Life Insurance and Life Annuity
There are many types of Life Insurance products (see Sherris Section 8.3), with the most common being:
- Whole of Life: Pays a benefit at death, regardless of when death occurs.
- Term Insurance: Pays a benefit at death, but only if death occurs in a certain period.
- Endowment: Pays a benefit at death if death occurs in a certain period, but also pays the benefit if the life is alive at the end of the period.
Term Insurance - Claim Payments
- $T(x)$ is the future lifetime (a continuous random variable) for a life $(x)$.
- $X$ is the age-at-death (also a continuous random variable).
\(T(x) = X - x\)
- Will require probability of this random variable taking values $0,1,\cdots,n-1$.
- $x+k$ age last birthday, with integer $x+k$.
Term Insurance - Benefit Payment
Assume the insured life is aged $x$ at purchase, the death benefit is of amount $S$, and it is paid at the end of the year of death, but only if death occurs within $n$ years of purchase. Assume an annual (effective) rate of interest $i$.
- If death occurs in the first year of the policy, when the life is aged $x$, then the present value of the death benefit would be:
- If death occurs when the life is aged $x+1$ then the present value of the benefit would be:
In general, if death occurs when the life is aged $x+k$ last birthday where $k=0,1,2,\ldots n-1$ then the present value of the benefit payment at the end of the year of death would be:
\[\text{PV[Payment]} = \begin{cases} Sv^{k+1} & \text{for } k=0,1,2,\ldots, n-1 \\ 0 & \text{for } k \geq n \end{cases}\]Term Insurance - EPV of Claim Payments
What are the associated probabilities with the values of the present value of the payment?
Let $K(x)$ be the discretized future lifetime random variable for age $x$ (i.e., $K(x) = \lfloor T(x) \rfloor$). Then:
\[\text{PV}[Payment] = \begin{cases} Sv^{k+1} & \text{with probability Pr} [K(x) = k] \\ & \text{for } k = 1,2,3,\ldots,n-1 \\ 0 & \text{with probability Pr} [K(x) \geq n] = ~_{n} p_x \end{cases}\]What is $\Pr[K(x) = k]$?
\Pr[K=k] = \Pr[k \leq T(x) < k+1] = _{k}p_{x}q_{x+k}
Expected Present Value of Benefit Payment
The Expected Present Value of the benefit payment for this term insurance is then:
\sum_{k=0}^{n-1} S v^{k+1} \cdot _{k}p_{x}q_{x+k} = S \cdot \sum_{k=0}^{n-1} v^{k+1} \cdot _{k}p_{x}q_{x+k} = S \cdot A_{x:\overline{n}|}^{1}
Where $A_{x:\overline{n}\rvert}^{1}$ is standard actuarial notation (for the expected PV of a term life insurance paying a benefit of 1, and covering the next $n$ years of a life aged $x$).
Example 8.5
Determine the expected present value of the claim payments for a 5-year term insurance on a life aged 20 with a sum insured of $100,000 using the following mortality probabilities and a 6% p.a. effective interest rate.
Mortality Probabilities
| Age | $q_{\text{age}}$ |
|---|---|
| 20 | 0.00192 |
| 21 | 0.00181 |
| 22 | 0.00160 |
| 23 | 0.00138 |
| 24 | 0.00118 |
Solution
To calculate the EPV of 1 payable on death within $n$ years:
A_{20:\overline{5}|}^{1} = \sum_{k=0}^{4}v^{k+1}\left( _{k}p_{20}q_{20+k}\right)
where:
- $v^{k+1} = \left( \frac{1}{1.06}\right)^{k+1}$
- $_{k}p_{20}$ is assumed as follows, with $_{0}p_{20}=1$
Detailed Calculations
| Age(x+k) | $q_{x+k}$ | $k$ | $v^{k+1}$ | $_{k}p_{x}$ | $_{k}p_{x} q_{x+k}$ |
|---|---|---|---|---|---|
| 20 | 0.00192 | 0 | 0.94340 | 1.00000 | 0.00192 |
| 21 | 0.00181 | 1 | 0.89000 | 0.99808 | 0.00181 |
| 22 | 0.00160 | 2 | 0.83962 | 0.99627 | 0.00159 |
| 23 | 0.00138 | 3 | 0.79209 | 0.99468 | 0.00137 |
| 24 | 0.00118 | 4 | 0.74726 | 0.99331 | 0.00117 |
The EPV of the claim payments for a sum insured of 100,000 is then:
100,000 \cdot A_{20:\overline{5}|}^{1} = 672.06
Life Annuities
- A life annuity is a stream of regular payments, paid as long as a life is alive.
