Matrix-Exponential Solutions With Constant Coefficients

Why Use a Matrix Exponential?

A multi-state reserve is a vector: its components are the reserves for the possible states at a given time. When transition intensities, interest, and cash-flow rates are constant, the coupled Thiele equations have constant coefficients. A matrix exponential then gives a compact solution that handles the state interactions and the terminal reserve condition together.

The key is to define the cash-flow signs and the matrix convention before calculating. In this tutorial, reserves are column vectors, and positive cash flow means an outgo from the insurer. A premium received by the insurer therefore enters as negative outgo. The examples use hypothetical contracts and amounts stated in thousands of dollars.

Definition: Consider a finite set of states, with transition intensity \(q_{ij}\) from state \(i\) to state \(j\), for \(i\ne j\). Let \(Q\) be the row transition-intensity matrix: \(Q_{ij}=q_{ij}\) for \(i\ne j\), and \(Q_{ii}=-\sum_{j\ne i}q_{ij}\). Let \(V(t)\) be the column vector of state reserves. If the force of interest is constant at \(\delta\), the continuous net outgo rate in state \(i\) is \(c_i\), and a lump sum \(b_{ij}\) is paid on transition \(i\to j\), define \(h_i=c_i+\sum_{j\ne i}q_{ij}b_{ij}\). Then the constant-coefficient Thiele system is \(V'(t)=A V(t)-h\), where \(A=\delta I-Q\). A payment is included only for transitions on which it is actually payable.

The vector \(h\) combines continuous outgo with the expected transition outgo per unit time. It may have negative components if premiums exceed outgo. The matrix \(A\) discounts future values and accounts for movement between states. For an absorbing state, the corresponding row of \(Q\) is zero; the state can still have its own continuous cash flow.

Deriving the Finite-Horizon Solution

Assume the state transition intensities and force of interest are constant over a finite interval \([t,T]\). Let \(g=V(T)\) be the specified terminal reserve vector. The solution is

V(t)=e^{-A(T-t)}g+\int_t^T e^{-A(s-t)}h\,ds.

Here \(e^M\) is the matrix exponential, defined by the convergent power series \(e^M=I+M+M^2/2!+\cdots\). The first term carries the terminal reserve back to time \(t\); the integral accumulates the future net outgo, weighted for interest and transitions. This formula does not require \(A\) to be invertible.

Shift to remaining time.
Set \(\tau=T-t\). The terminal condition is at the end of the interval, so the exponential uses the time remaining, not the elapsed time \(t\).

Propagate the terminal value.
The homogeneous system \(V'=AV\) has solution \(V(t)=e^{-A(T-t)}g\) when \(V(T)=g\).

Add future outgo.
Each outgo at time \(s\) contributes \(e^{-A(s-t)}h\,ds\) to the reserve at \(t\). Integrating from \(t\) to \(T\) gives the accumulated cash-flow term.

A direct justification verifies the formula. Write \(W(\tau)=V(T-\tau)\), so \(W(0)=g\) and \(W'(\tau)=-A W(\tau)+h\). Differentiating \(e^{A\tau}W(\tau)\) gives

\frac{d}{d\tau}\bigl(e^{A\tau}W(\tau)\bigr) =Ae^{A\tau}W(\tau)+e^{A\tau}(-AW(\tau)+h) =e^{A\tau}h.

Integrating from \(0\) to \(\tau\), then multiplying by \(e^{-A\tau}\), yields \(W(\tau)=e^{-A\tau}g+\int_0^\tau e^{-A(\tau-u)}h\,du\), which is the stated solution after changing variables. If \(A\) is invertible, the integral can also be written \(A^{-1}(I-e^{-A(T-t)})h\). For constant \(\delta>0\) and a finite-state generator \(Q\), \(A=\delta I-Q\) is invertible; the integral form is often simpler and remains the safer general expression.

Worked Examples

Worked Example: Symmetric Transitions and Unequal Outgo

A hypothetical contract has two states, \(A\) and \(B\), with transition intensity \(0.2\) per year in each direction. The force of interest is \(0.1\) per year. Continuous net outgo is \(2\) thousand dollars per year in state \(A\) and zero in state \(B\); there are no transition payments. The terminal reserves are zero after one year. Find the reserves at time 0.

Method. Form \(A=\delta I-Q\), then separate the common-state and contrast directions. This exploits the symmetry of the transition matrix without solving two coupled differential equations.

