Future Loss Random Variable

Study guide: Exam FAM

The Future Loss Random Variable ${}_tL$ of a long-term insurance or annuity is the present value, at time $t$ , of the insurer's future outgo less the present value of its future premium income, for a policy still in force at $t$ . At issue it is written $L_0$ . The net loss counts benefits and net premiums, and the gross loss adds expenses and uses the gross premium. Under the equivalence principle $E[L_0] = 0$ , and at a later time the expected future loss is the policy value ${}_tV$ (Dickson, Hardy & Waters, chapters 6–7).

$L_0 = S\,v^{K_x+1} - P\,\ddot{a}_{\overline{K_x+1}|}$

$L_0 = \left(S + \frac{P}{d}\right)v^{K_x+1} - \frac{P}{d}$

$\text{Var}(L_0) = \left(S + \frac{P}{d}\right)^2\left({}^2A_x - A_x^2\right)$

$E\left[{}_tL \mid T_x > t\right] = {}_tV$

Symbols: The blocks are for a fully discrete whole life insurance on $(x)$ with sum insured $S$ and annual premium $P$ . The second block uses $\ddot{a}_{\overline{k}|} = (1 - v^k)/d$ , which makes the loss a linear function of the single variable $v^{K_x+1}$ . At time $t$ , ${}_tL$ has the same form with $K_{x+t}$ in place of $K_x$ .
Other policies: For an $n$ -year endowment insurance, replace $K_x + 1$ by $\min(K_x+1, n)$ throughout, and the variance uses ${}^2A_{x:\overline{n}|} - A_{x:\overline{n}|}^2$ . A fully continuous whole life has $L_0 = \left(S + P/\delta\right)e^{-\delta T_x} - P/\delta$ . For an $n$ -year term insurance, $L_0 = S\,v^{K_x+1}\,\mathbb{1}\{K_x < n\} - P\,\ddot{a}_{\overline{\min(K_x+1,\,n)}|}$ , which is not linear in one variable, so its variance has no such shortcut. An annuity of $B$ a year bought for a single premium $G$ has $L_0 = B\,\ddot{a}_{\overline{K_x+1}|} - G$ .
With the equivalence premium: Setting $P = S A_x/\ddot{a}_x$ gives $L_0 = \left(S + P/d\right)\left(v^{K_x+1} - A_x\right)$ . The insurer loses money exactly when the benefit's present value exceeds its EPV, which means when the life dies early.

Loss Distribution of an Endowment Insurance The fully discrete 20-year endowment insurance of $100,000 on $(45)$ in Endowment Insurance has net premium $P = 2{,}966.62$ . On the SULT at $5\%$ , $A_{45:\overline{20}|} = 0.38385$ , ${}^2A_{45:\overline{20}|} = 0.14937$ and ${}_{19}p_{45} = 0.96010$ . Find the standard deviation of $L_0$ and $\Pr(L_0 > 0)$ .
Answer Here $P/d = 2{,}966.62(21) = 62{,}299.02$ .
 $\begin{align*} \text{Var}(Z) &= 0.14937 - 0.38385^2 \\ &= 0.0020292 \\ \text{SD}(L_0) &= (162{,}299.02)\sqrt{0.0020292} \\ &= 162{,}299.02(0.045046) \\ &= 7{,}311 \end{align*}$ 
The loss is positive when $v^{\min(K+1,\,20)} > 62{,}299.02/162{,}299.02 = 0.38385$ , the EPV itself.
 $\begin{align*} \min(K+1, 20) &< \frac{-\ln 0.38385}{\ln 1.05} \\ &= 19.62 \\ \Pr(L_0 > 0) &= \Pr(K_{45} \le 18) \\ &= 1 - 0.96010 \\ &= 0.03990 \end{align*}$ 
The insurer loses only if the insured dies in the first 19 years. A death in year 20 is paid at the same time as the maturity benefit, so it costs no more.