- Consider a life aged $x$. Call $P_k$ the payment they receive at time $k$, for $k=0,1,2,\ldots$
- The Expected Present Value of a life annuity due (paid in advance) of 1 per period to this life aged $x$ is:
\ddot{a}_{x} = \mathbb{E} [ \text{PV}(P_0) + \text{PV}(P_1) + \ldots + \text{PV}(P_{\omega-x-1})] = \sum_{k=0}^{\omega-x-1}v^k \cdot _{k}p_{x}
Furthermore:
\ddot{a}_{x} = 1 + \sum_{k=1}^{\omega -x-1} v^k \cdot _{k}p_{x} = 1 + \sum_{k=1}^{\omega -x-1} v \cdot v^{k-1} \cdot p_{x} \cdot _{k-1}p_{x+1} = 1 + v \cdot p_{x} \left[ \sum_{k=0}^{\omega -x-2} v^k \cdot _{k}p_{x+1} \right] = 1 + v \cdot p_{x} \cdot \ddot{a}_{x+1}
- Note that $\ddot{a}_{x} = 1 + a_x.$ Do you see why?
If the payments are restricted to $n$ payments maximum:
\ddot{a}_{x:\overline{n}|} = \sum_{k=0}^{n-1} v^k \cdot _{k}p_{x} = 1 + a_{x:\overline{n-1}|}
Principle of Equivalence
- The `actuarially fair’ Principle of Equivalence states that:
\text{EPV of premiums} = \text{EPV of claims} + \text{EPV of expenses}
- A complication in Life Insurance is that annual premiums $P$ are only paid while a life is alive, so that the expected present value of $n$ premiums $P$, for a life age $x$, is:
P \cdot \sum_{k=0}^{n-1} v^{k} \left( _{k}p_{x} \right) = P \cdot a_{x:\overline{n}|}
-
Note: premium payments are always made at the beginning of the cover, hence $a_{x:\overline{n}|}$
-
Present value of the premium due at age $x+k$ is:
\begin{cases}
Pv^{k} & \text{for } k=0,1,2,\ldots n-1 \\
0 & \text{for } k\geq n
\end{cases}
- For a life aged $x$, the probability that they will be alive at age $x+k$ and will pay the premium $_{k}p_{x}$
Recurrence Relations
Recurrence relations for the expected present value of benefits for an $n$ year term insurance on a life aged $x$. Let $_{t}B_{x:\overline{n}|}$ be the expected present value of benefits at age $x+t$:
If the life dies during the year, with probability $q_{x+t}$, then the benefit of $S$ is paid at the end of the year. The expected present value of the benefits will then be equal to $S$ since no future payments are made once the life has died.
-
If the life survives to the end of the year, with probability $p_{x+t}$, then the expected present value of the benefits will equal that for a life aged $x+t+1$, this is just $_{t+1}B_{x:\overline{n}|}$.
-
EPV at the start of the year - divide the end of year value by $1+i$:
_{t}B_{x:\overline{n}|} = \frac{q_{x+t}S + p_{x+t} \left( _{t+1}B_{x:\overline{n}|} \right)}{1+i}
-
For the $n$ year term insurance on a life aged $x$, at age $x+n$ the expected value of the benefit will be zero since the benefit is only paid up to age $x+n$.
-
At $t=n$ we have $_{n}B_{x:\overline{n}|}=0$.
Over the final year of the policy:
_{n-1}B_{x:\overline{n}|} = \frac{q_{x+n-1}S + p_{x+n-1} \cdot 0}{1+i} = \frac{q_{x+n-1}S}{1+i}
- Repeating the recurrence:
_{n-2}B_{x:\overline{n}|} = \frac{q_{x+n-2}S + p_{x+n-2} \frac{q_{x+n-1}S}{1+i}}{1+i} = q_{x+n-2} \frac{S}{1+i} + p_{x+n-2} q_{x+n-1} \frac{S}{(1+i)^{2}}
- At age $x$ we have:
_{0}B_{x:\overline{n}|} = \frac{q_{x}S + p_{x} \left( _{1}B_{x:\overline{n}|} \right)}{1+i} = S \sum_{k=0}^{n-1} v^k \left( _{k}p_{x} q_{x+k} \right)
Example 8.6 & 8.7
Use the Principle of Equivalence to determine the annual premium for a 5-year term insurance on a 20-year-old male with a sum insured of 100,000 using the following mortality probabilities and a 6% p.a. effective interest rate. Initial expenses are 0.5% of the sum insured and renewal expenses are 100 per premium payment.
