Corrected exercises: matrix models (201-SN4-RE)

$P=\begin{pmatrix}0.8 & 0.3\\0.2 & 0.7\end{pmatrix}$. First column: $0.8+0.2=1$; second column: $0.3+0.7=1$. The column convention puts in column $j$ the distribution of what LEAVES state $j$.

$x_{1}=Px_{0}=\begin{pmatrix}0.8\times 1+0.3\times 0\\0.2\times 1+0.7\times 0\end{pmatrix}=\begin{pmatrix}0.8\\0.2\end{pmatrix}$. Then $x_{2}=Px_{1}=\begin{pmatrix}0.8\times 0.8+0.3\times 0.2\\0.2\times 0.8+0.7\times 0.2\end{pmatrix}=\begin{pmatrix}0.64+0.06\\0.16+0.14\end{pmatrix}=\begin{pmatrix}0.70\\0.30\end{pmatrix}$. The total stays 1 at every step, and that is the check to make.

$p_{12}=0.3$ is the entry in row 1 and column 2: it is the probability of moving FROM state 2 TO state 1, in other words the share of B that joins A. The order of the indices is the most frequent source of error in the chapter: in the column convention, $p_{ij}$ reads “from $j$ to $i$”.

Because everything that leaves a state has to arrive somewhere: the total population is conserved. If a column summed to 0.9, 10% of that state would disappear at every step, and the total would decrease geometrically. Such a matrix is no longer stochastic and no longer describes a distribution, but it remains perfectly usable to model a population that loses members, through migration or mortality.

In the ROW convention we write the transpose, $P^{T}=\begin{pmatrix}0.8 & 0.2\\0.3 & 0.7\end{pmatrix}$, whose ROWS sum to 1. The state vector then becomes a ROW vector and we multiply on the right: $x_{1}=x_{0}P^{T}$. The two conventions are equivalent, but mixing them produces results that are wrong and plausible, which is the worst case. Choose a convention at the start of the problem and write it down in black and white.

$P^{2}=\begin{pmatrix}0.8 & 0.3\\0.2 & 0.7\end{pmatrix}\begin{pmatrix}0.8 & 0.3\\0.2 & 0.7\end{pmatrix}=\begin{pmatrix}0.64+0.06 & 0.24+0.21\\0.16+0.14 & 0.06+0.49\end{pmatrix}=\begin{pmatrix}0.70 & 0.45\\0.30 & 0.55\end{pmatrix}$. The columns still sum to 1, which is true of every power of a stochastic matrix.

$P^{2}x_{0}=\begin{pmatrix}0.70 & 0.45\\0.30 & 0.55\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0.70\\0.30\end{pmatrix}$, which indeed gives back $x_{2}$. The associativity of the matrix product guarantees that $P\left(Px_{0}\right)=\left(PP\right)x_{0}$: we can move forward step by step or all at once.

The entry in row 1 and column 2 of $P^{2}$, here 0.45, is the probability of being in A TWO steps after starting from B. It splits into two paths: start from B, stay in B, move to A, that is $0.7\times 0.3=0.21$; or start from B, move to A, stay in A, that is $0.3\times 0.8=0.24$. Total: 0.45. The matrix product therefore adds up every possible path automatically, which is exactly what makes it useful.

$x_{3}=\begin{pmatrix}0.8\times 0.70+0.3\times 0.30\\0.2\times 0.70+0.7\times 0.30\end{pmatrix}=\begin{pmatrix}0.56+0.09\\0.14+0.21\end{pmatrix}=\begin{pmatrix}0.65\\0.35\end{pmatrix}$. Then $x_{4}=\begin{pmatrix}0.52+0.105\\0.13+0.245\end{pmatrix}=\begin{pmatrix}0.625\\0.375\end{pmatrix}$. The values approach 0.60 and 0.40, and the gap to the limit is halved at every step: 0.20, then 0.10, then 0.05, then 0.025.

That the limit does NOT depend on the initial state. A chain whose entries are all strictly positive is called regular, and a regular chain has a unique steady-state vector to which every initial distribution converges. This is a major result: it means that the structure of the transitions, and it alone, determines the long-term distribution. The starting composition only affects how fast the limit is approached, not where it ends.

