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More generic insurance payout:

We can introduce indicator variable to model if an event happens at a particular interval to translate the expectation in insurance. It also allows us to show clear interpretation of calculation.

The indicator variable takes the following form:

I(A)={0 if A does not happen1 if A does happenI(A)=\begin{cases}0 & \text{ if }A \text{ does not happen}\\1 & \text{ if }A \text{ does happen}\end{cases}

Then we can expressed an insurance product that pays an individual $200\$200 if the individual die at first 55 years and pays $300\$300 if the individual die at the second 55 years, then we can express the above as

200ν5I(TX5)+300ν10(5TX10)200\nu^5 I(T_X\le 5) + 300\nu^{10} (5 \le T_X \le 10)

whose EPV will be 2005qxν5+300510qxν10200 {}_5q_x\nu^5 + 300 {}_{5|10}q_x \nu^{10}

We can also have the continuous form for indicator.

E[I(t<TXt+dt)]=Pr(t<TXt+dt)=Pr[TX>t]Pr[TXt+dtTX>t]=tpxμx+tdt\mathbb{E}[I(t< T_X \le t+dt)] = \Pr(t< T_X \le t+dt)\\=\Pr[T_X > t]\Pr[T_X \le t+dt | T_X > t] \\={}_tp_x \mu_{x+t} dt

Example.\textbf{Example.} Suppose an insurance is payable instantly with TXT_X amount. Then we will have

E[TXexp(δTX)]=0texp(δt)tpxμx+tdt\mathbb{E}[T_X \exp(-\delta T_X)] = \int_0^{\infty} t \exp(-\delta t) {}_tp_x \mu_{x+t} dt

We can denote it as (IˉAˉ)x(\bar{I}\bar{A})_x under actuarial notation, where the bar indicates continuous and the II indicates increasing.

If the policy is termed, then we will have

(IˉAˉ)x:n1=0ntexp(δt)tpxμx+tdt(\bar{I}\bar{A})_{x:\angl{n}}^1 = \int_0^n t \exp(-\delta t) {}_tp_x \mu_{x+t} dt

Note:\textbf{Note:} Below are a couple of exercises based on Actuarial Mathematics for Life Contingent Risks. The problems are adapted from the examples presented on pages 128 to 131. However, the solutions provided here have been written by me.

Example\textbf{Example}. There's an insurance policy issued to an individual with age (x)(x) for a term of nn years. The policy provides a death benefit of k+1k+1, which will be paid at the end of the year of death if the individual dies between ages x+kx+k and x+k+1x+k+1, excluding the right endpoint, where k=0,1,2,,n1k=0,1,2, \ldots, n-1.

(a) Calculate the Expected Present Value of the benefit using the first approach, which involves multiplying together the benefit amount, the discount factor, and the probability of payment, and then summing these values for each possible payment date.

(b) Calculate the formula for the variance of the present value of the benefit.

Note: In the provided context, (x)(x) refers to the age of the insured individual, and it is assumed to be a constant value throughout the calculations.

Solution:\textbf{Solution:}

(a)(a). the benefit amount here is k+1k+1 and the discount factor is νk+1\nu^{k+1} and the probability of the payment is Pr[kTX<k+1]=kqx\Pr[k \le T_X < k + 1] = {}_k|q_x so we will sum them up over the course of k=0,1,...,n1k=0,1,...,n-1

k=0n1(k+1)νk+1kqx\sum_{k=0}^{n-1}(k+1)\nu^{k+1}{}_k|q_x.

In actuarial science, we can denote the above as (IA)x:n1(IA)_{x:\angl{n}}^1.

If we send nn to infinity, then we will have

(IA)x=k=0(k+1)νk+1kqx(IA)_{x}=\sum_{k=0}^{\infty}(k+1)\nu^{k+1}{}_k|q_x.

(b).(b).

Let's express the payout as a random variable with curtate future lifetime random variable Z={(Kx+1)νKx+1Kxn10otherwisesZ=\begin{cases}(K_x+1)\nu^{K_x+1} & K_x \le n-1\\ 0 & \text{otherwises}\end{cases}

Then we will have the following as our second moment:

E[Z2]=k=0n1(k+1)2ν2k+2kqx\mathbb{E}[Z^2]=\sum_{k=0}^{n-1}(k+1)^2\nu^{2k+2} {}_{k|}q_x

So our variance will be Var[Z]=k=0n1(k+1)2ν2k+2kqx((IA)x)2\operatorname{Var}[Z] = \sum_{k=0}^{n-1}(k+1)^2\nu^{2k+2} {}_{k|}q_x - ((IA)_x)^2

Example.\textbf{Example}. Consider a whole life insurance policy that provides an increasing death benefit, which is payable at the end of the quarter year of the insured's death. If the insured's age at the policy's inception is denoted by (x)(x), the death benefit will be 11 if the insured dies in the first year of the contract, 22 in the second year, and so on. Now, let's derive an expression for the Expected Present Value (EPV) of the death benefit.

