Valuation with Age-Varying Decrement Probabilities

Why Age-Specific Decrement Probabilities Matter

In a pension plan, the probability that an active member retires, withdraws, or dies need not remain constant from one age to the next. A valuation must therefore update the number or probability of members still active at each age using that age's decrement probabilities. Applying a single probability across several years can misstate both the timing and the expected value of benefits.

The calculation has two linked parts. First, find the probability that a member is active at the start of each age interval. Second, multiply that probability by the probability of each cause of exit during the interval and by the corresponding benefit. Discount each payment according to when it is made. The probabilities in a multiple-decrement table are conditional on being active at the start of the relevant interval.

Definition: For a member active at exact age $x$, let $q_x^{(j)}$ be the probability of exit from active status during $[x,x+1)$ due to decrement cause $j$. These are multiple-decrement probabilities: each is measured while the other causes operate simultaneously. The probability of remaining active throughout the interval is $$ p_x^{(\tau)}=1-\sum_j q_x^{(j)}, $$ where $\tau$ denotes the total decrement. For successive ages, the probability of being active at the start of the interval beginning at age $x+k$, given active status at age $x$, is $$ {}_kp_x^{(\tau)}=\prod_{r=0}^{k-1}p_{x+r}^{(\tau)}, \qquad {}_0p_x^{(\tau)}=1. $$

Valuing Payments at Age-Specific Exit

Assume a member is active at age $x$, the annual effective interest rate is $i$, and $v=1/(1+i)$. Suppose a payment $B_j(x+k+1)$ is made at the end of the interval $[x+k,x+k+1)$ if the member exits during that interval for cause $j$. The payment may be a lump sum or a value assigned to a pension commencing at exit. Its amount can depend on the attained age at exit.

The probability of reaching the start of interval $k$ active and then exiting for cause $j$ during that interval is ${}_kp_x^{(\tau)}q_{x+k}^{(j)}$. Since payment is at the interval's end, it is discounted for $k+1$ years. Thus the expected present value over $n$ intervals is

\operatorname{EPV}_x = \sum_{k=0}^{n-1}v^{k+1}\,{}_kp_x^{(\tau)} \sum_j q_{x+k}^{(j)}B_j(x+k+1).

This expression assumes that the listed causes exhaust the exits being modelled, the member is active at age $x$, probabilities are conditional on active status at each interval's start, and all included payments occur at its end. It does not directly apply to payments made at the instant of exit within the year or to a continuing pension valued as a stream of future payments unless the relevant payment timing or pension value is incorporated into $B_j$.

Why the Valuation Formula Holds

The formula follows by partitioning possible outcomes according to the interval of exit and the cause of exit. Within interval $k$, a member must first survive all decrements in the preceding intervals, then experience cause $j$ in the current interval. The probability of this sequence is the product of the successive conditional probabilities:

P(\text{active at }x+k\text{ and cause }j\text{ in interval }k) = \left(\prod_{r=0}^{k-1}p_{x+r}^{(\tau)}\right)q_{x+k}^{(j)} = {}_kp_x^{(\tau)}q_{x+k}^{(j)}.

The present value in that outcome is $v^{k+1}B_j(x+k+1)$. Multiplying the present value by the probability of the outcome gives its expected present value. Outcomes with different exit intervals or causes are mutually exclusive, so adding their expected present values gives the total. This also explains why active survival must include every decrement cause: any exit removes the member from the active population at subsequent ages.

Read the table by attained age.
For each interval, identify every cause-specific probability conditional on being active at its start.

Update active survival.
Subtract the sum of all cause probabilities from one, then multiply the resulting active probabilities across earlier intervals.

Attach the appropriate benefit.
For each cause and age, use the benefit amount payable under that specific exit event.

Weight and discount.
Multiply benefit by active survival and the current cause probability; discount for the payment time and sum across intervals and causes.

Worked Valuations

Worked Example: Two Ages and Three Exit Causes

A hypothetical pension plan values payments for a member active at age 60. The annual effective interest rate is $5\%$. All benefits are paid at the end of the year of exit. The age-60 probabilities are $q_{60}^{(R)}=0.08$, $q_{60}^{(W)}=0.12$, and $q_{60}^{(D)}=0.02$, for retirement, withdrawal, and death. At age 61 they are $0.10$, $0.15$, and $0.03$, respectively. Retirement pays $\$80{,}000$, withdrawal pays $\$12{,}000$, and death pays $\$30{,}000$, regardless of the exit age. Find the expected present value of payments during these two years.

Method. Calculate active survival over the first year, then value each year's cause-specific payments at the end of that year.