The generator and reserve matrix are

Q=\begin{pmatrix}-0.2&0.2\\0.2&-0.2\end{pmatrix}, \qquad A=\begin{pmatrix}0.3&-0.2\\-0.2&0.3\end{pmatrix}.

Since there are no transition payments, \(h=(2,0)^{\mathsf T}\), and \(g=(0,0)^{\mathsf T}\). The vectors \((1,1)^{\mathsf T}\) and \((1,-1)^{\mathsf T}\) are eigenvectors of \(A\), with eigenvalues \(0.1\) and \(0.5\), respectively. Decompose the outgo vector as

\begin{pmatrix}2\\0\end{pmatrix} =1\begin{pmatrix}1\\1\end{pmatrix} +1\begin{pmatrix}1\\-1\end{pmatrix}.

For an eigenvalue \(\lambda\), constant outgo over one year with zero terminal reserve contributes \(\int_0^1 e^{-\lambda s}\,ds=(1-e^{-\lambda})/\lambda\). Thus the common component is \((1-e^{-0.1})/0.1=0.951625820\), and the contrast component is \((1-e^{-0.5})/0.5=0.786938681\). Therefore, using unrounded component values,

V_A(0)=0.951625820+0.786938681=1.738564500, \qquad V_B(0)=0.951625820-0.786938681=0.164687139.

To three decimals, the reserves are \(1.739\) thousand dollars in state \(A\) and \(0.165\) thousand dollars in state \(B\). The check is that the symmetric average reserve is \(0.951625820\), matching the common component; the difference between the state reserves is twice the contrast component. A common mistake is to use the transition intensity \(0.2\) as the contrast eigenvalue. The contrast eigenvalue is \(0.1+2(0.2)=0.5\), because transitions in either direction reduce the state difference.

Worked Example: A Lump Sum on Transition to an Absorbing State

A hypothetical policyholder in active state \(A\) moves to absorbing state \(C\) at intensity \(0.2\) per year. The force of interest is \(0.1\) per year. A benefit of \(5\) thousand dollars is paid on the transition, and continuous net outgo in state \(C\) is \(2\) thousand dollars per year. There is no other outgo. Both terminal reserves are zero after one year. Find the reserve in each state at time 0.

Method. Include the expected transition payment in \(h_A\), and then apply the matrix-exponential integral. The absorbing state still has its own reserve because it has ongoing outgo.

In state \(A\), the transition-payment contribution is \(q_{AC}b_{AC}=0.2(5)=1\) thousand dollars per year. Hence \(h=(1,2)^{\mathsf T}\), and

Q=\begin{pmatrix}-0.2&0.2\\0&0\end{pmatrix}, \qquad A=\begin{pmatrix}0.3&-0.2\\0&0.1\end{pmatrix}.

The absorbing-state reserve is \(V_C(0)=2\int_0^1e^{-0.1s}\,ds=2(1-e^{-0.1})/0.1=1.903251639\) thousand dollars. For the active-state row, the relevant row of the matrix exponential is

\bigl(e^{-As}\bigr)_{A,\cdot} =\bigl(e^{-0.3s},\,e^{-0.1s}-e^{-0.3s}\bigr).

Multiplying by \(h\) gives the integrand \(e^{-0.3s}+2(e^{-0.1s}-e^{-0.3s})=2e^{-0.1s}-e^{-0.3s}\). Therefore

V_A(0) =2\frac{1-e^{-0.1}}{0.1}-\frac{1-e^{-0.3}}{0.3} =1.903251639-0.863939264 =1.039312375

thousand dollars. To three decimals, \(V_A(0)=1.039\) and \(V_C(0)=1.903\) thousand dollars. The check is that \(V_C(0)\) equals the value of one year of outgo in the absorbing state, while \(V_A(0)\) includes both the transition benefit and future outgo after entering \(C\). The sign trap is to subtract the transition benefit: it is an insurer outgo, so it contributes positively to \(h_A\).

Worked Example: Nonzero Terminal Reserves in Both States

A hypothetical two-state contract has transition intensity \(0.1\) per year in each direction and force of interest \(0.05\) per year. Continuous net outgo is \(1.5\) thousand dollars per year in state \(A\) and \(2.5\) thousand dollars per year in state \(B\). There are no transition payments. At the end of two years, the reserves are \(4\) thousand dollars in state \(A\) and \(1\) thousand dollars in state \(B\). Calculate both reserves at time 0.

Method. Decompose both the cash-flow vector and terminal vector into the common and contrast eigenvectors. This keeps the terminal-value contribution distinct from the outgo contribution.

