Population neutrality
What is population neutrality?
An inequality measure satisfies population neutrality if it does not register a change in inequality when the population is scaled. More precisely, if a society is cloned a certain number of times, and these identical societies are combined to form one large society, inequality in the composite society is the same as inequality in the component societies.
Exercise
Is the rich/poor ratio an inequality measure that satisfies population neutrality?
Yes, the rich/poor ratio is an example of an inequality measure that satisfies population neutrality. If a society is cloned once, the population size will double but the average income of all deciles will remain the same. Thus, the ratio of the tenth decile to the first decile will also remain the same.
We can combine population neutrality, income neutrality, and the transfer principle to rank societies with different populations.
Exercise
Let \(X = (24, 76)\) and \(Y = (17, 34,49)\).
In this example, each society has the same total income (100), but different populations (2 and 3).
Can we use a combination of population neutrality, income neutrality, and the transfer principle to determine which society has more income inequality?
The answer is yes. Let’s systematically apply each principle to understand why.
Population neutrality
The goal is to clone each of the original societies so that they have the same population.

Choose a common multiple of the populations of the original societies.
You can always multiply the populations together. For example,
\[2 \times 3 = 6\]This will not always yield the smallest multiple, but in this case it does. There are many possible multiples that could be used.

Determine the number of times to clone the original society to reach the chosen common multiple.
In this case we will clone \(X\) twice and clone \(Y\) once to get:
\[\begin{align} X^{\prime} &= (24, 24, 24, 76, 76, 76) \\ Y^{\prime} &= (17, 17, 34, 34, 49, 49) \end{align}\]
Income neutrality

Now scale incomes to make total income identical across the societies.
Prior to this step, both \(X^{\prime}\) and \(Y^{\prime}\) have a population of 6, but now have different total incomes (300 and 200 respectively). To use income neutrality, we can double the incomes in \(X^{\prime}\) and triple the incomes in \(Y^{\prime}\) to get:
\[\begin{align} X^{\prime\prime} &= (48, 48, 48, 152, 152, 152) \\ Y^{\prime\prime} &= (51, 51, 102, 102, 147, 147) \end{align}\]
Transfer principle

Now that both societies have the same population and the same total income, we can apply the transfer principle:
Start with \(X^{\prime\prime}\). Take 50 away from the third richest person in \(X^{\prime\prime}\) and give 44 to the fourth richest and 3 each to the two poorest to get:
\[(48 + 3, 48 + 3, 48 + 44, 152 − 50, 152, 152) \rightarrow (51, 51, 92, 102, 152, 152)\]Now take 5 each from the two richest and give ten to the third richest to get:
\[(51, 51, 92 + 10, 102, 152 − 5, 152 − 5) \rightarrow (51, 51, 102, 102, 147, 147) = Y^{\prime\prime}\]By the transfer principle, \(X^{\prime\prime}\) has more income inequality than \(Y^{\prime\prime}\). By our other two principles, \(X\) has the same inequality as \(X^{\prime\prime}\) and \(Y\) has the same inequality as \(Y^{\prime\prime}\). Hence we conclude that \(X\) has more income inequality than \(Y\).
Exercise
Let \(A = (20, 30)\) and \(B = (10, 15, 35, 40)\).
In this example, each society has a different total income (50 and 100) and different populations (2 and 4).
Can we use a combination of population neutrality, income neutrality, and the transfer principle to determine which society has more income inequality?
The answer is yes. Let’s walk through the same process we followed in the last exercise to understand why.
Population neutrality
The goal is to clone each of the original societies so that they have the same population.

Choose a common multiple of the populations of the original societies.
You can always multiply the populations together. In this case, we use the least common multiple, which is 4. There are many possible multiples that could be used.

Determine the number of times to clone the original society to reach the chosen common multiple.
In this case we will clone \(A\) once and we won’t clone \(B\) (as it already has 4 people) to get:
\[\begin{align} A^{\prime} &= (20, 20, 30, 30) \\ B^{\prime} &= (10, 15, 35, 40) \end{align}\]
Income neutrality

Now scale incomes to make total income identical across the societies.
We can see that now both \(A^{\prime}\) and \(B^{\prime}\) have a population of 4, and the same total incomes (100). Thus, we do not need to apply any additional scaling factor. We move forward to the next step.
Transfer principle

Now that both societies have the same population and the same total income, we can apply the transfer principle.
Start with \(B^{\prime}\). Take 10 away from the richest person in \(B^{\prime}\) and give 10 to the poorest to get:
\[(10 + 10, 15, 35, 40−10) \rightarrow (20, 15, 35, 30)\]Now take 5 from the second richest person (second from the right) and give 5 to the second poorest person:
\[(20, 15 + 5, 35 − 5, 30) \rightarrow (20, 20, 30, 30) = A^{\prime}\]By the transfer principle, \(B^{\prime}\) has more income inequality than \(A^{\prime}\). By our other two principles, \(B\) has the same inequality as \(B^{\prime}\) and \(A\) has the same inequality as \(A^{\prime}\). Hence we conclude that \(B\) has more income inequality than \(A\).