Boundary Conditions for Transition Probability Matrices

Why the Boundary Condition Matters

A system of differential equations can have many solutions, but a transition-probability matrix is not just any solution: it must describe the probabilities of moving between states over a specified interval. The boundary condition identifies which solution corresponds to that interval. In particular, when the interval has zero length, the process has had no time to change state.

This idea is simple to state but easy to misapply. The forward equation holds the starting time fixed and varies the ending time; the backward equation holds the ending time fixed and varies the starting time. Their boundary conditions are therefore imposed at different ends of the time interval.

Definition: For a finite-state continuous-time Markov model, let \(P(s,t)\) be the transition-probability matrix from time \(s\) to time \(t\), where \(s\leq t\). Its entry \(P_{ij}(s,t)\) is the conditional probability that the process is in state \(j\) at time \(t\), given that it was in state \(i\) at time \(s\). Under the row convention, each row corresponds to a starting state and each column to an ending state. The zero-duration boundary condition is \(P(s,s)=I\), where \(I\) is the identity matrix.

The identity matrix has a one in each diagonal entry and zeroes elsewhere. Thus, over a zero-length interval, the probability of being in the same state is one, and the probability of being in a different state is zero. This condition applies at every time \(s\), not only at time zero.

Boundary Conditions for Forward and Backward Equations

Let \(Q(t)\) be the generator under the row convention: its off-diagonal entry \(q_{ij}(t)\) is the instantaneous transition intensity from state \(i\) to state \(j\), and each row sums to zero. When the intensities are sufficiently regular for the differential equations to hold, the forward equation varies the ending time:

\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t), \qquad P(s,s)=I.

Here \(s\) is held fixed. For each chosen starting time \(s\), solve forward from the endpoint \(t=s\), where the matrix is known to be \(I\). The backward equation instead varies the starting time while holding \(t\) fixed:

\frac{\partial}{\partial s}P(s,t)=-Q(s)P(s,t), \qquad P(t,t)=I.

The backward condition is written \(P(t,t)=I\) because the variable being moved is \(s\), and the fixed terminal time is \(t\). The two expressions \(P(s,s)=I\) and \(P(t,t)=I\) describe the same zero-duration fact; they emphasize the relevant endpoint for each differential equation. These equations use the stated row convention. Switching to a column convention changes matrix multiplication order, so it is not safe to transfer either equation unchanged.

Key takeaway: For a forward solution, fix \(s\) and impose the identity at \(t=s\). For a backward solution, fix \(t\) and impose the identity at \(s=t\). A condition such as \(P(0,t)=I\) is not the general boundary condition; it would assert that every interval starting at zero has no state changes.

Why the Identity Is the Correct Boundary

The boundary condition follows directly from the meaning of a transition probability. At zero elapsed time, the process remains in its current state with probability one. In symbols, for any states \(i\) and \(j\),

P_{ij}(s,s)= \begin{cases} 1, & i=j,\\ 0, & i\ne j. \end{cases}

These are exactly the entries of \(I\). The same conclusion can be seen from the forward equation over a very short interval. To first order, the transition matrix over an interval of length \(h\) is \(I+hQ(s)\), apart from terms that are smaller than \(h\). As \(h\) decreases to zero, this matrix tends to \(I\). The generator describes the initial rate of change away from that boundary; it does not replace the boundary itself.

This distinction matters when solving an equation containing arbitrary constants. The differential equation determines how a solution changes, while the endpoint condition determines which solution has the required transition-probability interpretation. After applying the condition, check that rows sum to one and entries are valid probabilities. Those checks are useful, but they do not substitute for the boundary condition.

Choose the equation and fixed endpoint.
For a forward equation, specify the starting time \(s\). For a backward equation, specify the ending time \(t\).

Write the identity at zero duration.
Use \(P(s,s)=I\) for the forward solution or \(P(t,t)=I\) for the backward solution.

Use the boundary to determine constants.
Apply the condition entry by entry or as a matrix equation; do not choose constants from a probability check alone.

Check the resulting matrix.
At zero duration it must be \(I\); over positive duration its entries must be nonnegative and its rows must sum to one.

