3 Life Annuities
3.1 Supplementary Concepts
The random variable \(Y\) represents the present value of a unit payable periodically for as long as the annuityholder is alive. For example, if a unit is paid at a continuous rate on survival, then \(Y=\bar{a}_{\enclose{actuarial}{T}}\), where \(T\) is the random variable of future lifetime at age \(x\).
The expected value of \(Y\) represents the actuarial present value of a unit payable at the moment of death. For example, for a coverage that spans the whole lifetime,
where the symbol \(\bar{a}_x\) conforms with the International Actuarial Notation.
- The following table shows common annuity types. In all cases, a benefit of \(1\) is payable continuously for as long as the annuityholder is alive.
| Type | \(Y\) | Range | \(\mathrm{E}[Y]\) |
|---|---|---|---|
| Life annuity | \(\bar{a}_{\enclose{actuarial}{T}}\) | \(T>0\) | \(\bar{a}_x= \int_0^\infty v^t\,_tp_x\,dt\) |
| \(n\)-year term annuity |
\(\bar{a}_{\enclose{actuarial}{T}}\) \(0\) |
\(T \leq n\) \(T > n\) |
\(\bar{a}_{x:{\enclose{actuarial}{n}}}= \int_0^n v^t\,_tp_x\,dt\) |
| \(n\)-year deferred annuity |
\(0\) \(\bar{a}_{\enclose{actuarial}{T}} - \bar{a}_{\enclose{actuarial}{n}}\) |
\(T \leq n\) \(T > n\) |
\(_{n|}\bar{a}_{x} = \int_n^\infty v^t\,_tp_x\,dt\) |
For the discrete case, benefits are typically paid at the beginning of each year while the annuityholder is alive. For example, if a unit is paid at the beginning of each year while \((x)\) is alive, then \(Y=\ddot{a}_{{\enclose{actuarial}{K+1}}}\), where \(K\) is the random variable of complete years of future lifetime at age \(x\).
Analogously to \(\#2\) above, the expected value of \(Y\) represents the actuarial present value of a unit payable at the beginning of each survival year. For example, for a coverage that spans the whole lifetime,
\[ \boxed{ \begin{gathered} \\ \quad \mathrm{E}[\,Y\,] = \mathrm{E}[\,\ddot{a}_{\enclose{actuarial}{K+1}}\,] = \sum_{k=0}^\infty v^k\,_kp_x\,=\ddot{a}_x \quad \\ \\ \end{gathered} } \]
where \(_{k|}q_x\) is the probability mass function of \(K=K(x)\) and the symbol \(\ddot{a}_x\) conforms with the International Actuarial Notation.
- The following table shows common annuity types. In all cases, a benefit of \(1\) is payable at the beginning of each year while \((x)\) is alive (unless otherwise stated).
| Type | \(Y\) | Range | \(\mathrm{E}[Y]\) | Commutation Functions |
|---|---|---|---|---|
| Life annuity | \(\ddot{a}_{\enclose{actuarial}{K+1}}\) | \(K = 0, 1,\cdots\) | \(\ddot{a}_x=\sum _{k=0}^\infty v^k \, _kp_x\) | \(\dfrac{N_x}{D_x}\) |
| \(n\)-year term annuity |
\(\ddot{a}_{\enclose{actuarial}{K+1}}\) \(\ddot{a}_{\enclose{actuarial}{n}}\) |
\(K = 0, 1, \cdots, n-1\) \(K = n, n+1, \cdots\) |
\(\ddot{a}_{x:{\enclose{actuarial}{n}}}=\sum_{k=0}^\infty v^k\,_kp_x\) | \(\dfrac{N_x-N_{x+n}}{D_x}\) |
| \(n\)-year deferred annuity |
\(0\) \(\ddot{a}_{\enclose{actuarial}{K+1}} - \ddot{a}_{\enclose{actuarial}{n}}\) |
\(K = 0, 1, \cdots, n-1\) \(K = n, n+1, \cdots\) |
\(_{n|}\ddot{a}_x=\sum _{k=n+1}^\infty v^k \, _kp_x\) | \(\dfrac{D_{x+n}}{D_x}\) |
- The “Commutation Functions” column in the table above represent calculation algorithms that are easily programmable. Consider how these commutation functions are used below:
where
\[ \boxed{ \begin{array}{rl} \\ \quad D_x &= v^x\,l_x \quad \\ \quad N_x &= \displaystyle \sum_{k=0}^{\infty} D_{x+k} \quad \\ \\ \end{array} } \]
3.2 Solved Exercises
Expected value of \(Y\)
Let \(Y\) be the present value random variable for a continuous annuity of \(1\) per year, payable for the lifetime of \((x)\).
