hat{\epsilon}$. In the present case, we also assume a second moment constraint, in addition to the first arising from a knowledge of the expected number of particles in the system, N (not the actual number of particles that must thus be assumed to vary between zero and infinity). On the basis of these two constraints, we proceed to apply MaxEnt for deriving the distribution

Let $p_{ij}$ be the probability of there being $j$ particles in state $i$ of the $n$ states. Since it is certain that the number of particles in any one state will be between zero and infinity, the normalizing constraints are given by the following equations

$\sum_{j=0}^\infty p_{ij} = 1$

The expected number of particles and the expected energy, which are the prescribed moments, are given by the following equations

$\sum_{i=1}^n \sum_{j=0}^\infty j p_{ij} = A$

$\sum_{i=1}^n \epsilon_i \sum_{j=0}^\infty j p_{ij} = B$

[...] the total entropy is

$- \sum_{i=1}^n \sum_{j=0}^\infty p_{ij} \ln(p_{ij}) $

[...] (solving for maximum entropy) yields

$p_{ij} = (1 - e^{-\lambda - \mu \epsilon_i}) e^{-j (\lambda + \mu \epsilon_i)}$

The expected number of particles in the $i$th state, $\bar{N_i}$, is given by

$ \bar{N_i} = \sum_{j=0}^\infty j p_{ij} = 1 / \left( e^{-\lambda + \mu \epsilon_i} -1 \right) $

The constraints can then be rephrased as (*)

$\sum_{i=1}^n \bar{N_i} = A$

$\sum_{i=1}^n \epsilon_i \bar{N_i} = B$

They then give a great example illustrating the distribution with product purchases

Example 3.2: There are three products whose costs are \$50, \$7.5, and \$1, respectively. If the number of the products required per month is 275 and the total cost average is $800, find the average number of each product required per month.

Solution 3.2: This is a problem on Bose-Einstein distribution. Here we are given

$\epsilon_1=50.0, \epsilon_2=7.5, \epsilon_3 = 1.0, A=275, B=800$

with solution





Before I read that section I would have solved the similar problem differently. Since we know 275 products were purchased, we can solve for the proportion of how many were in each category.

$n_1 + n_2 + n_3 = 275$

$\epsilon_1 n_1 + \epsilon_2 n_2 + \epsilon_3 n_3 = 800$

(Which is analogous to (*) above). Dividing by 275 yields

$p_1 + p_2 + p_3 = 1$

$\epsilon_1 p_1 + \epsilon_2 p_2 + \epsilon_3 p_3 = (800/275)$

Which when trying maximize the entropy

$- \sum_{i=1}^n p_{i} \ln(p_{i}) $

has a solution of

$\hat{p} = \{.0008,.2879,.7114\}$ and multiplying by 275 to get the estimated number




The answers are dramatically different. In what situations is BE distribution applicable, and which is the solving the proportions directly applicable? They yield different answers, so the assumptions are different - but as now I don't see a reason why I couldn't have just solved the proportions problem directly.

  • $\begingroup$ Are your customers bosons? If they are then you can use Bose-Einstein distribution :) $\endgroup$
    – Aksakal
    Feb 2, 2016 at 14:38
  • $\begingroup$ I think as many customers can purchase an individual product ('energy level') the customers are bosons to a good approximation. :) Later on they do develop an intermediate-statistics distribution assuming a maximum of m_i 'particles' per energy level i. They relate that distribution to wards in a hospital with different (finite) number of beds per ward. $\endgroup$
    – sheppa28
    Feb 2, 2016 at 14:56
  • 1
    $\begingroup$ Your "own" solution is but one of an infinite number of possible solutions of 2 linear equations in 3 unknowns plus nonnegativity constraints.. $\endgroup$ Feb 2, 2016 at 14:57
  • $\begingroup$ Ah yes, edited for clarity. The solution came from solving the maximum entropy criteria. $\endgroup$
    – sheppa28
    Feb 2, 2016 at 15:00

1 Answer 1


The Bose-Einstein distribution is used when you are trying to describe a distribution of discrete states that can be populated without restriction.

In your above example, each product and it's cost would be analogous to an energy state and it's energy value. This is assuming that you have a number of available products much greater than the total products sold. The total number of products, 275, would be the number of bosonic particles and your average cost would be the average energy of the system.

The requirements to use this distribution are:

  1. There are no restrictions as to how many individual particles can populate any given state. Thus, if you only had 50 of product A, this would not work as you could not populate product A with 275 purchases (which must be allowed). For situations where these restriction crop up, you might be interested in reading about Fermi-Dirac statistics.

  2. The states must be non-interacting. For example, if you were selling bananas, IPhones, and Turpentine, there is probably little correlation between the products (although you have obviously made some questionable business decision, but this is neither here nor there). On the other hand, if you sell IPhones, IPhone cases and bananas, it is likely that people who buy an Iphone will also buy a case. Thus, these events are correlated.

  • 1
    $\begingroup$ So, the customers are not bosons, but fermions. $\endgroup$
    – Aksakal
    Feb 2, 2016 at 18:31
  • $\begingroup$ The customer is nothing in this analogy. The purchase of a product is the "boson". $\endgroup$ Feb 2, 2016 at 18:31

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