Variation of Constants for Time-Dependent Systems

Why Variation of Constants Is Needed

A multi-state reserve system can change as transition intensities, interest, premiums, or expenses vary with time. The constant-coefficient matrix-exponential method is useful when the system matrix does not change over the period, but it is not generally valid to replace a time-varying matrix by its integral and take an ordinary matrix exponential. Variation of constants instead uses a state-transition matrix to transport reserve values through the changing system and to accumulate the effect of continuous cash flows.

The focus here is a linear system written for a column vector of state reserves. The system’s coefficients are treated as known functions of time, and a reserve vector is specified at a terminal time. Each example uses a hypothetical reserve system: the valuation force is $\delta=0$, and the displayed system matrix and cash-flow vector are the stipulated net Thiele coefficients. No separate mortality table, transition-intensity model, or claim-size distribution is required; those assumptions, if used to construct a system, have already been summarized in its coefficients.

Definition: On an interval $[0,T]$, a time-dependent linear reserve system has the form $$ \mathbf V'(t)=A(t)\mathbf V(t)+\mathbf g(t), \qquad \mathbf V(T)=\mathbf b. $$ Here $\mathbf V(t)$ is the column vector of reserves by state, $A(t)$ is a matrix of known coefficients, $\mathbf g(t)$ is a known vector of continuous source terms under the chosen Thiele sign convention, and $\mathbf b$ is the terminal reserve vector. The functions $A$ and $\mathbf g$ are assumed piecewise continuous, so the required integrals and solutions exist.

The exact interpretation of the entries of $A(t)$ and $\mathbf g(t)$ depends on how the Thiele equations have been arranged. In this tutorial, their signs are fixed by the displayed equation $\mathbf V'=A\mathbf V+\mathbf g$. A system written with the opposite sign for its cash-flow vector must use that sign consistently. The terminal condition means that the reserve is known at $T$ and is to be calculated at earlier times.

The State-Transition Matrix and the Terminal Formula

For the homogeneous system $\mathbf V'(t)=A(t)\mathbf V(t)$, define $\Phi(t,s)$ as the matrix that carries a vector at time $s$ to its homogeneous-system value at time $t$. It satisfies

\frac{\partial}{\partial t}\Phi(t,s)=A(t)\Phi(t,s), \qquad \Phi(s,s)=I.

The order of the arguments matters: $\Phi(t,s)$ moves values from time $s$ to time $t$, including when $t<s$. It obeys the composition rule $\Phi(t,r)\Phi(r,s)=\Phi(t,s)$. In general, a time-varying system does not have $\Phi(t,s)=\exp\!\left(\int_s^t A(u)\,du\right)$. That simplification is valid when the matrices $A(u)$ commute with one another for all relevant times; without commutation, the state-transition matrix preserves the order in which the changing matrices act.

Theorem: If $\mathbf V'=A\mathbf V+\mathbf g$ and $\mathbf V(T)=\mathbf b$, then, for $t\leq T$, $$ \mathbf V(t)=\Phi(t,T)\mathbf b-\int_t^T \Phi(t,s)\mathbf g(s)\,ds. $$ The negative sign is required for this terminal-value form under the stated differential-equation convention.

A short derivation explains both the transported forcing term and its sign. Let $U(t)$ be any nonsingular fundamental matrix satisfying $U'(t)=A(t)U(t)$. Then $\Phi(t,s)=U(t)U(s)^{-1}$. Write $\mathbf V(t)=U(t)\mathbf c(t)$. Differentiating and substituting into the system gives

U'(t)\mathbf c(t)+U(t)\mathbf c'(t) =A(t)U(t)\mathbf c(t)+\mathbf g(t), \qquad \mathbf c'(t)=U(t)^{-1}\mathbf g(t).

Integrate from $t$ to $T$, and use $U(T)\mathbf c(T)=\mathbf b$. This gives $\mathbf c(t)=\mathbf c(T)-\int_t^T U(s)^{-1}\mathbf g(s)\,ds$. Multiplying by $U(t)$ yields the theorem. In particular, the forcing term is subtracted because the known terminal value is transported backward across cash flows that would have contributed positively in a forward initial-value calculation.

Worked Applications

Worked Example: A Triangular System With a Continuous Source

A hypothetical two-state reserve system has $0\leq t\leq1$, force of interest $\delta=0$, and terminal vector $\mathbf V(1)=(4,1)^{\mathsf T}$. Its stipulated net Thiele system is

\mathbf V'(t)= \begin{pmatrix}0&t\\0&0\end{pmatrix}\mathbf V(t) +\begin{pmatrix}0\\2\end{pmatrix}.

