Being almost obsessively parsimonious, I once joked that if everyone’s spending habits were like mine the economy would collapse. Mathematician that I am, I wondered if that was actually true and whether it could be predicted mathematically. Now you can enjoy the result.
For simplicity, consider an economy with two kinds of people who differ in their wealth $w_i$, incomes, $N_i$, and expenses $E_i$ (the subscript $i$ being 1 or 2 to indicate one group or the other). The rate of change of each group’s wealth is its income minus expenses:
$$\displaystyle \frac{dw_i}{dt} = N_i – E_i$$
Assuming that the total wealth stays the same, the income of one group comes from the expenses of both groups.
$$\displaystyle N_i = A_{ij}E_j $$
where $A_{ij}$ is a matrix of the fractions (between 0 and 1) of each groups expenditures end up going to each group. For example, if group 1’s expenses go to other parts of group 1 60% of the time, then the other 40% of the time they go to group 2. If group 2 spends 30% of its expenses within itself with the other 70% going to group 1, the matrix $A_{ij}$ would be:
$$\displaystyle A_{ij} = \left[\begin{matrix} 0.6 & 0.7 \\ 0.4 & 0.3 \end{matrix}\right]$$
Note that the numbers in each column add up to 1, since all expenses must go to one group or the other.
To model expenditure, we’ll say that the total expenditure $E_i$ consists of a constant cost of living $L_i$ plus discretionary spending that depends on the profligacy $p_i \ge 0$ and total wealth $w_i$. A group with a smaller $p_i$ spends less of its disposable wealth per unit time than one with a larger $p_i$. For simplicity we’ll assume that $p_i$ does not depend on time or wealth.
$$\displaystyle E_i = L_i + p_i w_i $$
Now we can assemble the whole equation for the wealth rate of change:
$$\displaystyle \frac{dw_i}{dt} = A_{ij}E_j – E_i$$
Using matrix-vector notation instead of subscripts:
$$\displaystyle \frac{d\vec{w}}{dt} = (A-I)(\vec{L} + P\vec{w}) $$
where an arrow above a symbol indicates a vector and $P$ is a diagonal matrix of profligacies. This is a first-order, constant-coefficient system of ordinary differential equations.
Part 1: Constant Solutions
The first items of interest are the constant solutions. If there are any, they can be found by setting the time derivative to zero and solving the system for $\vec{w}$:
$$\displaystyle (A-I)P\vec{w} = -(A-I)\vec{L} $$
There is a small snag here because $A-I$ is not invertible. As noted before, the columns of $A$ add to 1, so:
$$\displaystyle \begin{bmatrix}1 & 1\end{bmatrix}(A-I) = 1 – 1 = 0$$
From which it follows that $A$ has a zero eigenvalue, hence it is not invertible. However, since the total wealth $w_1 + w_2 = W$ is constant, we can use that fact with one row of the system to get a system that can be solved for the steady-state $w_1$ and $w_2$:
$$\displaystyle \begin{bmatrix}(a_{11}-1)P_1 & a_{12}P_2 \\ 1 & 1\end{bmatrix}\begin{bmatrix}w_1 \\ w_2\end{bmatrix} = \begin{bmatrix}-(a_{11}-1)L_1 – a_{12}L_2 \\ W\end{bmatrix} $$
Using that $a_{11}-1 = -a_{21}$, the solution $\vec{w}^*$ is given by:
$$\displaystyle \begin{bmatrix}w_1^* \\ w_2^* \end{bmatrix} = \frac{1}{p_1a_{21} + p_2a_{12}}\begin{bmatrix}-a_{21}L_1 + a_{12}L_2 + a_{12}p_2W \\ a_{21}L_1 – a_{12}L_2 + a_{21}p_1W \end{bmatrix}$$
Using the $A$ matrix example from earlier with costs of living $L_1 = L_2 = 0.1$ using profligacies $p_1 = 0.2, p_2 = 0.1$ and total wealth $W=1$, we get constant solutions $w_1^* = 2/3, w_2^* = 1/3$. Even though group 1 is more profligate, at equilibrium it ends up with more of the wealth than group 2 because it spends most (60%) of its money within itself and because group 2 gives it most (70%) of its money. So our model can account for the interplay between thriftiness and financial insularity.