Let \(\delta_t=0.07\), \(t \geq 0\), and \(\mu_x=0.01\), \(x \geq 0\).
Calculate
- \(\bar{a}_x\)
- the standard deviation of \(\bar{a}_{\enclose{actuarial}{T}}\)
- the probability that \(\bar{a}_{\enclose{actuarial}{T}}\) will exceed \(\bar{a}_x\)
- We have constant forces of mortality and interest, which means that \({_t}p_x=e^{-\mu}\) and \(v^t=e^{-\delta}\).
Thus
\[\begin{align} \bar{a}_x &= \int_0^\infty v^t\,{_t}p_x\,dt \\ \\ &= \int_0^\infty e^{-\delta}\,e^{-\mu}dt \\ \\ &= \int_0^\infty e^{-t(\delta + \mu)}dt \\ \\ &= \left. {-\dfrac{e^{-t}}{\delta + \mu}} \right|_0^{\infty} \\ \\ &= \dfrac{1}{\delta + \mu} \\ \\ &= \dfrac{1}{0.07 + 0.01} \\ \\ &= 12.5 \\ \\ \end{align}\]
- To compute the variance of \(Y\), we use
\[\begin{align} \mathrm{Var}[\,Y\,] &= \mathrm{Var} \left[ \,\dfrac{1-v^T}{\delta}\, \right] \\ \\ &= \dfrac{1}{\delta^2}\mathrm{Var}[\,v^T\,] \\ \\ &= \dfrac{1}{\delta^2} \left[ {^2}\bar{A}_x - \bar{A}^2_x \right] \\ \\ \end{align}\]
where \({^2}\bar{A}_x\) is calculated at a constant force of interest \(2\delta\).
Thus
\[\begin{align} \mathrm{Var}[\,Y\,] &= \dfrac{1}{\delta^2} \left[ \dfrac{\mu}{2\delta+\mu} - \left( \dfrac{\mu}{\delta+\mu} \right)^2 \right] \\ \\ &= \dfrac{1}{{0.01}^2} \left[ \dfrac{0.01}{0.14+0.01} - \left( \dfrac{0.01}{0.07+0.01} \right)^2 \right] \\ \\ &= 510.41\overline{6} \end{align}\]
The standard devation is then \(\sqrt{510.41\overline{6}}=22.592\)
- To compute \(\mathrm{P}[\,Y>\mathrm{E}[\,Y\,]\,]\) we use the result of #1 above.
\[\begin{align} \mathrm{P}[\,Y>\mathrm{E}[\,Y\,]\,] &= \mathrm{P}[\,Y>\bar{a}_x\,] \\ \\ &= \mathrm{P}[\,Y>12.5\,] \\ \\ &= \mathrm{P} \left[ \,\dfrac{1-v^T}{\delta}>12.5\, \right] \\ \\ &= \mathrm{P}[\,v^T<1-12.5\delta\,] \\ \\ &= \mathrm{P} \left[ \,T>\dfrac{\log{1-12.5\delta}}{-\delta}\, \right] \\ \\ &= \mathrm{P}[\,T>29.70631\,] \\ \\ &= 1-{_{29.70631}p_x} \\ \\ \end{align}\]
Using the fact that \(\mu_x\) is constant, \(1-{_{29.70631}p_x}=1-\exp{(-29.70631\mu)}=0.257\), which is the probability being sought.
