Recovering Intensities from Transition Probability Derivatives

Why a Derivative, Not a Probability, Gives an Intensity

A transition probability describes the chance of moving between states over a specified interval. An intensity describes the instantaneous rate of a transition at a particular time, conditional on the process currently being in its starting state. The distinction matters: a probability is dimensionless and depends on the length of the interval, while an intensity has units of inverse time.

For a continuous-time multiple-state model, the generator records these instantaneous transition rates. Its off-diagonal entries are recovered from the right-hand derivatives of transition probabilities when the interval length tends to zero. Its diagonal entries describe the rate of leaving each state, with a negative sign.

Definition: Let $p_{ij}(s,t)$ be the probability that the process is in state $j$ at time $t$, given that it was in state $i$ at time $s$, where $s\le t$. For $i\ne j$, the transition intensity from $i$ to $j$ at time $t$ is $$ q_{ij}(t)=\lim_{h\downarrow 0}\frac{p_{ij}(t,t+h)}{h} =\left.\frac{\partial}{\partial u}p_{ij}(t,u)\right|_{u=t+}. $$ The generator $Q(t)=(q_{ij}(t))$ has diagonal entries $q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)$. Thus every row of $Q(t)$ sums to zero.

This definition assumes a continuous-time Markov model with well-defined, finite transition intensities. The derivative is a right derivative because the interval begins at $t$ and extends forward. For a finite-state model, the local transition matrix has the expansion $P(t,t+h)=I+hQ(t)+o(h)$, where $o(h)/h$ tends to zero as $h\downarrow0$. The expansion is local: it does not say that probabilities over a longer interval equal intensity multiplied by interval length.

Recovering the Generator from a Local Derivative

At zero interval length, the transition matrix is the identity: the process is in its starting state with probability one. Over a short interval of length $h$, a transition from $i$ to a different state $j$ has probability approximately $q_{ij}(t)h$. The probability of remaining in $i$ is approximately one minus the sum of the probabilities of leaving for all other states.

p_{ij}(t,t+h)=q_{ij}(t)h+o(h)\quad(i\ne j), \qquad p_{ii}(t,t+h)=1+q_{ii}(t)h+o(h).

Consequently, the right derivative of the local transition matrix at $h=0$ is the generator itself. Off-diagonal derivatives are nonnegative; diagonal derivatives are nonpositive. If a state has no possible exits, its entire generator row is zero.

Identify the interval.
Check whether the data describe $p_{ij}(t,t+h)$ as $h$ tends to zero, or a transition over a positive, finite horizon.

Read off off-diagonal rates.
For each distinct pair of states, the local derivative is $q_{ij}(t)$, with units such as per year.

Check the diagonal and row sums.
Set each diagonal entry to the negative sum of the other entries in its row. Confirm that each row sums to zero.

Worked Example: Reading Intensities from a Local Derivative Matrix

A hypothetical insurance model has states Active, Disabled, and Dead. At time $t$, the derivative of $P(t,t+h)$ with respect to $h$, evaluated at $h=0+$, is given per year as $$ D(t)= \begin{pmatrix} -0.12&0.08&0.04\\ 0.03&-0.11&0.08\\ 0&0&0 \end{pmatrix}. $$ Find the generator and interpret the entries.

Method. Since this is explicitly the derivative of a transition matrix over a vanishing interval, $D(t)=Q(t)$. The off-diagonal entries give the instantaneous rates directly. From Active, the rate to Disabled is $0.08$ per year and the rate to Dead is $0.04$ per year. From Disabled, the rates to Active and Dead are $0.03$ and $0.08$ per year, respectively.

The diagonal for Active is the negative total exit rate: $q_{\text{Active,Active}}=-(0.08+0.04)=-0.12$ per year. For Disabled it is $q_{\text{Disabled,Disabled}}=-(0.03+0.08)=-0.11$ per year. Dead has no exits, so its generator row is zero.