$P-I=\begin{pmatrix}0.8-1 & 0.3\\0.2 & 0.7-1\end{pmatrix}=\begin{pmatrix}-0.2 & 0.3\\0.2 & -0.3\end{pmatrix}$. The condition becomes $\left(P-I\right)x=0$, that is, a homogeneous system of two equations in two unknowns.

The two equations are $-0.2a+0.3b=0$ and $0.2a-0.3b=0$: they are proportional, so only one remains. From $0.2a=0.3b$ we get $b=\dfrac{2}{3}a$. With $a+b=1$: $a+\dfrac{2}{3}a=1$, so $\dfrac{5}{3}a=1$ and $a=0.6$, then $b=0.4$. The steady-state vector is $\begin{pmatrix}0.6\\0.4\end{pmatrix}$.

$P\begin{pmatrix}0.6\\0.4\end{pmatrix}=\begin{pmatrix}0.8\times 0.6+0.3\times 0.4\\0.2\times 0.6+0.7\times 0.4\end{pmatrix}=\begin{pmatrix}0.48+0.12\\0.12+0.28\end{pmatrix}=\begin{pmatrix}0.60\\0.40\end{pmatrix}$. The vector is indeed unchanged.

Because the columns of $P$ sum to 1, the sum of the rows of $P-I$ is zero: the rows are linearly DEPENDENT, so the determinant is zero and the homogeneous system has infinitely many solutions, all proportional to one another. This is exactly the situation of a rank-deficient homogeneous system. The constraint $a+b=1$ picks, on this line of solutions, the one that represents a distribution; without it, the vector $\begin{pmatrix}6\\4\end{pmatrix}$ or $\begin{pmatrix}600\\400\end{pmatrix}$ would do just as well.

The initial distribution is $\begin{pmatrix}0.8\\0.2\end{pmatrix}$, but it does not matter at all: since the chain is regular, the population tends to the steady-state distribution $\begin{pmatrix}0.6\\0.4\end{pmatrix}$, that is 3,000 people in A and 2,000 in B. Say “tends to”: equality is never reached exactly, the gap being halved at every step. After six steps, only $0.2\times 0.5^{6}\approx 0.003$ of gap is left, about fifteen people.

$A=\begin{pmatrix}0.2 & 0.3\\0.4 & 0.1\end{pmatrix}$, where column $j$ gives what one unit of sector $j$ consumes. The balance equation reads $x=Ax+d$: total output goes partly to internal exchanges, $Ax$, and partly to final demand, $d$.

$x-Ax=d$, so $\left(I-A\right)x=d$ with $I-A=\begin{pmatrix}1-0.2 & -0.3\\-0.4 & 1-0.1\end{pmatrix}=\begin{pmatrix}0.8 & -0.3\\-0.4 & 0.9\end{pmatrix}$.

Determinant: $0.8\times 0.9-\left(-0.3\right)\left(-0.4\right)=0.72-0.12=0.60$. Inverse of a 2 by 2 matrix: swap the diagonal entries, change the sign of the other two, and divide by the determinant. So $\left(I-A\right)^{-1}=\dfrac{1}{0.60}\begin{pmatrix}0.9 & 0.3\\0.4 & 0.8\end{pmatrix}=\begin{pmatrix}1.5 & 0.5\\0.\overline{6} & 1.\overline{3}\end{pmatrix}$.

$x=\left(I-A\right)^{-1}d=\begin{pmatrix}1.5 & 0.5\\0.\overline{6} & 1.\overline{3}\end{pmatrix}\begin{pmatrix}100\\60\end{pmatrix}=\begin{pmatrix}150+30\\66.\overline{6}+80\end{pmatrix}=\begin{pmatrix}180\\146.\overline{6}\end{pmatrix}$. So 180 units must be produced in sector 1 and about 146.67 in sector 2 to deliver 100 and 60.

Output of sector 1 used up inside the economy: $0.2\times 180+0.3\times 146.67=36+44=80$ units; $180-80=100$ are left, the final demand. Output of sector 2 used up inside: $0.4\times 180+0.1\times 146.67=72+14.67=86.67$; $146.67-86.67=60$ are left, the final demand. The balance is exact. This check is essential: it catches both inversion errors and mix-ups between the rows and the columns of $A$.

