Why Probability Conservation Needs a Proof
A continuous-time multiple-state model tracks probabilities as time passes. Those probabilities can move between states, but if the model accounts for every possible state, their total must remain one. The Kolmogorov forward equation gives the rates of change of the transition probabilities; the generator’s row sums explain why those changes cannot alter a row’s total.
The result depends on using a consistent matrix convention. Here, probability distributions are row vectors, and the transition probability matrix has entry \(P_{ij}(s,t)\) equal to the probability of being in state \(j\) at time \(t\), given state \(i\) at time \(s\). The forward equation is \(\partial P(s,t)/\partial t=P(s,t)Q(t)\). Under this convention, it is the row sums of the generator that are zero.
Assume the model is defined on a finite state space, the transition probabilities satisfy this forward equation, and the intensities are sufficiently regular for the equation and its solution to be well-defined. The generator condition expresses that all rates of leaving a state are represented by transitions to states in the model. If a model omits a possible destination, or its proposed matrix does not have zero row sums, this conservation argument does not apply as stated.
Deriving the Row-Sum Result
Let \(\mathbf{1}\) denote the column vector whose \(n\) entries are all one. The sum of row \(i\) of \(P(s,t)\) is the \(i\)th entry of \(P(s,t)\mathbf{1}\). The zero row sums of the generator are exactly the matrix statement \(Q(t)\mathbf{1}=\mathbf{0}\).
Thus \(P(s,t)\mathbf{1}\) is constant in \(t\). At the starting time, \(P(s,s)=I\), and \(I\mathbf{1}=\mathbf{1}\). Therefore the constant vector is \(\mathbf{1}\), proving that every row of \(P(s,t)\) sums to one. This argument works whether the intensities are constant or vary with time.
The same idea applies to an occupancy distribution. If \(\boldsymbol{\pi}(t)\) is a row vector of state probabilities, then its forward equation is \(\boldsymbol{\pi}'(t)=\boldsymbol{\pi}(t)Q(t)\). Consequently, \(\frac{d}{dt}\bigl(\boldsymbol{\pi}(t)\mathbf{1}\bigr)=\boldsymbol{\pi}(t)Q(t)\mathbf{1}=0\). A distribution that sums to one initially continues to sum to one. The initial condition matters: the equation preserves the initial total, whether or not that total is one.
With row probability vectors, use \(\partial P(s,t)/\partial t=P(s,t)Q(t)\).
Verify that each row of \(Q(t)\) sums to zero, or equivalently that \(Q(t)\mathbf{1}=\mathbf{0}\).
Apply the forward equation to \(P(s,t)\mathbf{1}\); its derivative is zero.
At \(t=s\), \(P(s,s)=I\), so every row sum starts at one and remains one.
Worked Applications
The proof is short, but applying it accurately requires careful attention to matrix products and initial conditions. In each example, rates are measured per year, and probability vectors multiply matrices from the left.
Worked Example: Checking an Instantaneous Change in a Three-State Model
A hypothetical policyholder can be in states 1, 2, or 3. At a particular time, the generator and occupancy distribution are \(Q=\begin{pmatrix}-0.20&0.20&0\\0.10&-0.40&0.30\\0.05&0.15&-0.20\end{pmatrix}\) per year and \(\boldsymbol{\pi}=(0.30,0.40,0.30)\). Calculate the instantaneous derivative of the occupancy distribution and check whether its components sum to zero.
Method. First check the generator’s row sums, then calculate \(\boldsymbol{\pi}Q\). The derivative of the total should be zero if the generator meets the required condition.
The three generator row sums are \(-0.20+0.20+0=0\), \(0.10-0.40+0.30=0\), and \(0.05+0.15-0.20=0\). Thus \(Q\mathbf{1}=\mathbf{0}\).
The first component of the product is \(0.30(-0.20)+0.40(0.10)+0.30(0.05)=-0.060+0.040+0.015=-0.005\). The second is \(0.30(0.20)+0.40(-0.40)+0.30(0.15)=0.060-0.160+0.045=-0.055\). The third is \(0.30(0)+0.40(0.30)+0.30(-0.20)=0+0.120-0.060=0.060\).
