Building an Alive-and-Dead Intensity Matrix

From a Mortality Assumption to a Two-State Model

A basic life-contingency model distinguishes two possible states: a person is alive or dead. Its intensity matrix records the instantaneous rates of movement between those states. Building the matrix correctly requires more than placing a mortality rate in one cell: the state order, the possible transitions, the diagonal entries, and the absorbing-state row must all agree with the model.

This tutorial considers a model in which death is the only way to leave the alive state and death is permanent. The mortality assumption may be constant or may vary with age or time. The model describes state transitions; it does not by itself specify an insurance benefit or its value.

Definition: Let the states be ordered as $A=\text{alive}$ and $D=\text{dead}$. The intensity matrix $Q(t)=(q_{ij}(t))$ uses row $i$ for the current state and column $j$ for the destination state. For distinct states, $q_{ij}(t)$ is the instantaneous intensity of transition from $i$ to $j$. Each diagonal entry is the negative of the total intensity of exit from that state, so each row sums to zero.

For an individual aged $x$ at the model origin, let $\mu_{x+t}$ denote the mortality intensity at attained age $x+t$. The notation makes the age basis explicit: after $t$ years, the individual is $t$ years older. If mortality is specified directly as a function of model time, write that intensity as $\mu(t)$ instead.

Key result: In the alive-and-dead model, death is the only possible exit from life, and there are no transitions out of death. With the order alive, dead, the intensity matrix is $$ Q(t)= \begin{pmatrix} -\mu_{x+t} & \mu_{x+t}\\ 0 & 0 \end{pmatrix}. $$ If mortality intensity is constant at $\mu$, replace $\mu_{x+t}$ by $\mu$ at every time.

Why the Matrix Has This Form

The off-diagonal entry in the alive row is the intensity of the only permitted transition, alive to dead. The diagonal entry must account for the total rate of exit from alive. Since that total is $\mu_{x+t}$, the diagonal is its negative. From the dead state there is no permitted exit, so both the exit intensity and the diagonal entry are zero.

The row-sum condition can also be seen from the infinitesimal definition. Over a short interval of length $h$, conditional on being alive at time $t$, the probability of death is $\mu_{x+t}h+o(h)$, where $o(h)/h$ tends to zero as $h$ tends to zero. The probability of remaining alive is therefore $1-\mu_{x+t}h+o(h)$. The first-order changes in these two probabilities add to zero:

\bigl(1-\mu_{x+t}h\bigr)+\mu_{x+t}h=1, \qquad q_{AA}(t)+q_{AD}(t)=-\mu_{x+t}+\mu_{x+t}=0.

This argument justifies the negative diagonal as the balancing entry for the row. The $o(h)$ terms matter for exact short-interval probabilities, but not for identifying the instantaneous rates in the intensity matrix. The corresponding dead row is $(0,0)$, which also sums to zero.

Fix the state order.
Write alive first and dead second. Rows will describe the current state; columns will describe the destination.

List permitted transitions.
In this model, alive to dead is permitted. Dead to alive is not permitted.

Enter the off-diagonal intensity.
Put the mortality intensity in the alive-to-dead position and zero in the dead-to-alive position.

Complete each row.
Set each diagonal entry to the negative sum of the other entries in its row. In particular, the entire dead row is zero.

Worked Constructions

Worked Example: Converting a Rate per Person-Year

A hypothetical model follows a person aged 48. At the instant being considered, the mortality intensity is 15 deaths per 1,000 alive person-years. Construct the intensity matrix in the order alive, dead, express the rate per year, and interpret its entries.

Method. Convert the stated rate to a decimal intensity, enter it as the alive-to-dead intensity, and use the row-sum condition for the diagonal. This avoids confusing a rate per 1,000 person-years with a probability of 15%.

Step 1: Convert the rate. Fifteen deaths per 1,000 alive person-years corresponds to $15/1000=0.015$ per year. Thus the mortality intensity at this instant is $0.015$ per year.

Step 2: Specify the permitted transitions. The person can move from alive to dead at intensity $0.015$ per year. There is no transition from dead to alive.