It is the TOTAL output of sector 1 needed to deliver one unit of final demand for sector 1. It is 1.5 and not 1: one and a half units must be produced to deliver one, because part of the output is used up along the way, by sector 1 itself and indirectly by sector 2, which supplies it.

It is the output of sector 2 needed to deliver one unit of final demand for SECTOR 1. It is $0.\overline{6}$: each unit delivered by sector 1 mobilizes two thirds of a unit of output from sector 2, whereas the technology table announced only 0.4 directly. The difference, 0.267, is the INDIRECT effect: sector 2 must also feed the extra output of sector 1 that it triggered itself.

Because output is always at least what is delivered: one unit delivered costs one unit produced, plus everything consumed to produce it. Formally, $\left(I-A\right)^{-1}=I+A+A^{2}+A^{3}+\cdots$ when the series converges, which is the case as soon as the economy is productive. The term $I$ is the direct delivery, $A$ the first-round consumption, $A^{2}$ the second-round one, and so on: every entry added is positive.

Since the solution is linear, it is enough to multiply the inverse by the new demand vector: $x=\begin{pmatrix}1.5 & 0.5\\0.\overline{6} & 1.\overline{3}\end{pmatrix}\begin{pmatrix}120\\60\end{pmatrix}=\begin{pmatrix}180+30\\80+80\end{pmatrix}=\begin{pmatrix}210\\160\end{pmatrix}$. This is the whole point of computing the inverse once: any demand scenario is then handled by a single product. The extra 20 units of demand required 30 units of output in sector 1 and 13.33 in sector 2.

Multiplier of sector 1: $1.5+0.\overline{6}=2.1\overline{6}$. It means that one unit of final demand addressed to sector 1 triggers 2.17 units of output in total across the economy. This is exactly the reasoning of economic impact studies: a project that orders one million dollars from a sector with a multiplier of 2.17 generates 2.17 million dollars of output. Be aware that such studies often overstate the effect, because they assume idle capacity and constant technical coefficients.

$L=\begin{pmatrix}0 & 1.5 & 1.2\\0.6 & 0 & 0\\0 & 0.4 & 0\end{pmatrix}$. The first ROW carries the fertility rates, the SUBDIAGONAL carries the survival rates. Every other entry is zero: nobody gets younger and nobody skips a class.

$x_{1}=Lx_{0}=\begin{pmatrix}1.5\times 80+1.2\times 50\\0.6\times 100\\0.4\times 80\end{pmatrix}=\begin{pmatrix}120+60\\60\\32\end{pmatrix}=\begin{pmatrix}180\\60\\32\end{pmatrix}$, total 272. Then $x_{2}=\begin{pmatrix}1.5\times 60+1.2\times 32\\0.6\times 180\\0.4\times 60\end{pmatrix}=\begin{pmatrix}90+38.4\\108\\24\end{pmatrix}=\begin{pmatrix}128.4\\108\\24\end{pmatrix}$, total 260.4.

No, not steadily: the total population goes from 230 to 272, then falls back to 260.4. These oscillations are TRANSIENT and come from the unbalanced initial age structure: the first period produces a wave of young individuals that takes two periods to work its way through. After several periods, the structure settles and growth becomes geometric, at the rate given by the dominant eigenvalue. Demographers know this phenomenon well as the POPULATION WAVE, and it explains, for example, the echoes of the baby boom.

Test $\lambda=1.08$: $1.08^{3}=1.259712$ and $0.9\times 1.08+0.288=0.972+0.288=1.260$. The difference is $-0.000288$, so $\lambda$ is very slightly greater than 1.08. Testing $\lambda=1.081$: $1.263214$ against $1.2609$, the difference changes sign. The dominant eigenvalue is therefore about 1.0801: in the long run, the population grows by about 8.0% per period, whatever its initial structure.

