Why Project Reserves by State?
A multi-state contract can have different future premiums and benefits depending on whether the insured is active, disabled, or in another state. A single reserve value for the whole contract therefore does not describe its future obligations in every condition. Numerical projection separates two related calculations: move probabilities forward to see how the insured population may occupy states, and calculate reserves backward to value future cash flows conditional on each current state.
The calculations require a consistent time convention. In this tutorial, reserves are measured immediately before any premium due at the start of a duration. Premiums are paid at that time, transition benefits are paid at the end of the duration, and the next reserve is measured at the next duration boundary. A reserve is future benefits less future premiums, conditional on the state at the valuation time. It can be negative when the present value of future premiums exceeds that of future benefits.
A transition benefit is associated with the particular movement \(i\to j\); it need not be the same for every destination. A state-dependent payment, by contrast, may be expressed through \(B_t(i,j)\) for all transitions into that state. A terminated or dead state usually has no future premiums or benefits, so its reserve is often zero, but the contract terms determine the terminal value and any payment on entry.
Forward Probability and Cash-Flow Projection
Write the probabilities of occupying the states at time \(t\) as a row vector \(\boldsymbol{\pi}_t\), and collect the transition probabilities in a matrix \(M_t\), whose \(i,j\) entry is \(p_{ij}^{(t)}\). The forward projection is
Each row of \(M_t\) must sum to 1, including transitions to absorbing states. The probability vector must also sum to 1 after each projection. Expected premiums at time \(t\) are \(\sum_i \pi_t(i)P_t(i)\); expected transition benefits paid at \(t+1\) are \(\sum_i\pi_t(i)\sum_j p_{ij}^{(t)}B_t(i,j)\). Discount these expected flows to the same valuation date when performing an aggregate present-value check.
This recursion assumes that the stated transition probabilities, benefits, premiums, and interest rate apply over the duration as specified. If premiums or benefits occur at a different time, their timing must be adjusted rather than inserted unchanged into this formula. For example, a premium paid at the end of the year belongs in the end-of-duration cash flow, not as an immediate deduction.
A Practical Projection Sequence
List every possible state, the valuation times, and when premiums and benefits are paid. State the terminal reserves.
Multiply the current probability vector by the duration's transition matrix. Check that its entries are nonnegative and sum to 1.
Apply the current state probabilities to premiums and transition probabilities to transition benefits. Keep beginning-of-duration and end-of-duration flows distinct.
Start with the terminal conditions, then apply the reserve recursion to each state at each earlier duration.
Compare the initial reserve with the present value of projected expected benefits less premiums. Investigate any difference before rounding.
Worked Projections
Worked Example: Two-State Term Cover
A hypothetical two-year contract has an active state \(A\) and an absorbing dead state \(Z\). The insured starts active. The death probability is 0.10 in each year, independently of duration; the effective annual interest rate is 5%. A premium of $30 is paid at the start of each year in which the insured is active. A death benefit of $200 is paid at the end of the year of death. All reserves at time 2 are zero. Find the projected state probabilities and the initial active-state reserve.
Method. Project probabilities forward, then use the terminal values to calculate reserves backward. The transition matrix for each year is
After year 1, \(\boldsymbol{\pi}_1=(0.90,0.10)\). After year 2, \(\boldsymbol{\pi}_2=(0.81,0.19)\). Each vector sums to 1. The probability of death during year 2 is \(0.90(0.10)=0.09\); cumulative death probability by time 2 is \(0.10+0.09=0.19\).
Let \(v=1/1.05=0.9523809524\). At time 1, the insured is active with one year remaining. Thus
The dead-state reserve is zero. At time 0, the premium is due immediately; at year-end the death benefit is paid on death, while survival leads to the active reserve at time 1:
To check, expected premiums have present value \(30+0.90(30)/1.05=55.71428571\). Expected benefits have present value \(20/1.05+18/1.05^2=35.37414966\). Benefits less premiums equal \(-20.34013605\), matching the recursion. Rounded to cents, \(V_0(A)=-\$20.34\).
Worked Example: Active, Disabled, and Dead States
Consider a hypothetical two-year contract with active state \(A\), disabled state \(D\), and absorbing dead state \(Z\). In each year, \(A\) moves to \(A\) with probability 0.80 and to \(D\) with probability 0.20. From \(D\), the probabilities of moving to \(A,D,Z\) are 0.25, 0.65, and 0.10. An active insured pays a $50 premium at the start of each year; a disabled insured pays none. At year-end, a $40 benefit is paid on any transition into \(D\), and a $300 benefit is paid on transition \(D\to Z\). There are no other benefits. The effective annual interest rate is 5%, and all time-2 reserves are zero. Find the projected probabilities and \(V_0(A)\).
