Deriving the Diagonal Intensity Entry

Why the Diagonal Entry Is Negative

An intensity matrix records the instantaneous rates of movement between states. Its off-diagonal entries describe transitions from one state to another. The diagonal entry has a different role: it records the rate of probability loss from the state itself. This entry is negative whenever there is a positive chance of leaving the state, and its magnitude equals the total rate of departure.

That negative sign is not a negative transition rate. It is a consequence of probability accounting over a short interval: the probability of still being in the starting state decreases by the probabilities of moving to other states. The derivation also requires care about what a transition probability means. A process may end in a state after multiple jumps even if there is no direct jump to that state.

Definition: Consider a finite-state, time-homogeneous continuous-time Markov chain with finite transition intensities. Let $p_{ij}(h)$ be the probability that the process is in state $j$ after an interval of length $h$, given that it was in state $i$ at the start. For distinct states, the transition intensity is $q_{ij}=\lim_{h\downarrow 0}p_{ij}(h)/h$. The diagonal intensity is $q_{ii}=\lim_{h\downarrow 0}(p_{ii}(h)-1)/h$.

The parameter $h$ is measured in the same unit as the intensities. For example, if rates are stated per year, then $h$ must be measured in years. In a finite-state model with finite rates, the probability of two or more jumps during a sufficiently short interval is of smaller order than $h$. Thus, a route requiring two jumps does not contribute to the first-order rate of reaching its endpoint.

Deriving the Diagonal Intensity

Suppose the process starts in state $i$. At the end of the short interval it must be in some state: either still in $i$, or in one of the other states. Consequently, the endpoint probabilities satisfy

p_{ii}(h)=1-\sum_{j\ne i}p_{ij}(h).

For each $j\ne i$, the small-interval expansion is $p_{ij}(h)=q_{ij}h+o(h)$. Here, $o(h)$ denotes a term whose ratio to $h$ tends to zero as $h$ tends to zero. This expansion applies whether or not there is a direct jump from $i$ to $j$. If that direct jump is forbidden, then $q_{ij}=0$, and $p_{ij}(h)=o(h)$; the endpoint probability need not be exactly zero, because the process could reach $j$ through multiple jumps.

Substitute the expansions into the endpoint probability identity. There are finitely many states, so the sum of the remainder terms is still $o(h)$. This gives

\begin{aligned} p_{ii}(h) &=1-\sum_{j\ne i}\bigl(q_{ij}h+o(h)\bigr)\\ &=1-h\sum_{j\ne i}q_{ij}+o(h). \end{aligned}

Using the definition of the diagonal intensity now gives the result:

Theorem: For a finite-state continuous-time Markov model with finite transition intensities, the diagonal entry for state $i$ is $$ q_{ii}=-\sum_{j\ne i}q_{ij}. $$ The sum includes every direct outgoing intensity from state $i$. A forbidden direct transition contributes zero.

The derivation explains both the sign and the magnitude. Each outgoing intensity is nonnegative, so their sum is nonnegative and $q_{ii}$ cannot be positive. The magnitude $-q_{ii}$ is the total first-order rate of leaving $i$. If no transition can leave $i$, the sum is zero and $q_{ii}=0$. This does not mean that the process has a negative probability of remaining in $i$; it is a rate describing the first-order change in that probability.

Calculating and Interpreting the Entry

When calculating a diagonal entry, first list the states reachable by a direct jump from the state of interest. Then add their intensities and put a minus sign in front of the total. Do not include rates into the state from elsewhere: those describe incoming movement, not departure from the current state.

Worked Example: Leaving an Active State

In a hypothetical long-term disability model, a member in the active state can move directly to disabled at intensity $0.08$ per year or directly to dead at intensity $0.02$ per year. There are no other direct exits from active. Find the active-state diagonal intensity and the first-order small-interval probability of remaining active over $h=0.01$ year.

Method. Add the direct outgoing intensities, then negate the total. The intensity units are per year, so the interval is expressed in years.

The total outgoing intensity is

0.08+0.02=0.10\text{ per year}.

Therefore the diagonal entry is

q_{\text{active, active}}=-(0.08+0.02)=-0.10\text{ per year}.

For a short interval, the probability of remaining active is

\begin{aligned} p_{\text{active, active}}(h) &=1+q_{\text{active, active}}h+o(h)\\ &=1-(0.10)(0.01)+o(0.01)\\ &=0.999+o(0.01). \end{aligned}

Thus, $0.999$ is the first-order approximation to the probability of remaining active, not an exact finite-interval probability. The first-order exit contributions are $0.08(0.01)=0.0008$ to disabled and $0.02(0.01)=0.0002$ to dead; together they are $0.001$, matching the first-order decrease from $1$ to $0.999$.

Check. The diagonal is negative, and its magnitude $0.10$ equals the sum of the two direct outgoing rates. The common mistake avoided here is treating the diagonal as one of the outgoing transition rates rather than as the negative of their total.

Worked Example: Several Competing Exits

Suppose a hypothetical care model has a state called moderate care. From this state, a person can recover to active at intensity $0.12$ per year, move to intensive care at intensity $0.05$ per year, or die at intensity $0.03$ per year. There are no other direct exits. Find the diagonal intensity and the first-order endpoint probabilities for an interval of $h=1/12$ year.

Method. Include all direct destinations exactly once. First calculate the total outgoing intensity:

0.12+0.05+0.03=0.20\text{ per year}.