The Loss at Time 10 on a Whole Life Policy The fully discrete whole life insurance of $100,000 on $(50)$ in Equivalence Principle has net premium $P = 1{,}111.99$ . On the SULT at $5\%$ , $A_{60} = 0.29028$ , ${}^2A_{60} = 0.10834$ and $\ddot{a}_{60} = 14.9041$ . Given that the policy is in force at time 10, find the mean and standard deviation of ${}_{10}L$ , valued just before the premium then due.
Answer
 $\begin{align*} E\left[{}_{10}L\right] &= 100{,}000(0.29028) - 1{,}111.99(14.9041) \\ &= 29{,}028 - 16{,}573.21 \\ &= 12{,}454.79 \\ \text{SD}\left({}_{10}L\right) &= (100{,}000 + 21P)\sqrt{0.10834 - 0.29028^2} \\ &= 123{,}351.79(0.155169) \\ &= 19{,}140 \end{align*}$ 
The mean is the net premium policy value ${}_{10}V$ , which also equals $100{,}000\left(1 - \ddot{a}_{60}/\ddot{a}_{50}\right)$ . At issue the loss had a mean of zero and a standard deviation of $15,229. Ten years later the expected future loss is positive, and its standard deviation is larger than at issue.

A Term Insurance Loss Has No Shortcut A fully discrete 2-year term insurance of $100,000 on $(70)$ has net premium $P = 1{,}049.80$ , from the SULT values $q_{70} = 0.010413$ and $q_{71} = 0.011670$ at $5\%$ . Find the standard deviation of $L_0$ .
Answer $L_0$ takes three values: death in year 1, death in year 2, or survival to the end of the term.
 $\begin{align*} 100{,}000v - P &= 94{,}188.30 \\ 100{,}000v^2 - P(1 + v) &= 88{,}653.34 \\ -P(1 + v) &= -2{,}049.61 \end{align*}$ 
Their probabilities are $0.010413$ , $0.989587(0.011670) = 0.011548$ and $0.978039$ . Since $E[L_0] = 0$ , the variance is the second moment.
 $\begin{align*} E\left[L_0^2\right] &= 0.010413(94{,}188.30)^2 + 0.011548(88{,}653.34)^2 \\ &\quad + 0.978039(2{,}049.61)^2 \\ &= 187{,}247{,}428 \\ \text{SD}(L_0) &= 13{,}684 \end{align*}$ 
The whole life formula $(S + P/d)\sqrt{{}^2A^{1} - (A^{1})^2}$ would give $16,613. It is wrong here, because the premiums stop at the end of the term whether or not the life is alive.

How it appears on the exam

FAM asks candidates to identify the future loss random variable for whole life, term life and endowment insurance and for term and whole life annuities on a single life, net or gross, and with annual or continuous premiums.

Moments and probabilities: A question can ask for $E[L_0]$ at a given premium, $\text{Var}(L_0)$ by the $(S + P/d)^2$ shortcut, or $\Pr(L_0 > 0)$ by turning the loss into a condition on $K_x$ or $T_x$ .

Premiums and policy values: The loss is the starting point for the equivalence principle and for the portfolio percentile principle, which needs $E[L_0]$ and $\text{SD}(L_0)$ as functions of $P$ . FAM's policy values are $E[{}_tL]$ for a policy in force, so the same random variable returns after issue.

Exam traps

Forgetting the premium in the variance: $\text{Var}(L_0)$ scales by $(S + P/d)^2$ , not $S^2$ . Leaving out $P/d$ gives the variance of the benefit alone, which is too small.

$d$ or $\delta$ : Annual premiums use $P/d$ , and continuous premiums use $P/\delta$ , with ${}^2\bar{A}$ and $\bar{A}$ in the variance. Using $d$ where $\delta$ belongs changes $P/d$ by about $2.5\%$ at $5\%$ interest.

The shortcut for term insurance: The formula $(S + P/d)^2({}^2A - A^2)$ holds only when the benefit and the premium annuity end at the same time. For term insurance, work from the distribution of $L_0$ or include the covariance.

The direction of $\Pr(L_0 > 0)$ : The loss is positive when the present value of the benefit is large, which happens when death is early. Solving $v^{K+1} > c$ gives an upper limit on $K$ , and the answer is a death probability ${}_mq_x$ , not a survival probability.