Worked Example: Selecting the Forward Solution in a Two-State Model

A hypothetical process has two states, \(0\) and \(1\). It moves from \(0\) to \(1\) at constant intensity \(0.4\) per year and from \(1\) to \(0\) at the same intensity. Find the transition matrix after \(1.25\) years. Use the forward equation and show how the boundary condition selects the solution.

Method. Fix the starting time \(s\), write the forward equations for the first-column probabilities, and use \(P(s,s)=I\) to determine their initial values. The rates are constant, measured per year, and the time interval is measured in years.

Write \(p_{00}=P_{00}(s,t)\) and \(p_{10}=P_{10}(s,t)\). The first-column forward equations are

\frac{dp_{00}}{dt}=-0.4p_{00}+0.4p_{10}, \qquad \frac{dp_{10}}{dt}=0.4p_{00}-0.4p_{10}.

At \(t=s\), the identity condition gives \(p_{00}(s,s)=1\) and \(p_{10}(s,s)=0\). Define \(d(t)=p_{00}-p_{10}\). Subtracting the equations gives

\frac{dd}{dt}=-0.8d, \qquad d(s)=1, \qquad d(t)=e^{-0.8(t-s)}.

The two probabilities sum to one: adding their differential equations gives a derivative of zero, and their sum at \(t=s\) is one. Therefore,

p_{00}+p_{10}=1,\qquad p_{00}=\frac{1+d}{2},\qquad p_{10}=\frac{1-d}{2}.

For \(t-s=1.25\), the exponent is \(-0.8(1.25)=-1\), so \(d=e^{-1}=0.3679\) to four decimal places. Hence \(p_{00}=0.6839\) and \(p_{10}=0.3161\), to four decimal places. By symmetry, the complete matrix is

P(s,s+1.25)= \begin{pmatrix} 0.6839 & 0.3161\\ 0.3161 & 0.6839 \end{pmatrix}.

Check. Each row sums to \(1.0000\), and the matrix tends to \(I\) when the elapsed time tends to zero. A common mistake is to use \(p_{00}(s,s)=p_{10}(s,s)=0.5\), perhaps because the long-run probabilities are equal. That does not describe a zero-length interval: it violates \(P(s,s)=I\) and would select the wrong solution.

Worked Example: Applying a Forward Boundary with Time-Varying Intensity

In a hypothetical two-state model, a person can move from state \(0\) to state \(1\), but not back. State \(1\) is absorbing. The intensity at time \(u\), measured in years from the model origin, is \(\lambda(u)=0.10+0.10u\) per year. Find \(P(1,3)\).

Method. Use the forward equation with the starting time fixed at \(s=1\). The generator is

Q(u)= \begin{pmatrix} -\lambda(u)&\lambda(u)\\ 0&0 \end{pmatrix}.

Let \(p_{00}(1,t)\) be the probability of remaining in state \(0\) through time \(t\), given state \(0\) at time \(1\). The forward equation gives \(dp_{00}/dt=-\lambda(t)p_{00}\). Its boundary value is \(p_{00}(1,1)=1\), from \(P(1,1)=I\). Separating variables and integrating from \(1\) to \(t\) gives

\int_1^t \frac{1}{p_{00}(1,u)}\frac{dp_{00}(1,u)}{du}\,du =-\int_1^t(0.10+0.10u)\,du, \qquad p_{00}(1,t)= \exp\!\left[-0.10(t-1)-0.05(t^2-1)\right].

At \(t=3\), the integrated intensity is \(0.10(3-1)+0.05(3^2-1)=0.20+0.40=0.60\). Thus \(p_{00}(1,3)=e^{-0.60}=0.5488\) to four decimal places. Since state \(1\) is absorbing, the probability of being in state \(1\) at time \(3\), starting from state \(0\), is \(1-0.5488=0.4512\). Starting from state \(1\), the process remains there. Therefore,

P(1,3)= \begin{pmatrix} 0.5488&0.4512\\ 0&1.0000 \end{pmatrix}.

Check. The matrix is the identity if the interval shrinks to zero, and both rows sum to one. A common error is to integrate the rate from \(0\) to \(3\). The interval begins at time \(1\), so the correct accumulated intensity is \(\int_1^3\lambda(u)\,du=0.60\), not \(\int_0^3\lambda(u)\,du\).