The equation becomes $\lambda^{3}=0.6\lambda+0.288$. Testing $\lambda=1$: $1$ against $0.888$, so the left-hand side already wins and the root is LESS than 1. Testing $\lambda=0.95$: $0.857$ against $0.858$, nearly equal. The eigenvalue is therefore about 0.95: the population would DECLINE by about 5% per period. The replacement threshold is obtained by setting $\lambda=1$, that is $1=0.6f+0.288$ for an adult fertility $f$, so $f=\dfrac{0.712}{0.6}\approx 1.19$: below 1.19 offspring per adult per period, the population decreases.

$\det\left(P-\lambda I\right)=\left(0.8-\lambda\right)\left(0.7-\lambda\right)-0.3\times 0.2=\lambda^{2}-1.5\lambda+0.56-0.06=\lambda^{2}-1.5\lambda+0.5$. Roots: $\lambda=\dfrac{1.5\pm \sqrt{2.25-2}}{2}=\dfrac{1.5\pm 0.5}{2}$, so $\lambda_{1}=1$ and $\lambda_{2}=0.5$. Check: the trace, 1.5, is the sum of the eigenvalues, and the determinant, 0.5, their product.

The steady-state vector satisfies $Px=x=1\times x$: this is exactly the definition of an eigenvector associated with the eigenvalue 1. Computing the steady state through $\left(P-I\right)x=0$ was therefore, without saying so, a search for an eigenspace.

The second eigenvalue governs the SPEED of convergence. Every initial condition decomposes along the two eigenvectors; the component associated with $\lambda_{1}=1$ stays constant, the one associated with $\lambda_{2}=0.5$ is multiplied by 0.5 at every step and fades away. This is exactly what we observed: the gap to the limit was halved at every step, from 0.20 to 0.10 then 0.05. The smaller $\left|\lambda_{2}\right|$ is, the faster the chain settles.

Because the columns of $P$ sum to 1, so the columns of $P-I$ sum to 0, so the sum of the ROWS of $P-I$ is the zero vector: the rows are linearly dependent and $\det\left(P-I\right)=0$. Now $\det\left(P-\lambda I\right)=0$ at $\lambda=1$ means precisely that 1 is an eigenvalue. The argument holds for every stochastic matrix, of any size.

For a Leslie matrix, the dominant eigenvalue $\lambda$ is the asymptotic GROWTH RATE of the population: $\lambda>1$ for a growing population, $\lambda=1$ at the replacement threshold, $\lambda<1$ for a declining population. Its eigenvector gives the stable AGE STRUCTURE to which the population converges, whatever its initial composition. These two results, rate and structure, are the core of mathematical demography, and they are obtained exactly as in the chapter on diagonalization.

FALSE in the COLUMN convention, which is the one used in this chapter. The left index is the row, so the DESTINATION, and the right index is the column, so the ORIGIN: $p_{12}$ reads “from state 2 to state 1”. Correct statement: 30% of state 2 moves to state 1. In the row convention the reading would be reversed, which is why the convention must be stated at the start of the problem.

FALSE for a REGULAR chain, that is, one with a power whose entries are all strictly positive. Such a chain has a unique steady state, reached whatever the initial distribution, which only affects the speed of convergence. Correct statement: the long-term distribution depends only on the transition matrix. It can depend on the start if the chain has several closed classes, a case excluded here.

FALSE. Producing consumes: enough must be produced to deliver AND to feed production itself. The matrix $\left(I-A\right)^{-1}$ measures exactly this extra amount, and its diagonal entry was 1.5 here. Correct statement: to deliver 100 units of sector 1 and 60 of sector 2, 180 and 146.67 units must be produced.

FALSE. A homogeneous system ALWAYS has the zero solution, but it has others as soon as the determinant is zero. Here the columns of $P$ sum to 1, so the rows of $P-I$ add up to zero and the determinant is zero: there are infinitely many solutions, all proportional. Correct statement: the system has a line of solutions, and the constraint that the sum equals 1 picks out a single one.

FALSE. The growth of the first periods is TRANSIENT: it depends on the initial age structure, and it oscillates, here $+18.3\%$ then $-4.3\%$. The stable growth rate is given by the dominant eigenvalue, about 1.080, that is 8.0% per period. Correct statement: the population converges to geometric growth of 8% per period, after a phase of oscillations caused by its initial age structure.

