Expected Time Spent at Each Care Level

Why Expected Time in a Care State Matters

For a long-term care model, the probability of being in a care state at a particular duration is useful, but it does not by itself describe how much time an insured is expected to spend receiving care. A person may enter care, recover, move to another care level, or die before the end of the period. To measure expected time in each state, we use the entire path of occupancy probabilities over the horizon.

This quantity is also an exposure measure. If a hypothetical contract pays a constant amount per year while the insured occupies a specified care state, expected time in that state can be multiplied by the payment rate to obtain expected undiscounted benefits. This is not a present value: discounting payments over time is a separate calculation.

Definition: Let $J(t)$ denote the insured's state at time $t$, and let $j$ be a care state. Over a fixed horizon $T$, the time spent in state $j$ is $$ \mathcal{T}_j(T)=\int_0^T \mathbf{1}_{\{J(t)=j\}}\,dt, $$ where the indicator is 1 when the insured is in state $j$ and 0 otherwise. Its expected value is the expected time in state $j$ over $[0,T]$.

The Occupancy-Integral Formula

Write $p_j(t)=P(J(t)=j)$ for the probability of occupying state $j$ at time $t$. The expected time in the state is the area under its occupancy-probability curve from time 0 to time $T$:

E[\mathcal{T}_j(T)]=\int_0^T p_j(t)\,dt.

This result applies whether the model has one care state or many, and whether the insured can move between care states. The probabilities must be based on the model's initial-state distribution and transition assumptions. For a continuous-time Markov chain with constant transition intensities, the occupancy probabilities can be obtained from the state-probability equations; once they are known, integrate each care state's probability separately.

Model conditions: The horizon $T$ is fixed, and $p_j(t)$ describes the probability of occupying state $j$ at each time. For a continuous-time model, transition intensities specify the rates of movement between states. The occupancy-integral formula itself does not require constant intensities or the Markov property, but the method used to obtain $p_j(t)$ depends on the chosen model.

The formula follows directly by writing time in the state as an integral of indicators. Because the indicator is nonnegative, expectation and integration can be interchanged:

Represent time in the state.
At each instant, the indicator contributes one unit of time per unit of duration if the insured is in state $j$, and contributes zero otherwise. Thus $\mathcal{T}_j(T)=\int_0^T \mathbf{1}_{\{J(t)=j\}}dt$.

Take expectations.
Interchanging expectation and integration gives $E[\mathcal{T}_j(T)]=\int_0^T E[\mathbf{1}_{\{J(t)=j\}}]dt$.

Use the indicator's expectation.
The expected value of an indicator is the probability of its event, so $E[\mathbf{1}_{\{J(t)=j\}}]=P(J(t)=j)=p_j(t)$. Therefore $E[\mathcal{T}_j(T)]=\int_0^T p_j(t)dt$.

This argument also explains why state occupancy is integrated over time rather than assessed only at the horizon. A person who is in care for most of the year but dies just before its end still contributes nearly a full year of care-state time.

Worked Example: One Care State With Mortality

A hypothetical insured begins in care state $C$. The only possible transition is from $C$ to death, with constant mortality intensity $0.40$ per year. Find the expected time in $C$ during the next 1.5 years. Give the answer to four decimal places.

Method. With a constant exit intensity of $0.40$, the probability of remaining in $C$ at time $t$ is $p_C(t)=e^{-0.40t}$. Integrating this occupancy probability gives

E[\mathcal{T}_C(1.5)] =\int_0^{1.5}e^{-0.40t}\,dt =\frac{1-e^{-0.40(1.5)}}{0.40}.

The exponent is $0.40(1.5)=0.60$, and $e^{-0.60}=0.5488116361$. Hence

E[\mathcal{T}_C(1.5)] =\frac{1-0.5488116361}{0.40} =\frac{0.4511883639}{0.40} =1.1279709098\text{ years}.

To four decimal places, the expected time is 1.1280 years. As a check, this is less than the 1.5-year horizon, as it must be. The probability of still being in care at the horizon is $0.5488$, but that endpoint probability is not the expected time in care.