Therefore \(\boldsymbol{\pi}'=(-0.005,-0.055,0.060)\) per year. Its component sum is \(-0.005-0.055+0.060=0\), as required. The initial occupancy probabilities also sum to \(0.30+0.40+0.30=1\), so their total remains one locally and, under the model, throughout its evolution.
Check and common mistake. The second and third product components must be calculated separately; checking only that the final three numbers sum to zero is not enough to verify the multiplication. For example, \((-0.005,-0.005,0.010)\) also sums to zero, but it is not the product. The corrected product is \((-0.005,-0.055,0.060)\).
Worked Example: Verifying Conservation in a Two-State Model
A hypothetical contract’s insured is either active (state 1) or disabled (state 2). The transition intensity from active to disabled is \(0.4\) per year, and the intensity from disabled to active is \(0.2\) per year. Using the two-state transition probabilities, verify the row sums at \(t=2\) years and calculate the derivative of each row at that time.
Method. The generator is \(Q=\begin{pmatrix}-0.4&0.4\\0.2&-0.2\end{pmatrix}\). For these constant intensities, the total rate is \(0.4+0.2=0.6\), and the stationary probabilities of states 1 and 2 are \(0.2/0.6=1/3\) and \(0.4/0.6=2/3\). The two-state transition matrix is \(P(0,t)=\begin{pmatrix}1/3+(2/3)e^{-0.6t}&(2/3)(1-e^{-0.6t})\\(1/3)(1-e^{-0.6t})&2/3+(1/3)e^{-0.6t}\end{pmatrix}\).
At \(t=2\), \(e^{-0.6t}=e^{-1.2}\approx0.3010\). The first row is approximately \((1/3+(2/3)(0.3010), (2/3)(1-0.3010))=(0.5340,0.4660)\). Its sum is \(0.5340+0.4660=1.0000\).
The second row is approximately \(((1/3)(1-0.3010), 2/3+(1/3)(0.3010))=(0.2330,0.7670)\). Its sum is \(0.2330+0.7670=1.0000\).
Now apply the forward equation \(P'=PQ\). For the first row, the derivative is \((0.5340(-0.4)+0.4660(0.2),\;0.5340(0.4)+0.4660(-0.2))=(-0.2136+0.0932,\;0.2136-0.0932)=(-0.1204,0.1204)\) per year.
For the second row, the derivative is \((0.2330(-0.4)+0.7670(0.2),\;0.2330(0.4)+0.7670(-0.2))=(-0.0932+0.1534,\;0.0932-0.1534)=(0.0602,-0.0602)\) per year. Each derivative row sums to zero. Rounded to four decimal places, the transition rows sum to \(1.0000\) and the derivative components cancel exactly at the displayed precision.
Check and common mistake. Each row describes probabilities conditional on a different starting state; the calculation must use the corresponding row of \(P\) in \(PQ\). Multiplying \(Q\) by a row vector on the right would use the wrong convention for this forward equation.
Worked Example: Conservation with Time-Varying Intensities
A hypothetical long-term care model has three states: independent, receiving care, and dead. Dead is absorbing. The intensity from independent to care at time \(t\) is \(0.10+0.02t\) per year; from care to independent it is \(0.20\); and from care to dead it is \(0.10\). At \(t=2\), the occupancy distribution is \((0.50,0.30,0.20)\). Calculate its instantaneous derivative and verify conservation.
Method. Evaluate all intensities at the specified time before forming the generator. At \(t=2\), the independent-to-care intensity is \(0.10+0.02(2)=0.14\). The care-state total exit intensity is \(0.20+0.10=0.30\). Thus \(Q(2)=\begin{pmatrix}-0.14&0.14&0\\0.20&-0.30&0.10\\0&0&0\end{pmatrix}\).