Check. The row sums are $-0.12+0.08+0.04=0$, $0.03-0.11+0.08=0$, and $0+0+0=0$. The off-diagonal entries are nonnegative and the diagonal entries are nonpositive, as required. The common mistake this method avoids is treating a derivative entry as a probability; these entries are rates per year.

Why the Row-Sum Condition Holds

For each starting state $i$, the probabilities of being in all possible states at time $t+h$ sum to one. Differentiating that identity at $h=0+$ shows why the generator rows sum to zero. This also explains why a diagonal derivative is determined by the rates to other states rather than being an additional transition intensity.

\sum_j p_{ij}(t,t+h)=1 \quad\Longrightarrow\quad \sum_j\left.\frac{\partial}{\partial h}p_{ij}(t,t+h)\right|_{h=0+}=0 \quad\Longrightarrow\quad q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t).

Equivalently, leaving state $i$ during a very short interval can occur through one of the possible destination states. The total instantaneous exit rate is the sum of those competing rates. The negative diagonal records the corresponding first-order decrease in the probability of remaining in $i$.

Finite-Horizon Probabilities Are Not Intensities

The derivative definition applies to an interval whose length tends to zero. A finite-horizon transition probability includes the accumulated effect of transitions over the whole interval. Its value is generally not a rate, and dividing it by the horizon generally does not recover the instantaneous intensity.

Worked Example: Inferring a Constant Rate from a Finite-Horizon Probability

In a hypothetical two-state model, state 0 is Active and state 1 is Dead. Death is absorbing, and the only transition is from Active to Dead. Assume the intensity is constant over time. The probability of being Dead two years after starting Active is $0.5$. Find the intensity from Active to Dead, and compare it with the finite-horizon probability divided by two.

Method. Under a constant intensity $\lambda$, the probability of remaining Active for $T$ years is $e^{-\lambda T}$. Therefore the probability of being Dead by time $T$ is $1-e^{-\lambda T}$. With $T=2$, $$ 0.5=1-e^{-2\lambda}, \qquad e^{-2\lambda}=0.5, \qquad \lambda=-\frac{\ln(0.5)}{2} =\frac{\ln 2}{2} \approx 0.3466\text{ per year}. $$

The finite-horizon probability divided by the horizon is $0.5/2=0.25$ per year. That is not the intensity: the model’s intensity is approximately $0.3466$ per year. The two quantities differ because the probability of death over two years accumulates as the chance of surviving Active decreases.

Check. Substituting the inferred rate gives $e^{-2\lambda}=e^{-\ln 2}=0.5$, so the death probability is $1-0.5=0.5$, as specified. The rate is rounded to four significant figures. The common mistake is to divide a finite-horizon probability by the horizon and call the result an intensity; that approximation is only a first-order local relation for a sufficiently short interval.

When the Initial Time Is Earlier Than the Endpoint

There is an important distinction between differentiating $P(t,t+h)$ at $h=0$ and differentiating $P(s,t)$ with respect to its endpoint $t$ while keeping $s$ fixed. In the first case, the derivative is the generator at the starting time. In the second case, the process may already have moved away from its initial state by time $t$, so the derivative is not generally the generator itself.

For a differentiable transition matrix in a time-inhomogeneous Markov model, the forward equation is $\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t)$. If $P(s,t)$ is invertible, this gives $Q(t)=P(s,t)^{-1}\frac{\partial}{\partial t}P(s,t)$. The inverse is essential: endpoint derivatives from an earlier starting time have already been weighted by the probabilities of being in each state at the endpoint.

Worked Example: Recovering the Current Rate from an Endpoint Derivative

A hypothetical two-state model has states Active and Dead, with Dead absorbing. At time $t_0$, the transition matrix from time 0 and its derivative with respect to the endpoint are $$ P(0,t_0)= \begin{pmatrix} 0.5&0.5\\ 0&1 \end{pmatrix}, \qquad \left.\frac{\partial}{\partial t}P(0,t)\right|_{t=t_0} = \begin{pmatrix} -0.1&0.1\\ 0&0 \end{pmatrix} $$ where the derivative is in years. Recover $Q(t_0)$.