Sequential Care States

When transitions between care levels are possible, each state's occupancy probability reflects both entry and exit. Consider a model with care states $A$ and $B$, and an absorbing death state. Suppose the insured begins in $A$, can deteriorate from $A$ to $B$, and can die from either care state. For constant intensities, the probability of being in $A$ is the probability of no transition out of $A$ by time $t$. The probability of being in $B$ accounts for every possible time of transition from $A$ to $B$.

Worked Example: Expected Time and Benefit Exposure Across Two Care Levels

A hypothetical insured starts in care state $A$. The transition intensity from $A$ to $B$ is $0.50$ per year, and the mortality intensity from $A$ is $0.10$ per year. In state $B$, the mortality intensity is $0.20$ per year, with no recovery or further care transition. Find expected time in each care state over one year. Then calculate expected undiscounted benefits if hypothetical continuous benefit rates are $20,000 per year in $A$ and $36,000 per year in $B$. Give the monetary result to the cent.

Step 1: Find occupancy probabilities. The total exit intensity from $A$ is $0.50+0.10=0.60$ per year, so

p_A(t)=e^{-0.60t}.

To be in $B$ at time $t$, the insured must move from $A$ to $B$ before time $t$, then remain in $B$ until time $t$. With transition intensity $0.50$ from $A$ and exit intensity $0.20$ from $B$, this gives

p_B(t)=\int_0^t 0.50e^{-0.60s}e^{-0.20(t-s)}\,ds =\frac{0.50}{0.60-0.20}\left(e^{-0.20t}-e^{-0.60t}\right) =1.25\left(e^{-0.20t}-e^{-0.60t}\right).

Step 2: Integrate the occupancy of $A$.

E[\mathcal{T}_A(1)] =\int_0^1 e^{-0.60t}\,dt =\frac{1-e^{-0.60}}{0.60} =\frac{1-0.5488116361}{0.60} =0.7519806065\text{ years}.

Step 3: Integrate the occupancy of $B$. Since $e^{-0.20}=0.8187307531$,

\begin{aligned} E[\mathcal{T}_B(1)] &=1.25\left(\int_0^1e^{-0.20t}\,dt-\int_0^1e^{-0.60t}\,dt\right)\\ &=1.25\left(\frac{1-0.8187307531}{0.20}-0.7519806065\right)\\ &=1.25(0.9063462345-0.7519806065)\\ &=1.25(0.1543656280) =0.1929570350\text{ years}. \end{aligned}

Step 4: Calculate expected benefits. A constant continuous benefit rate in a state is multiplied by expected time in that state. Thus expected undiscounted benefits are

\begin{aligned} 20{,}000(0.7519806065)+36{,}000(0.1929570350) &=15{,}039.61213+6{,}946.45326\\ &=21{,}986.06539. \end{aligned}

The expected benefit is $21,986.07 to the cent. The method avoids treating the two care-state probabilities as if they were fixed for the whole year: each is integrated over its changing occupancy. As a probability check at time 1, $p_A(1)=0.5488116361$, $p_B(1)=1.25(0.8187307531-0.5488116361)=0.3373988963$, and the remaining probability is death, approximately $0.1137894676$. These values sum to 1, subject to rounding.

Initial Distributions and Multiple Care States

The initial state need not be known with certainty. If an insured begins in state $j$ with probability $q_j$, the occupancy probability at a later time incorporates that initial distribution. Expected time in each state is still found by integrating its resulting occupancy probability. In a model with several care states, calculate the integral separately for each state; the total expected care time is the sum of those state-specific expected times.

Worked Example: A Mixed Initial State Distribution

A hypothetical cohort has initial probability $0.40$ of being in care state $A$ and $0.60$ of being in care state $B$. There are no transitions between care states. The only exit from $A$ is death, at constant intensity $0.50$ per year; the only exit from $B$ is death, at constant intensity $0.25$ per year. Find the expected time in each care state over two years, and the total expected time in care. Give times to four decimal places.