Step 3: Complete the rows. The alive row must sum to zero, so its diagonal is $-0.015$. The dead state has no exit, so its row is $(0,0)$. Therefore,

Q= \begin{pmatrix} -0.015 & 0.015\\ 0 & 0 \end{pmatrix} \quad\text{per year}.

Interpretation and check. While alive, the instantaneous intensity of death is $0.015$ per year. The first row sums to $-0.015+0.015=0$, and the second row sums to $0+0=0$. The intensity is a rate, not a statement that the probability of dying over any specified year is exactly $0.015$.

Worked Example: An Age-Dependent Mortality Intensity

A hypothetical model begins at age 60. Its mortality intensity at attained age $y$ is $\mu_y=0.006+0.0003(y-60)$ per year for ages from 60 through 70. Construct the intensity matrices at durations $t=0$, $t=4$, and $t=10$, where duration is measured in years from age 60.

Method. Evaluate the mortality formula at attained age $60+t$, then use that value in the alive row. This avoids substituting duration for age without accounting for the starting age.

Step 1: Identify attained ages. At durations $0$, $4$, and $10$, the attained ages are $60+0=60$, $60+4=64$, and $60+10=70$.

Step 2: Evaluate the intensities. At age 60, $\mu_{60}=0.006+0.0003(60-60)=0.006$. At age 64, $\mu_{64}=0.006+0.0003(64-60)=0.006+0.0012=0.0072$. At age 70, $\mu_{70}=0.006+0.0003(70-60)=0.006+0.003=0.009$.

Step 3: Construct the three matrices. At each time, the off-diagonal alive-to-dead entry is the calculated mortality intensity; the alive diagonal is its negative, and the dead row is zero.

Q(0)= \begin{pmatrix} -0.006 & 0.006\\ 0 & 0 \end{pmatrix}, \qquad Q(4)= \begin{pmatrix} -0.0072 & 0.0072\\ 0 & 0 \end{pmatrix}, \qquad Q(10)= \begin{pmatrix} -0.009 & 0.009\\ 0 & 0 \end{pmatrix}.

Check. The alive row sums to zero in each matrix: $-0.006+0.006=0$, $-0.0072+0.0072=0$, and $-0.009+0.009=0$. Each dead row also sums to zero. The intensity increases with attained age, as specified by the formula.

Worked Example: A Piecewise-Constant Mortality Assumption

A hypothetical two-year model starts with an alive individual. Mortality intensity is $0.010$ per year for $0\leq t<1$, and $0.018$ per year for $1\leq t\leq2$. Death is permanent. Write the intensity matrix for each time interval, state the matrix at $t=0.6$ and $t=1.4$, and verify the rows.

Method. Use the stated intensity separately on each interval, while retaining the same state order and transition structure. This avoids averaging different interval rates into a single constant rate.

Step 1: Assign each time to an interval. Since $0.6$ lies in $[0,1)$, its mortality intensity is $0.010$ per year. Since $1.4$ lies in $[1,2]$, its mortality intensity is $0.018$ per year.

Step 2: Write the interval matrices. For $0\leq t<1$, the alive diagonal is $-0.010$; for $1\leq t\leq2$, it is $-0.018$. The death row is zero in both intervals:

Q(t)= \begin{cases} \begin{pmatrix}-0.010&0.010\\0&0\end{pmatrix},&0\leq t<1,\\[6pt] \begin{pmatrix}-0.018&0.018\\0&0\end{pmatrix},&1\leq t\leq2. \end{cases}

Step 3: Read the requested times. Thus $Q(0.6)$ is the first matrix and $Q(1.4)$ is the second. At the boundary $t=1$, the stated interval convention assigns the intensity $0.018$ per year.

Check. The first alive row sums to $-0.010+0.010=0$; the second sums to $-0.018+0.018=0$. Both dead rows sum to zero. The interval change affects the mortality intensity, not the identity of the states or the absorbing nature of death.

Interpreting and Checking the Entries

The four entries have distinct roles. With the order alive, dead, $q_{AA}(t)$ is negative and balances the total exit intensity from life; $q_{AD}(t)$ is the mortality intensity; $q_{DA}(t)=0$ rules out recovery from death; and $q_{DD}(t)=0$ reflects that there is no exit from death. A zero diagonal for death does not mean that death is ignored. It means the total exit intensity from that state is zero.