The row sums are \(-0.14+0.14+0=0\), \(0.20-0.30+0.10=0\), and \(0+0+0=0\). Calculate the first derivative component: \(0.50(-0.14)+0.30(0.20)+0.20(0)=-0.070+0.060=-0.010\). The second is \(0.50(0.14)+0.30(-0.30)+0.20(0)=0.070-0.090=-0.020\). The third is \(0.50(0)+0.30(0.10)+0.20(0)=0.030\).
Therefore \(\boldsymbol{\pi}'(2)=(-0.010,-0.020,0.030)\) per year, and the derivative components sum to \(-0.010-0.020+0.030=0\). The given occupancy probabilities sum to \(0.50+0.30+0.20=1\), so their total is conserved. The fact that the intensities change with time does not change the argument: at every time, the generator has zero row sums.
Check and common mistake. Do not use the time-zero intensity \(0.10\) in place of the rate at \(t=2\). Also, the care-state diagonal must be \(-0.30\), the negative of the sum of both exit intensities, rather than \(-0.20\) or \(-0.10\).
What the Result Does and Does Not Establish
Probability conservation is a normalization result. It says that if the forward equation and the zero-row-sum condition hold, the total of each transition-probability row remains at its initial value. Since \(P(s,s)=I\), that value is one for every row of the transition matrix. For an occupancy distribution, the total remains whatever total its initial distribution had.
The theorem does not excuse checking the model’s assumptions. In the row convention, a proposed generator must have zero row sums. Its off-diagonal entries must also be nonnegative transition intensities. A matrix with zero row sums but a negative off-diagonal entry may preserve the algebraic row total while failing to represent valid transition rates. Likewise, if the state space omits an outcome, the diagonal may need to account for a transition out of the modeled states rather than make the row sum zero; the probabilities of the retained states then need not sum to one.
A numerical solution may show row sums such as \(0.9998\) or \(1.0002\) because of rounding or integration error. Such small discrepancies can be useful diagnostics, but they do not change the exact theorem. A substantial discrepancy suggests checking the matrix convention, the diagonal entries, the time argument in \(Q(t)\), and whether the initial condition was applied correctly.
A particularly subtle trap is that an incorrect matrix product can still have components that sum to zero. The conservation check is necessary, but it does not certify every component. Calculate each component using the stated row-vector convention, then check the total separately. This material is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use the row-vector convention and distinguish checking a total from checking each matrix-product component.
- A two-state model has generator \(Q=\begin{pmatrix}-0.30&0.30\\0.10&-0.10\end{pmatrix}\) per year and current occupancy distribution \((0.60,0.40)\). Find the derivative of the distribution and its component sum. (Answer: \((-0.14,0.14)\) per year; component sum \(0\).)
- For \(Q=\begin{pmatrix}-0.50&0.20&0.30\\0.10&-0.40&0.30\\0.20&0.10&-0.30\end{pmatrix}\) per year and \(\boldsymbol{\pi}=(0.20,0.50,0.30)\), calculate \(\boldsymbol{\pi}Q\) and check its sum. (Answer: \((0.01,-0.13,0.12)\) per year; component sum \(0\).)
- A time-varying generator is \(Q(t)=\begin{pmatrix}-(0.10+0.02t)&0.10+0.02t&0\\0.05&-0.05&0\\0&0&0\end{pmatrix}\) per year, with \(t\) in years. At \(t=5\), the occupancy distribution is \((0.40,0.40,0.20)\). Find its instantaneous derivative and verify conservation. (Answer: \((-0.06,0.06,0)\) per year; component sum \(0\).)
- A proposed generator is \(Q=\begin{pmatrix}-0.30&0.20&0.10\\0.10&-0.40&0.20\\0&0&0\end{pmatrix}\), and the current distribution is \((0.20,0.50,0.30)\). Does the zero-row-sum proof apply? Calculate the derivative’s component sum. (Answer: No; the second generator row sums to \(-0.10\), and the derivative component sum is \(-0.05\) per year.)
- A valid generator has zero row sums, but an occupancy row at the initial time is \((0.25,0.30,0.42)\). What total does the forward equation preserve, and is this a probability distribution as stated? (Answer: It preserves \(0.97\); no, the stated initial row is not normalized to one.)