$P=\begin{pmatrix}0.7 & 0.1 & 0.1\\0.2 & 0.8 & 0.1\\0.1 & 0.1 & 0.8\end{pmatrix}$. Columns: $0.7+0.2+0.1=1$; $0.1+0.8+0.1=1$; $0.1+0.1+0.8=1$. All three sum to 1.

$x_{1}=Px_{0}$ with $x_{0}=\begin{pmatrix}400\\350\\250\end{pmatrix}$. First component: $0.7\times 400+0.1\times 350+0.1\times 250=280+35+25=340$. Second: $0.2\times 400+0.8\times 350+0.1\times 250=80+280+25=385$. Third: $0.1\times 400+0.1\times 350+0.8\times 250=40+35+200=275$. Total: $340+385+275=1{,}000$, the population is conserved.

The system $\left(P-I\right)x=0$ reads $-0.3a+0.1b+0.1c=0$, $0.2a-0.2b+0.1c=0$ and $0.1a+0.1b-0.2c=0$. From the third: $c=\dfrac{a+b}{2}$. Substituting into the first: $-0.3a+0.1b+0.05\left(a+b\right)=0$, that is $-0.25a+0.15b=0$ and $b=\dfrac{5}{3}a$. Then $c=\dfrac{a+\frac{5}{3}a}{2}=\dfrac{4}{3}a$. The constraint gives $a+\dfrac{5}{3}a+\dfrac{4}{3}a=4a=1{,}000$, so $a=250$, $b\approx 416.7$ and $c\approx 333.3$.

First component: $0.7\times 250+0.1\times 416.\overline{6}+0.1\times 333.\overline{3}=175+41.\overline{6}+33.\overline{3}=250$. Second: $0.2\times 250+0.8\times 416.\overline{6}+0.1\times 333.\overline{3}=50+333.\overline{3}+33.\overline{3}=416.\overline{6}$. Third: $0.1\times 250+0.1\times 416.\overline{6}+0.8\times 333.\overline{3}=25+41.\overline{6}+266.\overline{6}=333.\overline{3}$. The vector is unchanged, and the third equation of the system serves as an independent check.

In the long run, about 250 places should be planned in program 1, 417 in program 2 and 333 in program 3, a marked redistribution compared with the current 400, 350 and 250: program 1 would lose 150 students and program 3 would gain 83. Three reservations are in order. The matrix assumes CONSTANT transition rates, which no decade guarantees. The model ignores students entering and leaving the CEGEP: it redistributes a closed population of 1,000 students, whereas recruitment and graduation change everything. Finally, convergence is slow: after three or four years the gap to the limit is still sizeable, and the CEGEP would have time to observe and adjust. The calculation therefore shows a TREND, not a facilities plan.

$I-A=\begin{pmatrix}0.75 & -0.20\\-0.30 & 0.85\end{pmatrix}$. Determinant: $0.75\times 0.85-\left(-0.20\right)\left(-0.30\right)=0.6375-0.06=0.5775$. Inverse: $\dfrac{1}{0.5775}\begin{pmatrix}0.85 & 0.20\\0.30 & 0.75\end{pmatrix}\approx \begin{pmatrix}1.4719 & 0.3463\\0.5195 & 1.2987\end{pmatrix}$.

$x=\left(I-A\right)^{-1}\begin{pmatrix}200\\0\end{pmatrix}=\begin{pmatrix}1.4719\times 200\\0.5195\times 200\end{pmatrix}\approx \begin{pmatrix}294.4\\103.9\end{pmatrix}$. So about 294.4 units of agriculture and 103.9 of manufacturing must be produced, 398.3 units in total.

The agency confuses FINAL DEMAND with OUTPUT. An order of 200 units delivered requires 294.4 units produced in the sector that receives the order, because part of the output is used up by the process itself, and it also drives 103.9 units in the other sector. Announcing 200 units of output understates the total effect by a factor of about 2, and this time the error goes towards an underestimate, which is rarer than the opposite in impact studies.