Solution:\textbf{Solution:}

Here our possible pay day will be 14\frac{1}{4} of the year or 12\frac{1}{2} of the year or 34\frac{3}{4} of the year of the end of the year. So our discount factor for these scenarios will be νk+1/4,νk+1/2,νk+3/4,νk+1\nu^{k+1/4}, \nu^{k+1/2}, \nu^{k+3/4}, \nu^{k+1} with the payout to be k+1k+1. The probability for each scenario will be k14qx and k+1414qx and k+1214qx and k+3414qx{}_{k|\frac{1}{4}}q_{x}\text{ and } {}_{k+\frac{1}{4}|\frac{1}{4}}q_{x} \text{ and } {}_{k+\frac{1}{2}|\frac{1}{4}}q_{x} \text{ and } {}_{k+\frac{3}{4}|\frac{1}{4}}q_{x}

So we can express the EPVEPV as i=0j=14(i+1)νi+j4i+j1414qx=i=0(i+1)j=03νi+j+14i+j414qx=i=0(i+1)νiipxj=03ν(j+1)/4j414qx+i=i=0(i+1)νiipxA(4)x+i:n1=i=0(i+1)iA(4)x:n1\sum_{i=0}^{\infty}\sum_{j=1}^4 (i+1)\nu^{i+\frac{j}{4}} {}_{i+\frac{j-1}{4}| \frac{1}{4}}q_x\\ = \sum_{i=0}^{\infty}(i+1)\sum_{j=0}^3 \nu^{i+\frac{j+1}{4}} {}_{i+\frac{j}{4}| \frac{1}{4}}q_x \\ = \sum_{i=0}^{\infty}(i+1)\nu^i{}_{i}p_x\sum_{j=0}^3 \nu^{(j+1)/4} {}_{\frac{j}{4}| \frac{1}{4}}q_{x+i} \\ = \sum_{i=0}^{\infty}(i+1)\nu^i{}_{i}p_x A^{(4)}{}_{x+i:\angl{n}}^1\\ = \sum_{i=0}^{\infty}(i+1) {}_{i|}A^{(4)}{}_{x:\angl{n}}^1

Instead of having arithemetic payout, let's consider geometric payout.

Example.\textbf{Example}. We introduces an nn-year term insurance policy issued to an individual at age (x)(x). The policy pays the death benefit at the end of the year in which the insured person passes away. The amount of the death benefit varies depending on the age at which death occurs. Specifically, if death occurs between ages xx and x+1x+1, the benefit is 1. If death occurs between ages x+1x+1 and x+2x+2, the benefit is 1+j1+j, and if death occurs between ages x+2x+2 and x+3x+3, the benefit is (1+j)2(1+j)^2, and so forth. Therefore, if death occurs between ages x+kx+k and x+k+1x+k+1, the death benefit is (1+j)k(1+j)^k for k=0,1,2,,n1k=0,1,2,\ldots,n-1. The objective is to derive a formula for the Expected Present Value (EPV) of this death benefit.

Solution:\textbf{Solution:}

The benefit amount here is (1+j)k(1+j)^k and the discount factor is νk+1\nu^{k+1} and the probability of the payment is Pr[kTX<k+1]=kqx\Pr[k \le T_X < k + 1] = {}_k|q_x so we will sum them up over the course of k=0,1,...,n1k=0,1,...,n-1

So we will have the following formula for our EPV:

E[Z]=k=0n1(1+j)kνk+1kqx=11+jk=0n1(ν(1+j))k+1kqx\mathbb{E}[Z] = \sum_{k=0}^{n-1} (1+j)^k \nu^{k+1} {}_k q_x = \frac{1}{1+j}\sum_{k=0}^{n-1} (\nu (1+j))^{k+1} {}_k q_x

So we will have the following: 11+i=1+j1+i    i=i+1j+11\frac{1}{1+i'} = \frac{1+j}{1+i} \implies i' =\frac{i+1}{j+1} -1

and so we can express E[Z]\mathbb{E}[Z] as 11+jAx:ni1\frac{1}{1+j} A_{x:\angl{n}i'}^1.

Example\textbf{Example} We discusses an insurance policy issued to an individual at age (x)(x), wherein the death benefit is determined as (1+j)t(1+j)^t, where tt represents the time of death measured from the beginning of the policy. The policy ensures that the death benefit is paid out immediately upon the occurrence of death. The example seeks to derive the expressions for the Expected Present Value (EPV) of the death benefit for both an nn-year term insurance and a whole life insurance.

(a) Calculate the EPV of the death benefit for the nn-year term insurance.

(b) Calculate the EPV of the death benefit for the whole life insurance.

Solution:\textbf{Solution:}

(a)(a) the benefit amount here is (1+j)t=exp(tln(1+j))(1+j)^t=\exp(t\ln(1+j)) and the discount factor is exp(δt)\exp(-\delta t) and the probability of the payment is Pr[tTX<t+dt]=tpxμx+tdt\Pr[t \le T_X < t + dt] = {}_{t}p_{x} \mu_{x+t}\, dt so we will sum them up over the course of

So we will have E[Z]=0nexp(ln(1+j)t)exp(δt)tpxμx+tdt=0nexp((ln(1+j)δ)t)tpxμx+tdt\mathbb{E}[Z]=\int_0^n \exp(\ln(1+j)t) \exp(-\delta t){}_{t}p_{x} \mu_{x+t}\, dt = \int_0^n \exp((\ln(1+j)-\delta) t){}_{t}p_{x} \mu_{x+t}\, dt

And so our new δ=δln(1+j)\delta' = \delta-\ln(1+j) and we also know that exp(δ)=ν=11+i    δ=ln(1+i)\exp(-\delta) = \nu = \frac{1}{1+i} \implies \delta = \ln(1+i) and so ln(1+i)=ln(1+i1+j)    i=i+1j+11\ln(1+i') = \ln(\frac{1+i}{1+j}) \implies i' = \frac{i+1}{j+1}-1 and so

E[Z]=Aˉx:ni1\mathbb{E}[Z] = \bar{A}_{x:\angl{n}i'}^1

(b)(b) we can always sending nn to infinity, then we will have E[Z]=(Aˉx)i\mathbb{E}[Z]= (\bar{A}_{x})_{i'}