Total Exit Intensity From a State

Why Total Exit Intensity Matters

A state model can allow several different transitions out of the same state. For example, a hypothetical income-protection model might allow a person who is active to become disabled or to die. Each possible destination has its own transition intensity. The total exit intensity measures the instantaneous rate of leaving the current state by any of those routes.

This quantity is useful for checking an intensity matrix and for comparing how quickly different states are left at a given instant. It is a rate, not a finite-period probability: its units are the inverse of the model's time unit, such as per year. The actual probability of a transition over a finite period generally requires more information than the intensity at one instant.

Definition: In a continuous-time Markov state model, let $q_{ij}(t)$ be the transition intensity at time $t$ from state $i$ to a distinct state $j$. The total exit intensity from state $i$ at time $t$ is $$ \lambda_i(t)=\sum_{j\ne i}q_{ij}(t). $$ It is the sum over every distinct destination state. Under the row-sum-zero intensity-matrix convention, $\lambda_i(t)=-q_{ii}(t)$.

From Destination-Specific Rates to the Total

An intensity matrix records the allowed transitions in its off-diagonal entries. Those entries are nonnegative rates; a zero means that the model assigns no direct transition along that route. The diagonal entry is not an intensity of moving to another state. It is the negative of the sum of the other entries in its row, so that each row sums to zero.

The sum defining $\lambda_i(t)$ must include all distinct destinations, not just the most likely destination or the destination of greatest interest in a particular calculation. A state's total exit intensity can therefore be determined by adding its off-diagonal row entries, or by taking the negative of its diagonal entry when the matrix is valid and uses the stated convention.

q_{ij}(t)\ge 0 \quad (j\ne i), \qquad q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t), \qquad \lambda_i(t)=\sum_{j\ne i}q_{ij}(t)=-q_{ii}(t).

A short derivation follows directly from the definition of transition intensity. Over a short interval of length $h$, conditional on being in state $i$ at time $t$, the probability of moving to a particular destination $j$ is $q_{ij}(t)h+o(h)$. The events of moving to different destinations are disjoint. In a continuous-time Markov model with finite transition intensities, the probability of multiple transitions during this short interval is $o(h)$. Thus the probability of leaving state $i$ during the interval is the sum of the destination-specific probabilities, up to a remainder that vanishes after division by $h$.

P(\text{leave }i\text{ during }[t,t+h)\mid X(t)=i) = \sum_{j\ne i}\bigl(q_{ij}(t)h+o(h)\bigr)+o(h) = \lambda_i(t)h+o(h).

Dividing by $h$ and taking the limit as $h$ decreases to zero gives the total exit intensity. This reasoning concerns the instantaneous rate. It does not say that the probability of leaving over a longer interval is exactly the intensity multiplied by the interval length.

Worked Example: Total Exit Rate from an Active State

A hypothetical three-state model has states Active, Disabled, and Dead. Its constant transition intensities, measured per year, are shown below. Find the total exit intensity from each state and identify which of Active and Disabled has the greater total exit intensity.

Q= \begin{pmatrix} -0.024 & 0.018 & 0.006\\ 0.040 & -0.060 & 0.020\\ 0 & 0 & 0 \end{pmatrix}, \qquad (\text{Active},\text{Disabled},\text{Dead}).

Method. Add the off-diagonal entries in each row. Then check that each sum is the negative of the diagonal entry.

From Active, the possible destinations are Disabled and Dead. Therefore, $ \lambda_{\text{Active}}=0.018+0.006=0.024 $ per year. The diagonal check gives $ -q_{\text{Active,Active}}=-(-0.024)=0.024 $ per year.

From Disabled, the possible destinations are Active and Dead. Therefore, $ \lambda_{\text{Disabled}}=0.040+0.020=0.060 $ per year. The diagonal check gives $ -q_{\text{Disabled,Disabled}}=-(-0.060)=0.060 $ per year.

The row for Dead has no off-diagonal transitions, so its total exit intensity is $0+0=0$ per year; the diagonal check also gives $-0=0$. Comparing Active and Disabled, $0.060/0.024=2.5$, so the total exit intensity from Disabled is 2.5 times the total exit intensity from Active.

Check. The row sums are $-0.024+0.018+0.006=0$, $-0.060+0.040+0.020=0$, and $0$. The totals agree with the diagonal entries and are nonnegative. The comparison is between instantaneous rates, not finite-period probabilities.

Common mistake avoided: counting only the transition from Active to Disabled would give $0.018$ per year, not the total exit intensity. The transition from Active to Dead must also be included.

Comparing Outgoing Rates

The total exit intensity gives one summary rate for all routes out of a state. Individual off-diagonal entries still matter: they describe the rates for specific destinations. Comparing those entries answers which particular route has the greater instantaneous intensity; comparing row totals answers which state has the greater total rate of departure. These are different comparisons.

For instance, a state can have a relatively large intensity for one particular transition but still have a smaller total exit intensity than another state if that other state has several outgoing routes. Conversely, the largest total exit intensity does not by itself identify which destination is most likely over a future period. That question involves how the transition rates operate over the period, rather than just a single instantaneous total.

When intensities vary with time, the same sum applies at each time $t$. The result is then a time-dependent total exit intensity, $\lambda_i(t)$. It is important to evaluate all outgoing rates at the same time before adding them. A total calculated at one time is not automatically the total at another.