Method. The matrix, with state order \(A,D,Z\), is
Starting from \(\boldsymbol{\pi}_0=(1,0,0)\), the first projection is \(\boldsymbol{\pi}_1=(0.80,0.20,0)\). The second is
The entries sum to 1 in both years. At time 1, one year remains. From \(A\), the only benefit-producing transition is \(A\to D\):
From \(D\), there is no premium. The expected year-end benefit is \(0.65(40)+0.10(300)=26+30=56\). Therefore
At time 0, the premium is $50. The end-of-year expected benefit is \(0.20(40)=8\), and the next reserve is \(V_1(A)\) or \(V_1(D)\) according to the destination state:
The aggregate check gives expected premiums \(50+0.80(50)/1.05=88.09523810\). Expected benefits are \(8/1.05+17.6/1.05^2=23.58276644\): the year-2 expected benefit is \(0.80(0.20)(40)+0.20[0.65(40)+0.10(300)]=6.4+11.2=17.6\). Benefits less premiums equal \(-64.51247166\). Rounded to cents, \(V_0(A)=-\$64.51\).
Worked Example: Time-Varying Mortality and a Maturity Benefit
A hypothetical two-year contract begins with an active insured and has an absorbing dead state. The death probabilities are 0.10 in year 1 and 0.20 in year 2. The effective annual interest rate is 4%. A premium of $20 is paid at time 0 and a further $25 is paid at time 1 if the insured is active. Death produces a $100 benefit at the end of the year of death. If the insured survives to time 2, a maturity benefit of $80 is paid then. There is no reserve after time 2. Calculate the reserve at times 1 and 0.
Method. Work backward from time 2. At time 1, the second-year expected benefit, conditional on being active, is \(0.20(100)+0.80(80)=20+64=84\). Hence
At time 0, year-1 death produces $100 with probability 0.10; survival with probability 0.90 leads to the time-1 active reserve. Thus
The forward active-state probabilities are 1.0000 at time 0, 0.9000 at time 1, and \(0.9000(0.8000)=0.7200\) at time 2. For a direct check, expected benefits have present value \(10/1.04+0.90(84)/1.04^2=9.61538462+69.89644970=79.51183432\). Premiums have present value \(20+0.90(25)/1.04=41.63461538\). Their difference is \(37.87721894\), agreeing apart from displayed precision. Rounded to cents, \(V_0(A)=\$37.88\).
Common Traps and Checks
A frequent error is to substitute the probability of entering a state for the probability of the transition that triggers a benefit. In the three-state example, the $300 payment is triggered only by \(D\to Z\), not by every insured who is disabled at the beginning of the year. Its contribution conditional on state \(D\) is \(0.10(300)\), not \(0.65(300)\) or \(0.20(300)\). The transition-specific benefit must be multiplied by the probability of that transition from the current state.
Another trap is mixing the forward and backward directions. Probabilities are projected from the initial distribution toward later durations; reserves are calculated from terminal values toward the valuation date. A further common mistake is to use a premium due at the start of a year as though it were paid at year-end. Under the convention here, it is subtracted immediately and is not discounted. A premium due only if the insured survives to the next duration instead belongs inside the discounted expected end-of-duration cash flow.
Check Your Understanding
For each problem, use the stated timing convention and round monetary answers to the nearest cent.
- A one-year active/dead contract has death probability 0.15, an effective annual interest rate of 4%, an immediate active-state premium of $18, and a $100 death benefit at year-end. The terminal reserves are zero. Find \(V_0(A)\). (Answer: \(-\$3.58\))
- A two-year active/dead contract has death probabilities 0.20 and 0.10 in years 1 and 2, respectively. Interest is 5% effective annually; a $25 premium is due at each year start while active, and death pays $120 at year-end. Terminal reserves are zero. Calculate \(V_1(A)\) and \(V_0(A)\). (Answer: \(V_1(A)=-\$13.57,\;V_0(A)=-\$12.48\))
- There are active, disabled, and dead states. The initial distribution is \((0.60,0.40,0)\). During one year, active moves to active or disabled with probabilities 0.70 and 0.30; disabled moves to active, disabled, or dead with probabilities 0.20, 0.70, and 0.10; dead remains dead. Find the distribution after one year. (Answer: \((0.5000,0.4600,0.0400)\))
- For one year, an active insured pays a $10 premium immediately. The year-end transition is active with probability 0.90 or disabled with probability 0.10. A $50 benefit is paid on transition to disabled. At time 1 the active reserve is $20 and the disabled reserve is zero. Interest is 5% effective. Find the initial reserve. (Answer: \(\$11.90\))
- A two-year active/dead contract has death probabilities 0.10 and 0.20, a 5% effective annual interest rate, and a $15 premium at each year start while active. Death pays $80 at year-end and all time-2 reserves are zero. Find \(V_1(A)\), then \(V_0(A)\). (Answer: \(V_1(A)=\$0.24,\;V_0(A)=-\$7.18\))