Method. The initial time is fixed at 0, so the supplied derivative is an endpoint derivative, not a zero-length local derivative. Use the forward equation and multiply on the left by $P(0,t_0)^{-1}$. The inverse is $$ P(0,t_0)^{-1}= \begin{pmatrix} 2&-1\\ 0&1 \end{pmatrix}, \qquad Q(t_0)= \begin{pmatrix} 2&-1\\ 0&1 \end{pmatrix} \begin{pmatrix} -0.1&0.1\\ 0&0 \end{pmatrix} = \begin{pmatrix} -0.2&0.2\\ 0&0 \end{pmatrix}. $$

Thus the instantaneous intensity from Active to Dead at $t_0$ is $0.2$ per year, and Dead is absorbing. Check: multiplying back gives $$ \begin{pmatrix} 0.5&0.5\\ 0&1 \end{pmatrix} \begin{pmatrix} -0.2&0.2\\ 0&0 \end{pmatrix} = \begin{pmatrix} -0.1&0.1\\ 0&0 \end{pmatrix}, $$ which is the supplied derivative. The common mistake is to read $0.1$ directly as the current intensity. It is an endpoint derivative entry, and the forward equation shows why the transition matrix must first be accounted for.

Common Exam Traps

The main issue is identifying exactly which derivative or probability the question provides. Check the time arguments and the units before interpreting any matrix entry.

Using a finite-interval probability as a rate. $p_{ij}(t,t+T)$ is a probability for a specified $T>0$; the local intensity comes from the limit as $T$ decreases to zero.
Forgetting the diagonal sign. A generator diagonal is nonpositive and is the negative sum of the rates to distinct destination states.
Confusing endpoint and local derivatives. $\left.\frac{\partial}{\partial u}P(t,u)\right|_{u=t+}=Q(t)$, whereas $\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t)$ when $s$ is fixed.
Dropping the time unit. If time is measured in years, generator entries are per year. Changing the time unit changes the numerical value of the rates.

Key takeaway: The generator is the derivative of the transition matrix at a zero-length interval. Off-diagonal entries are instantaneous transition intensities, diagonal entries make each row sum to zero, and positive-horizon probabilities must not be mistaken for those rates.

Check Your Understanding

For each problem, identify whether the information is local or finite-horizon before calculating.

At a specified time, the local derivative matrix for states $A$ and $B$ is $\begin{pmatrix}-0.15&0.15\\0.06&-0.06\end{pmatrix}$ per year. What are the intensities from $A$ to $B$ and from $B$ to $A$? (Answer: $0.15$ per year and $0.06$ per year, respectively.)
A two-state constant-intensity model has an absorbing destination state. The probability of reaching it from the starting state within three years is $0.7$. What is the intensity? (Answer: $-\ln(0.3)/3\approx0.4013$ per year.)
In a three-state model, the local derivatives from state $C$ to $A$ and $B$ are $0.02$ and $0.07$ per year. Find the diagonal generator entry for state $C$, and state the row-sum check. (Answer: $q_{CC}=-0.09$ per year; $0.02+0.07-0.09=0$.)
For two states, $P(0,t_0)=\begin{pmatrix}0.5&0.5\\0&1\end{pmatrix}$ and $\frac{\partial}{\partial t}P(0,t)|_{t=t_0}=\begin{pmatrix}-0.05&0.05\\0&0\end{pmatrix}$. What is the current intensity from state 0 to the absorbing state? (Answer: $0.10$ per unit of time, since $P^{-1}P'=\begin{pmatrix}-0.10&0.10\\0&0\end{pmatrix}$.)
A model has time-varying intensity from state 0 to an absorbing state. Only the probability of reaching that state over a positive five-year horizon is supplied. Can the current intensity at the endpoint be determined from that probability alone? (Answer: No; a finite-horizon probability alone does not identify the endpoint intensity without additional model information.)

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