The displayed coefficients fully specify this deterministic example; there is no separate mortality or claim-size distribution. Find $\mathbf V(0)$, using the state-transition matrix, and check the result in the differential equations.

Method. Let $A(t)=\begin{pmatrix}0&t\\0&0\end{pmatrix}$. For any $u,v$, $A(u)A(v)=0$, so the transition matrix is

\Phi(t,s)=I+\int_s^t A(u)\,du =\begin{pmatrix}1&(t^2-s^2)/2\\0&1\end{pmatrix}.

At $t=0$ and $s=1$, this becomes $\Phi(0,1)=\begin{pmatrix}1&-1/2\\0&1\end{pmatrix}$. Transport the terminal vector: $\Phi(0,1)\mathbf V(1)=(4-1/2,1)^{\mathsf T}=(7/2,1)^{\mathsf T}$. For the source term,

\Phi(0,s)\mathbf g(s) =\begin{pmatrix}1&-s^2/2\\0&1\end{pmatrix} \begin{pmatrix}0\\2\end{pmatrix} =\begin{pmatrix}-s^2\\2\end{pmatrix}, \qquad \int_0^1\Phi(0,s)\mathbf g(s)\,ds =\begin{pmatrix}-1/3\\2\end{pmatrix}.

Apply the terminal formula, subtracting this integral: $\mathbf V(0)=(7/2,1)^{\mathsf T}-(-1/3,2)^{\mathsf T}=(23/6,-1)^{\mathsf T}$.

Check. The second equation is $V_2'=2$, so $V_2(t)=2t-1$, which gives $V_2(1)=1$ and $V_2(0)=-1$. The first equation is $V_1'=tV_2=t(2t-1)$. Therefore $\int_0^1V_1'(t)\,dt=2/3-1/2=1/6$, and $V_1(0)=4-1/6=23/6$, as calculated.

Common mistake avoided. Using a plus sign before the source integral in the terminal formula would give $19/6$ for the first component and $3$ for the second. The sign is negative because this is a backward calculation from a terminal value.

Worked Example: Time-Varying Diagonal Coefficients

For a second hypothetical two-state system on $0\leq t\leq1$, let $\delta=0$, let the stipulated net coefficient matrix be $A(t)=\begin{pmatrix}t&0\\0&-t\end{pmatrix}$, and let $\mathbf g(t)=(e^{t^2/2},e^{-t^2/2})^{\mathsf T}$. There is no additional mortality or claim-size assumption: all needed deterministic inputs are displayed. The terminal vector is $\mathbf V(1)=(2e^{1/2},3e^{-1/2})^{\mathsf T}$. Find $\mathbf V(0)$.

The two diagonal equations are independent: $V_1'=tV_1+e^{t^2/2}$ and $V_2'=-tV_2+e^{-t^2/2}$. For the first equation, multiply by the integrating factor $e^{-t^2/2}$: $(e^{-t^2/2}V_1(t))'=1$. At $t=1$, $e^{-1/2}V_1(1)=2$. Integrating the transformed equation from $0$ to $1$ gives $2-e^0V_1(0)=1$, so $V_1(0)=1$.

For the second equation, multiply by $e^{t^2/2}$: $(e^{t^2/2}V_2(t))'=1$. At $t=1$, $e^{1/2}V_2(1)=3$. Thus $3-V_2(0)=1$, and $V_2(0)=2$. Therefore $\mathbf V(0)=(1,2)^{\mathsf T}$.

Check. The solutions are $V_1(t)=e^{t^2/2}(1+t)$ and $V_2(t)=e^{-t^2/2}(2+t)$. They give the specified terminal values at $t=1$, and at $t=0$ give $1$ and $2$. The integrating-factor method is the diagonal special case of variation of constants. A common mistake is to use the same exponential factor for both equations despite their opposite diagonal coefficients.

Worked Example: Piecewise Coefficients and Matrix Order

A hypothetical two-state system has terminal time $T=2$, force of interest $\delta=0$, and terminal vector $\mathbf V(2)=(2,1)^{\mathsf T}$. For $0\leq t<1$, take $A(t)=A_1=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $\mathbf g(t)=(1,0)^{\mathsf T}$. For $1\leq t\leq2$, take $A(t)=A_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ and $\mathbf g(t)=(0,1)^{\mathsf T}$. These are stipulated net coefficients, with no separate mortality or claim-size distribution. Find $\mathbf V(0)$.

