Probability of a random variable - Wealth Model Consider a group of people. Each person has a certain wealth. At step
Let's split the group of people into two groups. The people are assigned to either group at random.
The total wealth of group 1 is $m_1$. The total wealth of group two is $m_2$. The total wealth of the overall group is $m=m_1+m_2$
We say that $P(m)=P(m_1)P(m_2)$. Why is this so?
This question is based on an excerpt from a paper (https://arxiv.org/pdf/cond-mat/0001432.pdf
)

 A: Instead of the probability density $P(m)$ I would probably rather look at $\mathbb P\{A\ge m\}$ which is the probability that agent $A$ has at least $m$ amounts of money. For reals $n\le m$ it is clear that $\{A\ge m\}\subset\{A\ge n\}$ and therefore, we have the conditional probability
$$\tag{1}
\mathbb P\{A\ge m\,|A\ge n\}=\frac{\mathbb P\{A\ge m\}}{\mathbb P\{A\ge n\}}\,.
$$
In mathematics, the Boltzmann-Gibbs distribution is called exponential distribution.
It is the one that leads to
$$
\mathbb P\{A\ge m\}=e^{-m/T}
$$
and
$$\tag{2}
\mathbb P\{A\ge m\,|A\ge n\}=\frac{\mathbb P\{A\ge m\}}{\mathbb P\{A\ge n\}}=
\mathbb P\{A\ge m-n\}=e^{-(m-n)/T}\,.
$$
This expression reflects the fact that money is conserved and additive. In other words, when we know that $A$ has at least $n$ then the probability to have at least $m$ is equal to the probability that unconditionally he has at least $m-n\,.$
Things to consider

*

*The Boltzmann-Gibbs distribution maximises the entropy (see link) which is why it is called equilibrium distribution.


*The constant $C=1/T$ corresponds to the intensity $\lambda$ in the exponential distribution.


*The expression $P(m)=P(m_1)P(m_2)$ should not be confused with the independence of the events that agent $1$ has $m_1$ and agent $2$ has $m_2\,.$