Part 2: Transient Solutions
The next question is whether these constant solutions are approached by the non-constant solutions. For that we must reconsider the differential equation:
$$\displaystyle \frac{d\vec{w}}{dt} = (A-I)(\vec{L} + P\vec{w}) $$
which we will rewrite for convenience as:
$$\displaystyle \frac{d\vec{w}}{dt} = \vec{v} + M\vec{w} $$
so $\vec{v} = (A-I)\vec{L}$ and $M = (A-I)P$. The integrating factor technique gives the solution to this equation as:
$$\displaystyle \vec{w}(t) = e^{Mt}\vec{w}(0) + e^{Mt}\left( \int_0^t e^{-M\tau}\,d\tau \right)\vec{v} $$
So we will have everything we need from the matrix exponential $e^{Mt}$, which can be found from the eigenvalues and eigenvectors of $M$ if $M$ is diagonalizable. Letting $r=(a_{11}-1)p_1$ and $s = (a_{22}-1)p_2$, $M$ has the form:
$$\displaystyle M = \begin{bmatrix} r & -s \\ -r & s \end{bmatrix} $$
One eigenvector is $\vec{q}_1 = [s\,\,r]^T$ which has eigenvalue $\lambda_1 = 0$. Since the sum of the eigenvectors is the trace of $M$, the other eigenvalue must be $\lambda_2 = r+s$. It has the eigenvector $\vec{q}_2 = [1\,\,-1]^T$. $M$ will be diagonalizable as long as $r+s \neq 0$, which is the case whenever $a_{11} < 1$ or $a_{22}<1$. Since $a_{11}$ and $a_{22}$ are always $\le 1$ anyway, we need to avoid them both being equal to 1. That case corresponds to one in which money is not spent between groups, which is not very interesting since the solution is just the initial wealths remaining constant. So we will ignore that case and proceed assuming $M$ is diagonalizable. Note that in that case $r+s < 0$. Also, since both eigenvalues are real we know the solution will not be oscillatory.
Knowing the eigenvalues and eigenvectors, we can write $M$, $e^{Mt}$, and the integral of $e^{Mt}$ in terms of $Q = [\vec{q}_1 \,\, \vec{q}_2]$ as:
$$\displaystyle M = Q\begin{bmatrix}0 & 0 \\ 0 & r+s \end{bmatrix}Q^{-1} $$
$$\displaystyle e^{Mt} = Q\begin{bmatrix}1 & 0 \\ 0 & e^{(r+s)t} \end{bmatrix}Q^{-1} $$
$$\displaystyle \int_0^t e^{-M\tau}\, d\tau = Q\begin{bmatrix} t & 0 \\ 0 & \frac{1 – e^{-(r+s)t}}{r+s}\end{bmatrix} Q^{-1}$$
So the solution $\vec{w}(t)$ is:
$$\displaystyle \vec{w}(t) = Q\begin{bmatrix}1 & 0 \\ 0 & e^{(r+s)t} \end{bmatrix}Q^{-1}\vec{w}(0) + Q\begin{bmatrix}1 & 0 \\ 0 & e^{(r+s)t} \end{bmatrix}\begin{bmatrix} t & 0 \\ 0 & \frac{1 – e^{-(r+s)t}}{r+s}\end{bmatrix} Q^{-1}\vec{v}$$
$$\displaystyle \vec{w}(t) = Q\begin{bmatrix}1 & 0 \\ 0 & e^{(r+s)t} \end{bmatrix}Q^{-1}\vec{w}(0) + Q\begin{bmatrix} t & 0 \\ 0 & \frac{e^{(r+s)t}-1}{r+s}\end{bmatrix} Q^{-1}\vec{v}$$
We know the eigenvector matrix and can find its inverse:
$$\displaystyle Q = \begin{bmatrix}s & 1 \\ r & -1\end{bmatrix} $$
$$\displaystyle Q^{-1} = \frac{1}{r+s}\begin{bmatrix}1 & 1 \\ r & -s \end{bmatrix} $$
With some work, we end up with the solution as:
$$\displaystyle \begin{aligned}\vec{w}(t) &= \frac{1}{r+s}\begin{bmatrix} s + re^{(r+s)t} & s – se^{(r+s)t} \\ r-re^{(r+s)t} & r + se^{(r+s)t} \end{bmatrix}\vec{w}(0) \\ &+ \frac{1}{r+s}\begin{bmatrix} st + \frac{r}{r+s}\left( e^{(r+s)t}-1 \right) & st – \frac{s}{r+s}\left( e^{(r+s)t}-1 \right) \\ rt – \frac{r}{r+s}\left( e^{(r+s)t}-1 \right) & rt + \frac{s}{r+s}\left( e^{(r+s)t}-1 \right) \end{bmatrix}\vec{v} \end{aligned}$$