The probability of remaining active from age 60 to 61 is $1-0.08-0.12-0.02=0.78$. The age-61 probabilities sum to $0.28$, so the age-61 probability of remaining active is $1-0.10-0.15-0.03=0.72$. The latter is not needed for payments restricted to the first two years, but confirms that the age-61 probabilities are consistent.

At age 60, the expected payment conditional on being active at age 60 is $0.08(80{,}000)+0.12(12{,}000)+0.02(30{,}000)=6{,}400+1{,}440+600=\$8{,}440$. Its present value is $8{,}440/1.05=\$8{,}038.095238\ldots$.

At age 61, the conditional expected payment is $0.10(80{,}000)+0.15(12{,}000)+0.03(30{,}000)=8{,}000+1{,}800+900=\$10{,}700$. Only $0.78$ of the original members are active at age 61, so the expected payment measured from age 60 is $0.78(10{,}700)=\$8{,}346$. Its present value is $8{,}346/(1.05)^2=\$7{,}570.068027\ldots$.

Adding the two present values gives $\$8{,}038.095238\ldots+\$7{,}570.068027\ldots=\$15{,}608.163265\ldots$. The expected present value is $\$15{,}608.16$, rounded to the nearest cent. As a check, each interval's weighted expected payment uses probabilities that are conditional on active status at that interval's start.

Worked Example: A Second Exit Cause Changes Year-Two Survival

A hypothetical plan pays $\$10{,}000$ at the end of a year if an active member retires during that year. There are no other payments. The annual effective interest rate is $5\%$. In the first year, the retirement probability is $0.10$ and the withdrawal probability is $0.10$. In the second year, conditional on still being active, the retirement probability is $0.20$ and the withdrawal probability is $0.10$. Death probabilities are zero in both years. Find the expected present value over two years for a member active at the outset.

Method. Use the total probability of any exit—not only retirement—to update active membership. The first-year active probability is $1-0.10-0.10=0.80$.

The year-one retirement payment contributes $10{,}000(0.10)/1.05=\$952.380952\ldots$. For a year-two retirement payment, the member must remain active through year one and then retire during year two. Its contribution is $10{,}000(0.80)(0.20)/(1.05)^2=\$1{,}451.247166\ldots$.

The total is $\$952.380952\ldots+\$1{,}451.247166\ldots=\$2{,}403.628118\ldots$, or $\$2{,}403.63$, rounded to the nearest cent.

Common mistake and correction. Using $0.90$ as the year-two active probability ignores the $0.10$ withdrawal probability. That incorrect calculation gives $10{,}000(0.10)/1.05+10{,}000(0.90)(0.20)/(1.05)^2=\$2{,}585.034013\ldots$, or $\$2{,}585.03$. The correction is to subtract both first-year exits: $1-0.10-0.10=0.80$. This produces the correct value of $\$2{,}403.63$.

Worked Example: Age-Dependent Pension Values and Exit Payments

A hypothetical plan values three years for a member active at age 62. The annual effective interest rate is $4\%$, and payments occur at each year's end. Retirement probabilities at ages 62, 63, and 64 are $0.05$, $0.08$, and $0.12$. Withdrawal probabilities are $0.10$, $0.10$, and $0.08$; death probabilities are $0.02$, $0.03$, and $0.04$. At retirement, the member receives an annual pension of $\$2{,}000$, $\$2{,}200$, or $\$2{,}400$, respectively. For this valuation, the supplied pension-value factors at those retirement ages are 8, 9, and 10. Treat the pension value at retirement as the annual pension multiplied by the supplied factor. Withdrawal pays $\$5{,}000$, $\$6{,}000$, or $\$7{,}000$, respectively, and death pays $\$10{,}000$ at every age. Find the expected present value.

Method. First convert each retirement pension to its value at retirement, then use the age-specific multiple-decrement probabilities and recursively updated active survival.

The retirement values are $2{,}000(8)=\$16{,}000$, $2{,}200(9)=\$19{,}800$, and $2{,}400(10)=\$24{,}000$. The active probabilities for the three intervals are $1-0.05-0.10-0.02=0.83$, $1-0.08-0.10-0.03=0.79$, and $1-0.12-0.08-0.04=0.76$. Therefore the probabilities active at ages 62, 63, and 64 are $1$, $0.83$, and $0.83(0.79)=0.6557$.

In the first interval, the expected payment conditional on being active at age 62 is $0.05(16{,}000)+0.10(5{,}000)+0.02(10{,}000)=800+500+200=\$1{,}500$. Its present value is $1{,}500/1.04=\$1{,}442.307692\ldots$.