Worked Example: Using the Terminal Condition in a Backward Equation

Consider a hypothetical process with states \(0\) and \(1\), where state \(0\) moves to absorbing state \(1\) at constant intensity \(0.20\) per year. Use the backward equation to find \(P(2,5)\).

Method. Fix terminal time \(t=5\) and vary the starting time \(s\). The generator is

Q= \begin{pmatrix} -0.20&0.20\\ 0&0 \end{pmatrix}.

Let \(p(s)=P_{00}(s,5)\). From \(\partial P(s,5)/\partial s=-QP(s,5)\), the first-column equation is \(dp/ds=0.20p\). The backward boundary condition is imposed at \(s=5\), not \(s=0\): \(p(5)=P_{00}(5,5)=1\). Solving the scalar equation from this terminal value gives

p(s)=C e^{0.20s}, \qquad 1=C e^{0.20(5)}, \qquad p(s)=e^{-0.20(5-s)}.

At \(s=2\), \(p(2)=e^{-0.20(3)}=e^{-0.60}=0.5488\) to four decimal places. The probability of being in state \(1\) at time \(5\), starting from state \(0\) at time \(2\), is \(1-0.5488=0.4512\). Starting in state \(1\) gives probability one of ending in state \(1\). Thus,

P(2,5)= \begin{pmatrix} 0.5488&0.4512\\ 0&1.0000 \end{pmatrix}.

Check. At \(s=5\), the result is \(I\); at \(s=2\), both rows sum to one. The common mistake is to impose \(p(0)=1\) because the equation is being solved in \(s\). The relevant endpoint is \(s=t=5\), since that is where the elapsed time is zero.

Common Boundary-Condition Errors

Using the wrong endpoint. The forward equation starts at \(t=s\); the backward equation ends at \(s=t\). Identify the variable being changed before applying a condition.
Confusing a transition matrix with an initial distribution. \(P(s,s)=I\) is a statement about conditional transitions. It does not say that the process has a particular initial distribution across states.
Checking only row sums. A matrix with rows summing to one can still fail the required endpoint condition or contain negative entries. Check the identity at zero duration as well.
Using the wrong matrix order. With the row convention used here, the forward equation multiplies \(P\) by \(Q\) on the right and the backward equation multiplies \(Q\) by \(P\) on the left.
Integrating over the wrong interval. For a transition from \(s\) to \(t\), accumulated rates apply over that interval, not automatically from time zero.

Key takeaway: A Kolmogorov differential equation describes how transition probabilities change; the identity boundary condition makes its solution the transition matrix for the intended interval. Apply it at the zero-duration endpoint, then verify the resulting probabilities.

Check Your Understanding

For each problem, identify the appropriate endpoint condition and use it to find or assess the requested transition probabilities.

A forward equation is being solved for \(P(2,t)\), with \(t\geq2\). What is its boundary value? A backward equation is being solved for \(P(s,5)\), with \(s\leq5\). What is its boundary value? (Answer: \(P(2,2)=I\) and \(P(5,5)=I\), respectively.)
A two-state process has a transition from state \(0\) to absorbing state \(1\) at constant intensity \(0.25\) per year. Find \(P(0,4)\). (Answer: \(\begin{pmatrix}0.3679&0.6321\\0&1.0000\end{pmatrix}\), to four decimal places.)
In a two-state one-way model, state \(0\) moves to absorbing state \(1\) at intensity \(\lambda(u)=0.05+0.05u\) per year, where \(u\) is measured in years from the model origin. Find \(P(2,4)\). (Answer: \(\begin{pmatrix}0.6703&0.3297\\0&1.0000\end{pmatrix}\), to four decimal places.)
A two-state process moves in either direction at constant intensity \(0.30\) per year. Use the forward equation and its boundary condition to find the probability of remaining in the starting state after two years. (Answer: \((1+e^{-1.2})/2=0.6506\), to four decimal places.)
For a one-way model with intensity \(0.20\) per year from state \(0\) to absorbing state \(1\), a candidate backward solution for \(P_{00}(s,5)\) is \(e^{-0.20s}\). Does it satisfy the required boundary condition, and what is the correct value of \(P_{00}(2,5)\)? (Answer: No; at \(s=5\) it is \(e^{-1}\), not \(1\). The correct value is \(e^{-0.20(5-2)}=0.5488\), to four decimal places.)

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