Bounds for Continuous and Discrete Insurance Values

Why Bounds Are Useful

A death benefit paid at the moment of death is generally discounted for less time than the same benefit paid at the end of the policy year of death. That timing difference makes the continuous insurance value at least as large as the year-end value when interest is nonnegative. But the difference cannot be arbitrarily large: the two payment times are less than one year apart. The resulting bounds are useful both for understanding values and for detecting calculation errors.

The comparison is meaningful only when the two contracts use the same mortality model and the same death benefit in each policy year. The only difference is payment timing. The results below concern death benefits; a different benefit pattern, such as a payment that changes continuously within a year, requires a separate comparison.

Definition: Let $T_x$ be the complete future lifetime of a life aged $x$, and let $K_x=\lfloor T_x\rfloor$ be the curtate future lifetime. A death in policy year $k+1$ means $k\leq T_x<k+1$. Suppose the benefit paid for a death in that year is $B_k\geq 0$. The continuous value pays $B_k$ at the moment of death; the year-end value pays that same $B_k$ at time $k+1$. Both values use the same survival model and an effective annual interest rate $i\geq 0$, with $v=1/(1+i)$.

The Bounds

For a death at time $t$ in policy year $k+1$, the continuous payment has present value $B_kv^t$, while the year-end payment has present value $B_kv^{k+1}$. Since $k\leq t\leq k+1$, and $v^t$ decreases as $t$ increases when $i\geq 0$, the continuous payment’s present value lies between $B_kv^{k+1}$ and $B_kv^k$. Also, $v^k=(1+i)v^{k+1}$. Thus the present value for a death in that policy year is no more than $1+i$ times its year-end counterpart.

Taking expectations over all possible years of death preserves these inequalities. For a unit whole-life death benefit, the standard notation is $A_x$ for payment at the end of the year of death and $\bar{A}_x$ for payment at the moment of death.

\text{Year-end value}\ \leq\ \text{continuous value}\ \leq\ (1+i)(\text{year-end value})

A_x\leq \bar{A}_x\leq (1+i)A_x

The same bounds apply to term insurance and to benefits that vary by policy year, provided the amount $B_k$ is the same for the two payment timings and is nonnegative. For a term contract, include only the policy years during which death is covered. If $i=0$, the bounds coincide: the timing of payment does not change its present value. The inequalities as stated are not guaranteed for a negative interest rate, because the discount factor would no longer decrease with time.

Why the result holds: The bounds do not require a particular distribution of death within a policy year. They follow from the fact that the continuous payment time is between the start and end of that year. The mortality model determines how likely each year of death is, but it cannot move the payment outside that interval.

Worked Applications

The examples use a constant force of mortality $\lambda$, so $S_0(x+t)=e^{-\lambda t}$ for $t\geq 0$. This assumption is specified for calculation; the bounds themselves do not depend on constant-force mortality.

Worked Example: Checking a Whole-Life Value

A hypothetical whole-life contract pays $1,000 at death. The effective annual interest rate is $i=0.10$, and the force of mortality is constant at $\lambda=\ln 2$ per year. Calculate the year-end and continuous values, then check the bounds.

Method. Calculate the year-end value by summing the discounted probability of death in each policy year. Calculate the continuous value by integrating the discounted death density. Then compare both values with the bounds.

Year-end value. The one-year survival probability is $e^{-\lambda}=1/2$, so the probability of death in each successive year is $0.5(0.5)^k$, for $k=0,1,2,\ldots$. Here $v=1/1.10=10/11$. For a unit benefit,

A_x=\sum_{k=0}^{\infty}0.5(0.5)^k v^{k+1} =\frac{0.5v}{1-0.5v} =\frac{5/11}{1-5/11} =\frac{5}{6}.

Thus the year-end value for $1,000 is $1000(5/6)=\$833.33$.

Continuous value. Under the stated constant-force model, the density of $T_x$ is $\lambda e^{-\lambda t}$. The present value is discounted by $e^{-\delta t}$, where $\delta=\ln(1.10)=0.0953102$. Therefore,

\bar{A}_x =\int_0^\infty e^{-\delta t}\lambda e^{-\lambda t}\,dt =\frac{\lambda}{\lambda+\delta} =\frac{0.6931472}{0.6931472+0.0953102} \approx 0.879118.

The continuous value for $1,000 is approximately $1000(0.879118)=\$879.12$, rounded to the nearest cent.

Check. The bounds in dollars are

\$833.33\leq \$879.12\leq 1.10(\$833.33)=\$916.67.

The calculated value is inside the required range. The check also confirms the expected direction: paying at death, rather than at year-end, gives the larger value.

Worked Example: A Two-Year Contract with Changing Benefits

A hypothetical two-year term contract pays $1,000 for a death in the first policy year and $2,000 for a death in the second. The effective annual interest rate is $10\%$, and the force of mortality is $\lambda=\ln 2$ per year. Calculate both values and verify the bounds.

Method. First calculate the year-end value from the death probability and payment time in each covered year. Then integrate the discounted death density separately over each year, using that year’s benefit amount.