Entry	Meaning in this model	Value
$q_{AA}(t)$	Negative total intensity of exit from alive	$-\mu_{x+t}$
$q_{AD}(t)$	Intensity of transition from alive to dead	$\mu_{x+t}$
$q_{DA}(t)$	Intensity of transition from dead to alive	$0$
$q_{DD}(t)$	Negative total intensity of exit from dead	$0$

A compact verification is to check both the transition assumptions and the arithmetic. The off-diagonal entries must be nonnegative, forbidden transitions must be zero, and every row must sum to zero. In this model those conditions give the absorbing-state row $(0,0)$ as well as the alive row $(-\mu,\mu)$.

Common Mistake: Giving the Dead State an Exit

A frequent incorrect construction is to use a negative mortality intensity in the dead-state diagonal, for example

Q_{\text{wrong}}= \begin{pmatrix} -\mu & \mu\\ 0 & -\mu \end{pmatrix}.

This matrix has a dead row sum of $-\mu$, not zero. It therefore violates the row-sum condition and incorrectly assigns a nonzero exit intensity to death without identifying any destination. The correction is to use $q_{DD}=0$, because the model has no transition out of death:

Q_{\text{correct}}= \begin{pmatrix} -\mu & \mu\\ 0 & 0 \end{pmatrix}.

Another error is to put $+\mu$ on the alive diagonal. That would make the alive row sum $2\mu$, rather than zero. The diagonal is the negative total exit intensity, not another transition rate. Finally, the value of $\mu$ must be expressed in the time unit used by the model. A rate per month cannot be inserted as though it were a rate per year without conversion.

Key takeaway: With states ordered alive, dead, a mortality-only model has first row $(-\mu,\mu)$ and second row $(0,0)$. The mortality intensity may depend on age or time, but the signs, row-sum rule, and absorbing-state row follow from the model assumptions.

Check Your Understanding

For each problem, use the state order alive, dead. Assume death is permanent and no other transition is possible.

A hypothetical mortality intensity is $0.011$ per year at the time of observation. Write the intensity matrix and report both row sums. (Answer: $\begin{pmatrix}-0.011&0.011\\0&0\end{pmatrix}$; row sums $0,0$.)
A model starts at age 55 and has $\mu_y=0.004+0.0002(y-55)$ per year. What is the intensity matrix at duration 5 years? (Answer: attained age 60, $\mu_{60}=0.005$, so $Q(5)=\begin{pmatrix}-0.005&0.005\\0&0\end{pmatrix}$.)
A hypothetical rate is reported as 9 deaths per 1,000 alive person-years. Convert it to an annual intensity and construct the matrix. (Answer: $0.009$ per year; $Q=\begin{pmatrix}-0.009&0.009\\0&0\end{pmatrix}$.)
Mortality intensity is $0.007$ per year for $0\leq t<2$, and $0.013$ per year for $2\leq t\leq4$. Give $Q(1.5)$ and $Q(3)$, and state the intensity at $t=2$ under the stated convention. (Answer: $Q(1.5)=\begin{pmatrix}-0.007&0.007\\0&0\end{pmatrix}$; $Q(3)=\begin{pmatrix}-0.013&0.013\\0&0\end{pmatrix}$; at $t=2$, the intensity is $0.013$ per year.)
A proposed matrix is $\begin{pmatrix}-0.016&0.016\\0&-0.016\end{pmatrix}$. Identify the invalid row and give its correction under the alive-and-dead assumptions. (Answer: The dead row is invalid because it sums to $-0.016$; replace its second entry by zero, giving $\begin{pmatrix}-0.016&0.016\\0&0\end{pmatrix}$.)

Entry	Meaning in this model	Value
\(q_{AA}(t)\)	Negative total intensity of exit from alive	\(-\mu_{x+t}\)
\(q_{AD}(t)\)	Intensity of transition from alive to dead	\(\mu_{x+t}\)
\(q_{DA}(t)\)	Intensity of transition from dead to alive	\(0\)
\(q_{DD}(t)\)	Negative total intensity of exit from dead	\(0\)

What You'll Learn