\(\\\)
3.3 Supplementary Exercises
3.1
Show that
\(\bar{a}_{x}=\int_{0}^{\infty}v^{t}\,{_{t}p_{x}\,dt}=\int_{0}^{\infty}a_{\enclose{actuarial}{n}} \cdot\,{_{t}p_{x}\mu_{x+t}\,dt}\).
3.2
Show that \(\bar{A}_{x}=1-\delta \bar{a}_{x}\)
3.3
Let \(\mu_{x}=0.04\) for all \(x\) and \(\delta=0.06\).
Compute
\(\bar{a}_{x}\)
the standard deviation of \(Y\)
the probability that \(Y\) exceeds \(\mathrm{E}[Y]\)
3.4
Let \(s(x)=1-\dfrac{x}{80}\), \(0 \leq x \leq 80\) and \(\delta=0.06\).
Compute \(\mathrm{Pr}[\,Y<\mathrm{E}[Y]\,]\) for a person aged \(40\).
3.5
Let \(\mu_{x}=0.02\) for all \(x\) and \(\delta=0.10\).
Compute \(\mathrm{Pr}[\,Y > \bar{a}_{x}\,]\).
3.6
Let \(\mu_{x}=0.02\) for all \(x\) and \(\delta=0.04\).
Compute the probability that \(_{10|}\bar{a}_{x}\) is insufficient to provide a deferred annuity of \(1\) per year starting at age \(30\).
3.7
A continuous annuity contract with payments of \(10\) per year is sold to each of \(100\) independent individuals.
Compute the minimum amount an initial fund must have so that the annuity writer has \(95,%\) confidence of making all payments on time.
Use the normal approximation with \(\Phi^{-1}(0.95)=1.645\), \(\delta=0.06\), and \(\mu_{x}=0.04\) for all \(x\).
3.8
Let \(Y\) be the present value of a continuous whole life annuity of \(1\) issued at age \(x\), with \(\mu_{x}=0.02\) for all \(x\) and \(\delta=0.04\).
Compute the \(95\)th percentile of the distribution of \(Y\).
3.9
A continuous whole life annuity of \(1\) is issued to an individual aged \(x\). Let \(\mu_{x}=0.2\) for all \(x\) and \(\delta=0.1\).
Compute the required loading (as a percentage of the expected present value) such that the insurer has \(75\%\) confidence that the premium is sufficient to cover the annuity payments.
3.10
Let \(s(x)=1-\dfrac{x}{100}\), \(0 \leq x \leq 100\) and \(\delta=0.03\).
Compute the median of the distribution of \(Y\) for a continuous whole life annuity of \(100\) issued to a person aged \(50\).
3.11
Let
- \(s(x)=1-\dfrac{x}{100}\), \(0 \leq x \leq 100\)
- \(i=0.05\)
- \(Y\) the random variable for a temporary annuity-due of \(K\) years for \(K=0, 1, ..., 5\), and equal to a 5-year temporary annuity-due for \(K=6, 7, ...\)
Compute \(\mathrm{E}[Y]\) for a person aged \(25\).
3.12
Compute the standard deviation of the random variable \(Y\) defined in exercise #68, using the same age and interest rate.
3.13
Let
- \(Y\) be the present value of a temporary annuity-due of \(K\) years for \(K=0, 1, ..., 10\) and \(0\) otherwise
- \(d=0.10\)
- \(_{10}p_{x}=0.6\)
- \(_{11}p_{x}=0.5\)
- \(0.4\) the single net premium of an 11-year endowment insurance of \(1\) with death benefit payable at the end of the year of death
Compute \(\mathrm{E}[Y]\) for a person aged \(25\).