Here \(A\) has diagonal entries \(0.15\) and off-diagonal entries \(-0.1\). Its eigenvalues are \(0.05\) in the common direction \((1,1)^{\mathsf T}\), and \(0.25\) in the contrast direction \((1,-1)^{\mathsf T}\). The decompositions are

h=\begin{pmatrix}1.5\\2.5\end{pmatrix} =2\begin{pmatrix}1\\1\end{pmatrix} -0.5\begin{pmatrix}1\\-1\end{pmatrix}, \qquad g=\begin{pmatrix}4\\1\end{pmatrix} =2.5\begin{pmatrix}1\\1\end{pmatrix} +1.5\begin{pmatrix}1\\-1\end{pmatrix}.

For the common direction, the terminal contribution is \(2.5e^{-0.05(2)}=2.262093545\), and the outgo contribution is \(2(1-e^{-0.05(2)})/0.05=3.806503279\). Their sum is \(6.068596824\). For the contrast direction, the terminal contribution is \(1.5e^{-0.25(2)}=0.909795990\), and the outgo contribution is \(-0.5(1-e^{-0.25(2)})/0.25=-0.786938681\). Their sum is \(0.122857309\). Combining the directions gives

V_A(0)=6.068596824+0.122857309=6.191454133, \qquad V_B(0)=6.068596824-0.122857309=5.945739515.

To six decimals, the reserves are \(6.191454\) and \(5.945740\) thousand dollars. As a check, their average is \(6.068596824\), the common component, and half their difference is \(0.122857309\), the contrast component. A frequent mistake is to propagate the terminal vector with the outgo integral or to omit it; the nonzero terminal reserve must be carried back separately through \(e^{-A(T-t)}g\).

Practical Checks and Scope

The matrix-exponential solution is especially useful when several states interact but the assumptions are constant over the valuation interval. Before using it, confirm that state order is consistent in \(Q\), \(h\), \(g\), and \(V\); that each generator row sums to zero; and that the time interval is measured in the same units as the intensities and force of interest. The formula does not directly cover time-varying coefficients; those require a different solution method.

Check the signs. Continuous insurer outgo and transition benefits enter \(h\) positively; premiums received enter negatively.
Check the boundary. Use the reserve vector at the end of the interval as \(g\). A zero terminal vector is an assumption, not an automatic consequence of the model.
Check the direction. When solving backward from \(T\), the terminal propagation is \(e^{-A(T-t)}\), not \(e^{A(T-t)}\).
Check units. If cash flows are in thousands of dollars, reserves computed from the system are also in thousands of dollars.

These checks are useful in multi-state reserve calculations because a correct matrix calculation can still answer the wrong valuation question if the vector order, cash-flow signs, horizon, or terminal condition is misread. The approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, find the stated reserve. Amounts are in thousands of dollars unless stated otherwise.

Two states switch between each other at intensity \(0.1\) per year. The force of interest is \(0.1\), there is no outgo, and both terminal reserves after one year are \(3\). Find either time-0 reserve. (Answer: \(2.715\) thousand dollars in each state.)
A state has no transitions, force of interest \(0.04\), continuous net outgo \(1\) per year, and zero terminal reserve after three years. Find its time-0 reserve. (Answer: \(2.827\) thousand dollars.)
Two states switch at intensity \(0.1\) in each direction; the force of interest is \(0.1\). Continuous net outgo is \(2\) in the first state and \(0\) in the second. There are no transition payments, and both terminal reserves are zero after one year. Find the two reserves. (Answer: \(1.815565\) thousand dollars in the first state and \(0.087687\) thousand dollars in the second.)
A state \(A\) moves to absorbing state \(C\) at intensity \(0.2\). The force of interest is \(0.1\); there is no continuous outgo, and a \(5\)-thousand-dollar benefit is paid on transition. Both terminal reserves are zero after one year. Find the reserve in \(A\). (Answer: \(0.863939\) thousand dollars.)
Two states switch at intensity \(0.1\) in each direction; the force of interest is \(0.05\). Continuous net outgo is \(1.5\) in state \(A\) and \(2.5\) in state \(B\), with no transition payments. Terminal reserves after two years are \(4\) in \(A\) and \(1\) in \(B\). Find both time-0 reserves. (Answer: \(6.191454\) thousand dollars in state \(A\) and \(5.945740\) thousand dollars in state \(B\).)

What You'll Learn