Worked Example: Comparing Three Possible Exits

A hypothetical policy model has an In-force state with direct transitions to Surrender, Claim, and Death. At a specified instant, the respective intensities are $0.012$, $0.020$, and $0.004$ per year. Find the total exit intensity, compare the three route-specific rates, and give the first-order approximation to the probability of any exit during the next $0.01$ year.

Method. First add every off-diagonal rate from In-force. For the short-period approximation, use the small-interval result $P(\text{any exit})=\lambda_i(t)h+o(h)$, so the first-order term is $\lambda_i(t)h$.

The total is $ \lambda_{\text{In-force}}=0.012+0.020+0.004=0.036 $ per year. Claim has the greatest route-specific intensity, since $0.020>0.012>0.004$. Its intensity is $0.020/0.012=5/3$ times the surrender intensity, and the surrender intensity is three times the death intensity.

The stated interval is $h=0.01$ year. The first-order approximation to the probability of any exit is $0.036(0.01)=0.00036$, or $0.036\%$. This is an approximation, not an exact finite-interval probability.

Check. The total $0.036$ per year is greater than each individual rate and has the same time unit. Multiplying by $0.01$ year gives a dimensionless probability approximation. The small-interval formula alone does not provide an exact probability over the full interval.

Common mistake avoided: using only the Claim intensity would produce the first-order term $0.020(0.01)=0.00020$, omitting the Surrender and Death routes. The correct total-exit approximation uses all three rates.

Worked Example: A Time-Varying Total at a Specified Instant

In a hypothetical long-term care model, a person in the Care state can recover to Independent or move to Dead. At time $t$, measured in years from the model's starting date, the intensities are $q_{\text{Care,Independent}}(t)=0.016+0.002t$ and $q_{\text{Care,Dead}}(t)=0.008+0.001t$, both per year. Find the total exit intensity at $t=4$, compare it with the total at $t=0$, and check the matrix diagonal at $t=4$.

Method. Evaluate each destination-specific intensity at the requested time, then sum the results.

At $t=4$, the recovery intensity is $0.016+0.002(4)=0.024$ per year. The death intensity is $0.008+0.001(4)=0.012$ per year. Hence $ \lambda_{\text{Care}}(4)=0.024+0.012=0.036 $ per year.

At $t=0$, the two intensities are $0.016$ and $0.008$ per year, so $ \lambda_{\text{Care}}(0)=0.016+0.008=0.024 $ per year. The total has increased by $0.036-0.024=0.012$ per year. Relative to its value at $t=0$, this is an increase of $0.012/0.024=0.5$, or 50%.

At $t=4$, the diagonal entry for Care must be the negative total, $q_{\text{Care,Care}}(4)=-0.036$ per year. The row then sums to $-0.036+0.024+0.012=0$.

Check. Both destination rates were evaluated at $t=4$ before adding. The result agrees with the negative diagonal and is larger than the total at $t=0$, as the specified rates increase with $t$.

Common mistake avoided: adding one rate at $t=4$ to the other rate at $t=0$ would mix two different instants. A time-dependent total must use rates evaluated at the same time.

Checks and Common Traps

For a valid intensity matrix with the row convention used here, each off-diagonal entry is nonnegative and each row sums to zero. These properties provide a direct check on a calculated total exit intensity. If the off-diagonal entries in a row sum to $0.031$ per year, for example, its diagonal entry must be $-0.031$ per year. A discrepancy means that a rate may have been omitted or that the matrix was transcribed incorrectly.

A particularly common error is to read the diagonal as if it were a transition intensity. Its negative is the total exit intensity; the diagonal itself is nonpositive and is not a rate to a destination. Another error is to include the diagonal in the sum over destinations. The definition sums over $j\ne i$, so the diagonal is excluded.

Keep the distinction between rate and probability clear. A total exit intensity of $0.04$ per year does not mean that the probability of leaving during any chosen period is $0.04$. For a short interval of length $h$, the leading approximation is $\lambda_i(t)h$; for a longer interval, the model's transition behavior over time must be taken into account.

Key takeaway: To find the instantaneous rate of leaving a state, add every off-diagonal intensity in that state's row. The result is nonnegative, has the model's inverse-time units, and equals the negative diagonal entry.

Check Your Understanding

For each problem, calculate the total exit intensity before making any requested comparison or approximation.

A hypothetical Recovery state has transitions to Active at $0.027$ per year and to Dead at $0.003$ per year. What is its total exit intensity? (Answer: $0.030$ per year.)
A row of an intensity matrix has entries $0.014$, $0.009$, and diagonal entry $-0.023$, with the two positive entries leading to distinct states. Verify the total exit intensity and row sum. (Answer: $0.023$ per year; row sum $0$.)
A hypothetical state has intensities to Withdrawal, Disability, and Death of $0.010$, $0.025$, and $0.005$ per year. Find the total and state how many times the Disability intensity is the Withdrawal intensity. (Answer: $0.040$ per year; $2.5$ times.)
At a specified instant, two transition intensities from a state are $0.018$ and $0.006$ per year. Give the first-order approximation to the probability of any exit during the next $0.02$ year. (Answer: $0.00048$, or $0.048\%$.)
At time $t$ years, a state's only two exit intensities are $0.012+0.001t$ and $0.006+0.002t$ per year. Find its total exit intensity at $t=3$, and the required diagonal entry at that instant. (Answer: $0.027$ per year; diagonal entry $-0.027$ per year.)

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