3.14
Let
- \(s(x)=1-\dfrac{x}{100}\), \(0 \leq x \leq 100\)
- \(\delta=0.06\)
- \(Y\) the present-value random variable of a discrete whole life annuity-due of \(1\)
Compute \(\mathrm{Pr}[\,Y<\mathrm{E}[Y]\,]\) for a person aged \(53\).
3.15
\(s(x)=1-\dfrac{x}{100}\), \(0 \leq x \leq 100\) and \(\delta=0.03\).
Compute the median of the distribution of \(Y\), the present-value random variable of a whole life annuity-due of \(100\), issued to a person aged \(50\).
3.16
Compute the single net premiums of the following discrete annuities of \(1\) for a person aged \(45\), using the Illustrative Life Table.
- whole life annuity-due
- whole life annuity-immediate
- 5-year temporary annuity-due
- 5-year temporary annuity-immediate
- 5-year deferred whole life annuity-due
- 5-year deferred whole life annuity-immediate
3.17
An individual aged \(50\) receives payments of \(100x\) at ages \(60\), \(70\), and \(80\). Assuming \(l_x={(100-x)}^{0.5}\), \(0 \leq x \leq 100\) and \(i=0.05\).
Compute the actuarial present value of these payments at age \(50\).3.18
Under the assumptions of a constant force of mortality, \(\mu_{x}=\mu\) for all \(x\), and a constant force of interest, \(\delta=0.06\), compute
- \(\bar a_x\)
- the standard deviation of a continuous annuity for \(T\) years (where \(T\) is the future lifetime at age \(x\))
- the probability that a continuous annuity for \(T\) years exceeds \(\bar a_x\)
3.19
Assuming a constant force and \(A_x=2a_x\), compute \(\mu\).
3.20
Calculate \(i\), given
- \(a_x=12.3456\)
- \(A_x=0.61129\)
3.21
Let
- \(a_{{x+1\,:\,}{\enclose{actuarial}{9}}}=7.6071\)
- \(a_{{x\,:\,}{\enclose{actuarial}{9}}}=7.6214\)
- \(p = x=0.99647\)
- \(v=0.97087\)
Compute the annual premium at age \(x\) of a \(10\)-year term insurance with death benefit payable at the end of the year of death.
3.22
What is the value of \(50{,}000\) at age \(60\) accumulated to age \(70\) with interest at \(6\%\) and survivorship defined by the Illustrative Life Table?
3.23
A person wishes to purchase an life annuity payable to his son, now age \(10\), at the end of each year while he is alive, with the first payment made at the end of his \(21^{\text{st}}\) year of life.
Calculate the present value of the annuity assumning that survivorship is defined by the Illustrative Life Table.
3.24
Assuming that mortality follows De Moivre’s Law with \(\omega=105\) and \(i=7\%\), caclulate \(a_{{30\,:\,}{\enclose{actuarial}{5}}}.\)
3.25
Assuming that survivorship is defined by the Illustrative Life Table, compute the present value of a \(5{,}000\) annuity issued to \((65)\) if the first payment is to be made at age \(75\).
3.26
A person inherits at age \(75\) a fund of \(750{,}000\) that will be paid as a life annuity with the first payment made at age \(76\).
If survivorship is defined by the Illustrative Life Table, compute the annual payment.
3.27
Compute \(10{,}000\cdot a_{{30\,:\,}{\enclose{actuarial}{5}}}\).
3.28
At age \(90\) a person purchases a whole life insurance policy with annual premiums of \(800\).
If survivorship is defined by the Illustrative Life Table, compute the present value of the premiums.
3.29
A person receives a retirement fund of \(500{,}000\) at age \(65\).
The person has two purchase options:
- A five-year immediate annuity certain;
- A five-year immediate life annuity.
If survivorship is defined by the Illustrative Life Table, calculate the difference of the present value of both annuities.
3.30
If survivorship is defined by the Illustrative Life Table and \(i=6\%\), calculate the present value of a life annuity of \(10{,}000\) issued at age \(75\), in which the first fifteen payments are guaranteed.