In the second interval, the conditional expected payment is $0.08(19{,}800)+0.10(6{,}000)+0.03(10{,}000)=1{,}584+600+300=\$2{,}484$. Weighting by the $0.83$ active probability gives $0.83(2{,}484)=\$2{,}061.72$. The present value is $2{,}061.72/(1.04)^2=\$1{,}906.176035\ldots$.

In the third interval, the conditional expected payment is $0.12(24{,}000)+0.08(7{,}000)+0.04(10{,}000)=2{,}880+560+400=\$3{,}840$. Weighting by the $0.6557$ active probability gives $0.6557(3{,}840)=\$2{,}517.888$. Its present value is $2{,}517.888/(1.04)^3=\$2{,}238.393263\ldots$.

The total is $\$1{,}442.307692\ldots+\$1{,}906.176035\ldots+\$2{,}238.393263\ldots=\$5{,}586.876991\ldots$. Thus the expected present value is $\$5{,}586.88$, rounded to the nearest cent. The calculation uses the age-specific pension amounts and the probability of reaching each age active; neither is replaced by a constant across the three years.

Interpreting the Result and Avoiding Traps

Age-varying probabilities matter because the value at each age depends on both the benefit payable at that age and the probability of still being an active member then. A later age can have a larger pension value but contribute less to the original member's expected present value because fewer members remain active and the payment is discounted for longer.

In an actual valuation, the benefit definition determines what belongs in $B_j$. For example, a retirement benefit can be represented by the value of the pension commencing at retirement if that value is given or has been calculated on the applicable basis. Do not multiply by an annuity factor a second time if $B_j$ already represents the value of the pension. Conversely, if the benefit is a periodic payment stream and no pension value has been provided, a single lump-sum payment formula is not sufficient: the stream's timing and value must also be accounted for.

Omitting an exit cause from active survival. Active status ends upon any modelled decrement, not only the cause that generates the benefit currently being valued.
Using unconditional probabilities as if they were conditional. A table probability for age $x$ applies to members active at age $x$. Multiply by the probability of reaching age $x$ active to obtain the probability measured from the valuation age.
Using the wrong discount period. A payment at the end of the second year is discounted by $v^2$, not $v$. A payment at exit within the year would require a timing assumption different from the year-end convention used here.
Holding probabilities or benefits fixed without support. Use the values specified for each attained age. Do not assume age-invariance merely to simplify the recursion.

For a finite table, a useful check is that each age's cause probabilities are nonnegative and sum to at most one. The remaining active probability must also be nonnegative. The active probability at a later age cannot exceed the active probability at the preceding age, since it is multiplied by a probability no greater than one. Finally, each cause-specific expected payment should be nonnegative when both the benefit and probabilities are nonnegative.

Key takeaway: For each age interval, multiply the age-specific cause probability by the probability of being active at its start, apply the benefit triggered by that cause, and discount to the valuation date. Update active survival by including every decrement cause at every prior age.

Check Your Understanding

Use year-end payments and the stated annual effective interest rate unless the problem says otherwise.

A member is active at age 58. In the next year, retirement, withdrawal, and death probabilities are $0.07$, $0.11$, and $0.02$. Retirement pays $\$15{,}000$; the other causes pay nothing. At $i=4\%$, find the present value of the expected payment. (Answer: $\$1{,}009.62$)
At $i=5\%$, a hypothetical plan pays $\$5{,}000$ on retirement at the end of either of two years. Year one has retirement probability $0.10$ and other-exit probability $0.15$. In year two, conditional on remaining active, retirement probability is $0.12$. Find the two-year expected present value. (Answer: $\$884.35$)
There is no interest. Retirement pays $\$4{,}000$ at each year-end. In year one, retirement and other-exit probabilities are $0.04$ and $0.06$; in year two they are $0.10$ and $0.10$; in year three the retirement probability is $0.15$. Find the expected present value of retirement payments over three years. (Answer: $\$952.00$)
At $i=5\%$, year-one retirement and other-exit probabilities are $0.06$ and $0.14$, and year-two probabilities conditional on active status are $0.09$ and $0.11$. Retirement payments are $\$12{,}000$ in year one and $\$16{,}000$ in year two. Find the expected present value. (Answer: $\$1{,}730.61$)
A table gives first-year retirement probability $0.08$, withdrawal probability $0.07$, and death probability $0.03$. What is the probability of being active at the start of year two, and why is it not $0.92$? (Answer: $0.82$; withdrawal and death also remove members from active status.)

What You'll Learn