Method. On $[1,2]$, $A_2^2=0$, so $\Phi(1,2)=I-A_2=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$. Transport the terminal vector to time $1$: $\Phi(1,2)\mathbf V(2)=(2,-1)^{\mathsf T}$. For $1\leq s\leq2$, $\Phi(1,s)=\begin{pmatrix}1&0\\1-s&1\end{pmatrix}$, and hence $\Phi(1,s)\mathbf g(s)=(0,1)^{\mathsf T}$. Its integral over $[1,2]$ is $(0,1)^{\mathsf T}$. The terminal formula now gives $\mathbf V(1)=(2,-1)^{\mathsf T}-(0,1)^{\mathsf T}=(2,-2)^{\mathsf T}$.

On $[0,1]$, $\Phi(0,1)=I-A_1=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$, so $\Phi(0,1)\mathbf V(1)=(4,-2)^{\mathsf T}$. For $0\leq s\leq1$, $\Phi(0,s)=\begin{pmatrix}1&-s\\0&1\end{pmatrix}$, which gives $\Phi(0,s)\mathbf g(s)=(1,0)^{\mathsf T}$. Its integral is $(1,0)^{\mathsf T}$, and consequently $\mathbf V(0)=(4,-2)^{\mathsf T}-(1,0)^{\mathsf T}=(3,-2)^{\mathsf T}$.

Check. On the second interval, $V_1'=0$ and $V_2'=V_1+1=3$; from $(2,1)$ at time $2$, these equations give $(2,-2)$ at time $1$. On the first interval, $V_2' = 0$ and $V_1'=V_2+1=-1$; from $V_1(1)=2$, this gives $V_1(0)=3$, while $V_2(0)=-2$. The result checks directly. The matrix order matters: the transition across both intervals is $\Phi(0,1)\Phi(1,2)$, not the reversed product.

Practical Checks and Exam Traps

Variation of constants turns a changing reserve system into two calculations: transport the terminal vector, then subtract the transported source accumulated between the valuation time and the terminal time. In an actuarial model, the source and matrix coefficients come from the chosen Thiele equations; the formula does not replace the need to derive those equations correctly.

Check the direction. $\Phi(t,s)$ maps from $s$ to $t$. For a terminal-value problem with $t<T$, both the terminal transport and the source integral run backward in this sense.
Check the source sign. For $\mathbf V'=A\mathbf V+\mathbf g$, the terminal-value integral is subtracted. Re-derive the sign if the differential equation has been written with a different source convention.
Check the matrix order. For successive intervals, multiply the transition matrices in chronological composition order, such as $\Phi(0,1)\Phi(1,2)$. Matrix multiplication is generally not commutative.
Check the endpoints. Substituting $t=T$ into the terminal formula must give $\mathbf V(T)=\mathbf b$, because the integral then has length zero and $\Phi(T,T)=I$.

These checks are useful in multi-state reserve problems because a sign error, a reversed transition, or an incorrectly assumed matrix exponential can produce plausible-looking values while failing to solve the specified differential equations. When feasible, substitute the calculated reserve path or endpoint into the original system as an independent check.

Check Your Understanding

For each problem, use the stated system convention and calculate backward from the terminal vector.

On $0\leq t\leq1$, let $\mathbf V'=\begin{pmatrix}0&2t\\0&0\end{pmatrix}\mathbf V+(0,1)^{\mathsf T}$, with $\mathbf V(1)=(3,2)^{\mathsf T}$. Find $\mathbf V(0)$. (Answer: $(4/3,1)^{\mathsf T}$)
On $0\leq t\leq1$, let $\mathbf V'=\begin{pmatrix}2t&0\\0&0\end{pmatrix}\mathbf V+(0,3)^{\mathsf T}$, with $\mathbf V(1)=(2e,4)^{\mathsf T}$. Find $\mathbf V(0)$. (Answer: $(2,1)^{\mathsf T}$)
On $0\leq t<1$, let $A=\begin{pmatrix}0&0\\2&0\end{pmatrix}$ and $\mathbf g=(0,1)^{\mathsf T}$; on $1\leq t\leq2$, let $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $\mathbf g=(1,0)^{\mathsf T}$. If $\mathbf V(2)=(1,3)^{\mathsf T}$, find $\mathbf V(0)$. (Answer: $(-3,8)^{\mathsf T}$)
For $\mathbf V'=\begin{pmatrix}0&t\\0&0\end{pmatrix}\mathbf V$ on $0\leq t\leq2$, with $\mathbf V(2)=(5,1)^{\mathsf T}$, find $\mathbf V(0)$. (Answer: $(3,1)^{\mathsf T}$)
For the system in question 1, verify the first component by solving the two scalar differential equations backward from $t=1$. (Answer: $V_2(t)=t+1$, $V_1(t)=4/3+2t^3/3+t^2$, so $V_1(1)=3$ and $V_1(0)=4/3$.)

What You'll Learn