The second matrix has factors $st$ and $rt$ in its entries, which would grow unboundedly as time increases. This does not make sense for this problem, because the total amount of wealth should stay constant and no group’s wealth should become negative. However, the vector $v$ that that matrix multiplies has a structure that will result in the cancellation of these factors:
$$\displaystyle \vec{v} = (A-I)\vec{L} = \begin{bmatrix} a_{11}-1 & a_{12} \\ a_{21} & a_{22}-1 \end{bmatrix}\vec{L} = \begin{bmatrix} -a_{21} & a_{12} \\ a_{21} & -a_{12} \end{bmatrix}\vec{L} $$
Each element is the negative of the other element in the column. Therefore:
$$\displaystyle \begin{aligned}\vec{w}(t) &= \frac{1}{r+s}\begin{bmatrix} s + re^{(r+s)t} & s – se^{(r+s)t} \\ r-re^{(r+s)t} & r + se^{(r+s)t} \end{bmatrix}\vec{w}(0) \\ &+ \frac{1}{r+s}\begin{bmatrix} -a_{21} & a_{12} \\ a_{21} & -a_{12} \end{bmatrix}(e^{(r+s)t} – 1)\vec{L} \end{aligned} $$
Making some simplifications:
$$\displaystyle \vec{w}(t) = \frac{1}{r+s}\left( \begin{bmatrix} s & s \\ r & r \end{bmatrix} + e^{(r+s)t}(A-I)P\right)\vec{w}(0) + \frac{e^{(r+s)t} – 1}{r+s}(A-I)\vec{L}$$
Since $r+s < 0$ all the exponentials are decaying so as $t\to\infty$ we get
$$\displaystyle \lim_{t\to\infty} \vec{w}(t) = \frac{W}{r+s}\begin{bmatrix}s \\ r \end{bmatrix} + \frac{1}{r+s}(I-A)\vec{L} $$
Where $W$ is the total wealth at time 0. This result matches the constant solutions from Part 1. Putting this result in terms of the variables laid out at the beginning:
$$\displaystyle \lim_{t\to\infty} \frac{\vec{w}(t)}{W} = \frac{1}{a_{12}p_2 + a_{21}p_1} \left( \begin{bmatrix} a_{12}p_2 \\ a_{21}p_1 \end{bmatrix} + \begin{bmatrix} -a_{21}p_1 & a_{12}p_2 \\ a_{21}p_1 & -a_{12}p_2 \end{bmatrix} \begin{bmatrix}L_1/(Wp_1) \\ L_2/(Wp_2) \end{bmatrix}\right)$$
Part 3: Wrapping Up
As it stands, our model allows negative wealth which does not make sense. We’re also interested in the conditions under which one group approaches having all the wealth. We can answer both questions by examining $w_1(t)$. As $t\to\infty$:
$$\displaystyle \lim_{t\to\infty} w_1(t) = \frac{a_{12}p_2}{a_{12}p_2 + a_{21}p_1} \left( 1 + \frac{L_1}{Wp_1} + \frac{L_2}{Wp_2} \right) – \frac{L_1}{Wp_1} $$
To be realistic this limit must be between 0 (group 1 approaches zero wealth) and 1 (group 1 approaches total wealth). This condition is satisfied when:
$$\displaystyle \frac{\frac{L_1}{Wp_1}}{1 + \frac{L_1}{Wp_1} + \frac{L_2}{Wp_2}} < \frac{a_{12}p_2}{a_{12}p_2 + a_{21}p_1} < \frac{1 + \frac{L_1}{Wp_1}}{1 + \frac{L_1}{Wp_1} + \frac{L_2}{Wp_2}} $$
The middle quantity is in $[0,1]$ anyway, but the realistic range for it is narrower. We also see that group 1’s strategy to maximize wealth is to make that quantity as large as possible, which can be done by reducing $a_{21}$ (the fraction of expenditures they give to group 2) or reducing its own profligacy $p_1$. These are perfectly commonsensical results, so it’s both good and unsurprising that the model recovers them. The advantage to doing all this work is that you could quantify their size if you knew the parameters that form the model.
Returning to the first paragraph, what would happen to a group with zero profligacy? It would violate the realism condition outlined above, but leaving that aside it would result in that group eventually acquiring all the wealth. In reality the model we used here would break down before that point is reached, but I find it intriguing.