Conditional on starting in $A$, the probability of remaining there at time $t$ is $e^{-0.50t}$. Including the initial probability $0.40$, the unconditional occupancy probability is $p_A(t)=0.40e^{-0.50t}$. Therefore

\begin{aligned} E[\mathcal{T}_A(2)] &=0.40\int_0^2e^{-0.50t}\,dt\\ &=0.40\left(\frac{1-e^{-1}}{0.50}\right)\\ &=0.40\left(\frac{1-0.3678794412}{0.50}\right)\\ &=0.40(1.2642411176) =0.5056964470\text{ years}. \end{aligned}

For state $B$, $p_B(t)=0.60e^{-0.25t}$. Since $0.25(2)=0.50$ and $e^{-0.50}=0.6065306597$,

\begin{aligned} E[\mathcal{T}_B(2)] &=0.60\int_0^2e^{-0.25t}\,dt\\ &=0.60\left(\frac{1-e^{-0.50}}{0.25}\right)\\ &=0.60\left(\frac{1-0.6065306597}{0.25}\right)\\ &=0.60(1.5738773612) =0.9443264167\text{ years}. \end{aligned}

The total expected time in care is $0.5056964470+0.9443264167=1.4500228637$ years, or 0.5057 years in $A$, 0.9443 years in $B$, and 1.4500 years in care in total. The total is below the two-year horizon, as required. The initial probabilities are included in both occupancy curves; omitting them would instead calculate conditional expected times given a particular starting state.

Common Exam Trap: Endpoint Probability Is Not Expected Time

A frequent error is to multiply the probability of occupying a state at the end of the horizon by the length of the horizon. In the one-state example, that incorrect calculation would be $1.5e^{-0.60}=1.5(0.5488116361)=0.8232174542$ years. It is not the expected time in care: it uses only the probability at the endpoint and ignores the probabilities at all earlier times. The correct calculation is the area under the full curve, $(1-e^{-0.60})/0.40=1.1279709098$ years.

Another trap is using a transition intensity as though it were the total rate of leaving a state. In the two-care-state example, the total exit intensity from $A$ is $0.50+0.10=0.60$, not $0.50$: death is also an exit. Using only the deterioration intensity would overstate the time spent in $A$ and distort the probability of reaching $B$.

Key takeaway: Expected time in a care state over a fixed horizon is the integral of that state's occupancy probability over the horizon. For a constant continuous benefit rate $b_j$ payable while in state $j$, the corresponding expected undiscounted benefit is $b_jE[\mathcal{T}_j(T)]$. Check that expected times are nonnegative and no greater than the horizon, and that any total across mutually exclusive care states is consistent with the model.

Check Your Understanding

For each problem, integrate the stated occupancy probability or derive it from the given constant transition intensities. Times are in years; round time results to four decimal places and monetary results to the cent.

A hypothetical insured starts in care state $C$, with death as the only exit and constant intensity $0.20$ per year. Find expected time in $C$ over three years. (Answer: 2.2559 years)
An insured starts in care state $A$, with constant total exit intensity $0.40$ per year. Find expected time in $A$ over one year. (Answer: 0.8242 years)
An insured starts in $A$. The intensity from $A$ to $B$ is $0.30$ per year, the mortality intensity from $A$ is $0.10$, and the mortality intensity from $B$ is $0.20$. There are no other transitions. Find expected time in each state over two years. (Answer: $E[\mathcal{T}_A(2)]=1.3767$ years; $E[\mathcal{T}_B(2)]=0.4076$ years)
An insured begins in $A$ with probability $0.25$ and in $B$ with probability $0.75$. There are no transitions between care states; death intensities are $0.30$ in $A$ and $0.10$ in $B$. Find expected time in each state over one year. (Answer: $E[\mathcal{T}_A(1)]=0.2160$ years; $E[\mathcal{T}_B(1)]=0.7137$ years)
An insured starts in $A$. The transition intensity from $A$ to $B$ is $0.40$, the mortality intensity from $A$ is $0.10$, and the mortality intensity from $B$ is $0.20$, all per year. Over one year, hypothetical continuous benefit rates are $10,000 per year in $A$ and $15,000 per year in $B$. Find the expected undiscounted benefit. (Answer: $10,257.54)

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