Hence the diagonal entry for moderate care is

q_{\text{moderate, moderate}}=-0.20\text{ per year}.

For $h=1/12$ year, the first-order contributions for each endpoint are

\begin{aligned} p_{\text{moderate, active}}(h) &=0.12\left(\frac{1}{12}\right)+o(h) =0.0100+o(h),\\ p_{\text{moderate, intensive}}(h) &=0.05\left(\frac{1}{12}\right)+o(h) =0.0041667+o(h),\\ p_{\text{moderate, dead}}(h) &=0.03\left(\frac{1}{12}\right)+o(h) =0.0025+o(h),\\ p_{\text{moderate, moderate}}(h) &=1-0.20\left(\frac{1}{12}\right)+o(h) =0.9833333+o(h). \end{aligned}

The displayed decimals are first-order terms; the $o(h)$ terms have not been set to zero. As a check, the first-order contributions sum to $0.0100+0.0041667+0.0025+0.9833333=1.0000$, with the apparent rounding in that sum understood. The unrounded terms also sum to one at first order.

Check. The interval has been converted to years before multiplying by rates per year. Using $h=1$ merely because the interval is one month would produce an incorrect scale. The diagonal magnitude is the full $0.20$, not just the death intensity.

Forbidden Direct Jumps and Endpoint Probabilities

The distinction between a direct transition and an endpoint event is important in a multi-state model. A zero off-diagonal intensity means there is no first-order direct jump along that route. It does not, by itself, mean the process cannot end in that state during a positive interval. If two jumps can take the process from $i$ to $j$, the endpoint probability can be positive, but under the finite-rate assumptions its contribution is $o(h)$.

Worked Example: A Forbidden Direct Transition

Consider a hypothetical model with states active $A$, disabled $D$, long-term care $L$, and dead $Z$. From $A$, direct transitions are allowed only to $D$ at intensity $0.04$ per year and to $Z$ at intensity $0.01$ per year. A direct transition from $A$ to $L$ is forbidden. From $D$, a transition to $L$ is allowed. Find $q_{AA}$, and explain whether $p_{AL}(h)$ must be zero.

Method. Use only the direct outgoing intensities from $A$ to calculate its diagonal entry. The direct intensity to $L$ is zero, so

\begin{aligned} q_{AA} &=-(q_{AD}+q_{AL}+q_{AZ})\\ &=-(0.04+0+0.01)\\ &=-0.05\text{ per year}. \end{aligned}

However, $p_{AL}(h)$ need not be zero. The process can move from $A$ to $D$, then from $D$ to $L$, within the same interval. This requires at least two jumps, so its probability is of order $o(h)$ as $h$ tends to zero. Therefore $q_{AL}=\lim_{h\downarrow0}p_{AL}(h)/h=0$, even though $p_{AL}(h)$ may be positive for a nonzero interval.

Check. The diagonal depends on the first-order rates of direct exits from $A$; the possible two-jump route to $L$ does not add a first-order exit intensity. The common mistake avoided is to claim that a forbidden direct jump forces the endpoint probability to be exactly zero.

Common Errors to Avoid

Using a positive sign. If the outgoing rates total $0.07$ per year, the diagonal is $-0.07$ per year, not $+0.07$. The probability of remaining in the state decreases to first order.
Leaving out an exit. Sum all direct outgoing rates, including recovery, transfer, and death when those transitions are allowed. The diagonal is not based only on the most important or most likely exit.
Including incoming rates. A transition into $i$ from another state does not contribute to $q_{ii}$. The calculation concerns departures from $i$, conditional on starting in $i$.
Mixing time units. Multiply a rate per year by an interval in years. Convert months or days before using a small-interval approximation.
Confusing a direct jump with an endpoint. A forbidden direct route has zero intensity, but an endpoint reached through multiple jumps may have a nonzero probability over a positive interval. Its contribution is smaller than first order.

The diagonal entry is therefore a compact summary of how quickly probability leaves a state at the start of an interval. Its magnitude is useful for checking whether all exits have been counted, while its sign reflects the decline in the probability of remaining in that state. These calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each model, calculate the requested diagonal intensity. Keep the rate units consistent and distinguish direct jumps from endpoint probabilities.

A hypothetical annuity model has two direct exits from an eligible state: entry to a claim state at $0.025$ per year and death at $0.015$ per year. What is the diagonal intensity for the eligible state? (Answer: $-0.040$ per year.)
From a temporary disability state, direct recovery occurs at $0.18$ per year, transition to permanent disability at $0.06$ per year, and death at $0.02$ per year. Find the diagonal entry. (Answer: $-0.26$ per year.)
A hypothetical long-term-care state has direct exits to recovery at $0.09$ per year and death at $0.04$ per year. Find the first-order probability of remaining in that state over $h=0.05$ year. (Answer: $1-0.13(0.05)=0.9935+o(0.05)$.)
From active $A$, direct transitions are allowed to disabled $D$ at $0.03$ per year and dead $Z$ at $0.01$ per year. There is no direct transition from $A$ to care $C$, but $A\to D\to C$ is possible. Find $q_{AA}$, and state whether $p_{AC}(h)$ must be zero. (Answer: $q_{AA}=-0.04$ per year; no, $p_{AC}(h)$ can be positive through multiple jumps, but is $o(h)$.)
A model gives rates of $0.08$ per year from a state to recovery and $0.04$ per month from that state to death. Find the diagonal intensity in per-year units. (Answer: $-(0.08+12(0.04))=-0.56$ per year.)

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