| age | $q_{\text{age}}$ |
|---|---|
| 20 | 0.00192 |
| 21 | 0.00181 |
| 22 | 0.00160 |
| 23 | 0.00138 |
| 24 | 0.00118 |
Solution
- We must find the EPV of premiums, so we must find:
a_{20:\overline{5}|} = \sum_{k=0}^{n-1} v^k \left( _{k}p_{20} \right)
| age(x+k) | $q_{x+k}$ | k | $v^k$ | $_{k}p_x$ |
|---|---|---|---|---|
| 20 | 0.00192 | 0 | 1.00000 | 1.00000 |
| 21 | 0.00181 | 1 | 0.94340 | 0.99808 |
| 22 | 0.00160 | 2 | 0.89000 | 0.99627 |
| 23 | 0.00138 | 3 | 0.83962 | 0.99468 |
| 24 | 0.00118 | 4 | 0.79209 | 0.99331 |
- Let $P$ denote the annual premium. EPV is:
P a_{20:\overline{5}|} = P \times 4.45021.
- EPV of the claims payment is:
100000 a_{20:\overline{5}|} = 100000 \times 0.0067206 = 672.06.
- Expected present value of the sum of initial and renewal expenses is:
0.005 \times 100000 + 100 a_{20:\overline{5}|} = 500 + 100 \times 4.45021 = 945.02.
- Equating [1] = [2] + [3] (principle of equivalence):
P = \frac{672.06 + 945.02}{4.45021} = 363.37.
Valuation of Policy Liabilities
Policy value of a liability for a life insurance policy is the
\text{EPV of future claims and expenses} - \text{EPV of future premiums}.
Denote the value of this reserve by $_{t}V_{x:\overline{n}|}$ the EPV of the policy liability
- at age $x+t$ (CONDITIONAL on survival to age $x+t$)
- for the term insurance on a life aged $x$ for a term of $n$ years
- with sum insured payable at the end of the year of death
Term life insurance
We have:
{}_{t}V_{x:\overline{n}|} = S \cdot {}_{t+1} A^1_{x+t:\overline{n-t}|} - (P-E) \cdot \ddot{a}_{x+t:\overline{n-t}|}
(by definition)
= v \left[ q_{x+t} \cdot S + p_{x+t} \cdot {}_{t+1}V_{x:\overline{n}|} \right] -(P-E)
(noting that there are only two possible outcomes over the next year)
Note that we must have:
{}_{0}V_{x:\overline{n}|} = {}_{n}V_{x:\overline{n}|} = 0.
Example 8.10
An insurance company sells 5 year term insurances on 20 year old males with a sum insured of 100,000 for a premium of 363.37. Assume the following mortality probabilities, a 6% p.a. effective interest rate, initial expenses of 0.5% of the sum insured and renewal expenses of 100 per premium payment.
| Age | $q_{x}$ |
|---|---|
| 20 | 0.00192 |
| 21 | 0.00181 |
| 22 | 0.00160 |
| 23 | 0.00138 |
| 24 | 0.00118 |
Determine the expected value of the policy liability for this term insurance as at the end of each year for the five years of the policy term using a recursive formula.
Solution
The recurrence formula:
{}_{t}V_{x:\overline{n}|} = \frac{q_{x+t} S + p_{x+t} \left[ {}_{t+1}V_{x:\overline{n}|} \right]}{(1+i)} - (P-E)
Summarized in the following table:
| Age | k | $q_x$ | $_{k}V_{x:\overline{n}|}$ |
|---|---|---|---|
| 20 | 0 | 0.00192 | 0.00 |
| 21 | 1 | 0.00181 | -443.68 |
| 22 | 2 | 0.00160 | -372.80 |
| 23 | 3 | 0.00138 | -276.43 |
| 24 | 4 | 0.00118 | -152.05 |
| 25 | 5 | – | 0.00 |
Sign of the Reserves
- In the example, the expected value of the policy liability is negative!
- The liability is effectively an asset for the life insurance company (How? Why?)
- Since initial expenses are effectively loaned to the policyholder, this is like an asset that is repaid from the premium - this makes a policy look like an asset in the early years.
- Also the probability of death is decreasing over this age range - level premiums are less than the expected cost of death cover in the early years.
- The liability is effectively an asset for the life insurance company (How? Why?)
- Example 8.8 has positive reserves for low expenses.
- In general, the magnitude of expenses is small enough that reserves are positive.
References
Introduction Level
- Principles of Actuarial Science by Michael Sherris. 2010 Edition. Publisher: Cengage Learning. ISBN: 0170188213, APN: 9780170188210. Coverage includes Chapter 8. Ideal for those beginning their journey in actuarial studies.
Advanced Level
- Actuarial Mathematics for Life Contingent Risks by Dickson, C.M.D., Hardy, M.R., Waters, H.R. 2020 Edition. Publisher: Cambridge University Press. ISBN: 978-1-108-47808-3. Covers Chapters 1-8, focusing on deep actuarial methods and applications.