Year-end value. The probability of death in year 1 is $1/2$; the probability of death in year 2 is $(1/2)(1/2)=1/4$. With $v=10/11$,

D=1000\left(\frac12\right)v+2000\left(\frac14\right)v^2 =1000\left(\frac{5}{11}\right)+2000\left(\frac{25}{484}\right) =\$867.77.

The displayed total is rounded to the nearest cent; the unrounded value is approximately $867.7686.

Continuous value. Set $\delta=\ln(1.10)$ and $s=\lambda+\delta=0.7884574$. The discounted density is $\lambda e^{-st}$. The common integral factor is

\frac{\lambda}{s} =\frac{0.6931472}{0.7884574} \approx 0.879118.

For year 1, integrate over $0\leq t<1$. Since $e^{-s}=1/2.2=5/11$, the contribution is

1000\frac{\lambda}{s}(1-e^{-s}) =1000(0.879118)\left(1-\frac{5}{11}\right) \approx \$479.52.

For year 2, integrate over $1\leq t<2$. Since $e^{-2s}=(5/11)^2=25/121$, the contribution is

2000\frac{\lambda}{s}(e^{-s}-e^{-2s}) =2000(0.879118)\left(\frac{5}{11}-\frac{25}{121}\right) \approx \$435.93.

Adding the contributions gives a continuous value of approximately $\$479.52+\$435.93=\$915.45$.

Check. The bounds are

\$867.77\leq \$915.45\leq 1.10(\$867.77)=\$954.55.

The result is valid for the changing benefits because each year’s amount is applied identically under both payment timings.

Worked Example: Using the Bound to Check a Term Value

A hypothetical two-year term insurance pays $1,000 at death. The effective annual interest rate is $4\%$, and the force of mortality is constant at $\lambda=-\ln(0.75)$ per year, giving annual survival probability $0.75$. A calculation reports a continuous value of $450. Is that value possible?

Method. Compute the year-end value from the two annual death probabilities. The bounds then determine the full range in which the continuous value must lie; no continuous integral is needed to reject an impossible result.

Here $v=1/1.04=25/26$. The probabilities of death in years 1 and 2 are $0.25$ and $0.75(0.25)=0.1875$, respectively. Thus the year-end value is

D=1000\left(0.25v+0.1875v^2\right) =1000\left(0.25\frac{25}{26}+0.1875\frac{625}{676}\right) \approx \$413.74.

The continuous value must satisfy

\$413.74\leq \bar{D}\leq 1.04(\$413.74)=\$430.29.

The reported $450 exceeds the upper bound, so it cannot be correct under the stated contract, interest rate, and mortality model. The bound detects an error without identifying which step caused it. The continuous value must be recalculated.

Common Mistakes and Practical Checks

The bound is a quick diagnostic, not a substitute for matching the two contracts being compared. Before using it, confirm the payment amount for each policy year, the covered years, and the payment time under each valuation. A maturity benefit paid on survival is not a death benefit; if the same maturity payment is included in both contracts, it contributes the same discounted amount to both values and can be added to each side of the comparison.

Reversing the payment-time comparison. A common incorrect claim is that continuous value is at most the year-end value because it is “paid earlier.” Earlier payment has a larger present value when $i\geq 0$, so the correct lower bound is the year-end value.
Using the wrong upper factor. The upper bound is $(1+i)$ times the year-end value, not $1/(1+i)$ times that value. The earliest payment in policy year $k+1$ is at time $k$, and $v^k=(1+i)v^{k+1}$.
Comparing different benefits. If benefit amounts differ between the two contracts, the result may not apply. Check that each death in policy year $k+1$ has the same amount $B_k$ under both payment timings.
Including uncovered years. A term contract has no death benefit after its term unless the contract specifies one. Do not include later death probabilities in its value.

For a computed value, a result below the year-end value or above $(1+i)$ times that value signals a likely error in payment timing, discounting, the covered period, or the benefit amounts. Passing the check does not prove that a calculation is correct, but failing it proves that the stated assumptions and result are inconsistent.

Check Your Understanding

Use the same-benefit and same-mortality comparison throughout. Amounts in the answers are rounded to the nearest cent.

A year-end death-benefit value is $700 and the effective annual interest rate is $5\%$. Give the allowable range for the continuous value. (Answer: $700.00 to $735.00, inclusive.)
A one-year term benefit is $1,200. The effective annual interest rate is zero, and the force of mortality is constant at $\lambda=\ln 2$. Find the continuous and year-end values. (Answer: Both are $600.00.)
A one-year term benefit is $1,000, the annual death probability is $0.20$, and $i=0.10$. Find the year-end value and the bounds for the continuous value. (Answer: Year-end value $181.82; continuous value is between $181.82 and $200.00, inclusive.)
A two-year term benefit is $500, the annual death probability is $0.40$, and $i=0.05$. Assume a constant force of mortality $\lambda=-\ln(0.60)$, so the annual survival probability is $0.60$. Calculate the year-end value and the bounds for the continuous value. (Answer: Year-end value $299.32; continuous value is between $299.32 and $314.29, inclusive.)
A two-year term pays $500 for a death in year 1 and $1,500 for a death in year 2. The annual death probability is $0.25$, and $i=0.04$. Find the year-end value and the bounds for the continuous value. (Answer: Year-end value $380.22; continuous value is between $380.22 and $395.43, inclusive.)

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