Multiplier of agriculture: $1.4719+0.5195\approx 1.9913$. Multiplier of manufacturing: $0.3463+1.2987\approx 1.6450$. AGRICULTURE is therefore the sector with the stronger knock-on effect: each unit of final demand addressed to it triggers nearly two units of output in the region, against 1.65 for manufacturing. The reason can be read in the matrix: agriculture consumes more, 0.25 and 0.30 against 0.20 and 0.15, so it draws more heavily on the productive system.

“An order of 200 units of final demand addressed to agriculture would drive about 294 units of agricultural output and 104 units of manufacturing output, nearly 398 units in total. The multiplier of agriculture is estimated at 1.99, against 1.65 for manufacturing. These estimates assume constant technical coefficients and available production capacity; they therefore hold only for moderate changes in demand and an unchanged production structure.” Three corrections: output is distinguished from final demand, the effect on the other sector is quantified, and the assumptions of the model are stated.

$\det P=0.56-0.06=0.5$, and $P^{-1}=\frac{1}{0.5}\begin{pmatrix}0.7 & -0.3\\-0.2 & 0.8\end{pmatrix}=\begin{pmatrix}1.4 & -0.6\\-0.4 & 1.6\end{pmatrix}$.

The columns do sum to $1$: $1.4-0.4=1$ and $-0.6+1.6=1$. But $P^{-1}$ has NEGATIVE entries: they are not probabilities, and $P^{-1}$ is not a transition matrix.

$P^{-1}\begin{pmatrix}0.70\\0.30\end{pmatrix}=\begin{pmatrix}0.98-0.18\\-0.28+0.48\end{pmatrix}=\begin{pmatrix}0.8\\0.2\end{pmatrix}$: this is exactly $x_{1}$ of exercise 1, whose image was $x_{2}$.

$P^{-1}\begin{pmatrix}0.9\\0.1\end{pmatrix}=\begin{pmatrix}1.26-0.06\\-0.36+0.16\end{pmatrix}=\begin{pmatrix}1.2\\-0.2\end{pmatrix}$: a negative share and a share greater than $1$. The matrix arithmetic works, but no real distribution leads to $\begin{pmatrix}0.9\\0.1\end{pmatrix}$ in one year: this state cannot be the result of one step of the chain.

If the initial distribution is $(a, 1-a)$ with $0\leq a\leq 1$, the share of A after one step is $0.8a+0.3(1-a)=0.3+0.5a$, between $0.3$ and $0.8$. No distribution outside this interval has a predecessor, hence the result of d). A Markov chain “forgets”: it brings distributions closer together, and we cannot always go back.

The first row gives the births: $1$ per young insect, $1.5$ per adult. The $0.5$ at the bottom left is the survival of the young who become adults, and the $0$ at the bottom right expresses the death of the adults after the period.

$x_{1}=\begin{pmatrix}100\\50\end{pmatrix}$, $x_{2}=\begin{pmatrix}100+75\\50\end{pmatrix}=\begin{pmatrix}175\\50\end{pmatrix}$, $x_{3}=\begin{pmatrix}175+75\\87.5\end{pmatrix}=\begin{pmatrix}250\\87.5\end{pmatrix}$. Totals: $100$, $150$, $225$, $337.5$.

$\det(L-\lambda I)=\lambda^{2}-\lambda-0.75=(\lambda-1.5)(\lambda+0.5)$: eigenvalues $1.5$ and $-0.5$.

For $1.5$: the second row gives $0.5x_{1}=1.5x_{2}$, so $x_{1}=3x_{2}$, eigenvector $(3, 1)$. In the stable structure, the young make up $\frac{3}{4}=75\%$ of the population, which is multiplied by $1.5$ in each period.

$\frac{100}{150}\approx 0.667$, $\frac{175}{225}\approx 0.778$, $\frac{250}{337.5}\approx 0.741$: the share oscillates around $0.75$ while getting closer to it, because the component along the eigenvector of $-0.5$ changes sign at each period and is halved while the total grows by a factor of $1.5$: the gap to $0.75$ is divided by $3$ at each period. Here the total, for its part, is multiplied by $1.5$ from the very first period: $x_{0}=25(3, 1)+25(1, -1)$, and the second eigenvector has components that add up to zero.