Notes on Life Insurance (Part One)

We will use the following:

Makeham's Law, $\mu_x=A+B c^x$,

$$A=0.00022, \quad B=2.7 \times 10^{-6}, \quad c=1.124$$

So we know the following from the financial mathematics:

$$(1+i)(1-d) = 1$$, where $d$ is the discount for the interests. We can assume that our interest is risk-free. And $1+i = e^{\delta}$ so $1-d = e^{-\delta}$.

For the nominal interests for $n$ period within the paid period, we will have $(1+\frac{i^{(n)}}{n})^n = 1+i$, and so $$i^{(n)} = n \times [(1+i)^{1/n}-1]$$

and $$d^{(m)}= n \times[1-(1-d)^{1/n}].$$

The present value of $1$ unit of payment that is payable immediately upon death is a random variable $Z$.

$$Z = \left(\frac{1}{1+i}\right)^{T_x}.= e^{-\delta T_X}$$.

The Actuarial Present Value is the expected value of $Z$, which is $\mathbb{E}_{T_X}[Z]$, which we can denote as $\bar{A}_x$ and here bar represents payable immediately upon death.

We know that $T_x$ has a pdf $f_{T_x}(t) = {}_tp_x \mu_{x+t}$ and so

$$\bar{A}_x = \int_{0}^{\infty} \exp(-\delta t){}_tp_x \mu_{x+t}\, dt$$

We can also compute the second moment of $Z$. We will have the following:

$$\mathbb{E}[Z^2] = \mathbb{E}[\exp(-2\delta T_x)] = \int_0^{\infty} \exp(-2\delta t) {}_tp_x \mu_{x+t}={ }^2 \bar{A}_x,$$

Here, the superscript is the coefficient of the rate of interest.

And we can represent our variance as

$$\operatorname{Var}[Z] = {}^2 \bar{A}_x - (\bar{A}_x)^2$$

And for a constant $S$, $\mathbb{E}[SZ]= S \bar{A}_x$ and $\operatorname{Var}[SZ] = S^2 ({}^2 \bar{A}_x - (\bar{A}_x)^2).$

For any value $x$, we want to find the cdf of $Z$, we notice that $\Pr[Z \le x] = \Pr[\exp(-\delta T_X) \le x] = \Pr[ T_X \ge -\frac{1}{\delta}\ln x] = {}_{-\frac{\ln x}{\delta}}p_{x}.$

Now, instead of considering payment payable upon death. We consider payment payable at the end of the year of death. Then we will have $$Z = \left(\frac{1}{1+i}\right)^{K_n+1}=\nu^{K_n+1}$$

$A_x=\mathbb{E}[Z] = \sum_{i=0}^{\infty}\nu^{i+1} {}_{i}{}_{|}q_x = \nu q_x + \nu^2 {}_1{}_{|}q_x + \nu^3 {}_2{}_{|}q_x+...$

Notice that we have stripped off the overhead bar for $A_x$ because we are considering payment at the end of the year of death.

Now similarly, we will have

$${}^2 A_x= \sum_{k} \nu^{2(k+1)} {}_{k|}q_x$$

Now consider whole life annuity under the case of $1/m$-th.

Instead of having it to be paid at the end of the year, we can have it paid half a year, or even monthly or weekly. Then we will first consider $K_{x}^{(m)}$, where denotes that there are $m$ payments within the fiscal period, as the latest payment period where our insured individual lives. Then we will have $$K_x^{(m)}=\frac{1}{m} \lfloor m T_x\rfloor$$.

Example: Consider $T_x = 29.2039$, then $K_x = 29$, $K_x^{(4)} = 29, K^{(12)}_x = 29.16 $...

And the pmf for $K_x^{(m)}$ will be

$$\Pr[K_x^{(m)} =k ] = \Pr[\lfloor m T_x \rfloor =km ] = \Pr[km \le m T_x \le km +1 ] = \Pr[k \le T_x \le k + \frac{1}{m}] = {}_{k|\frac{1}{m}}q_x$$

Now if our payment is paid at the end of the period when our insurance individual's death, then we will have the following:

$$A_x^{(m)} = \sum_{k} \nu^{\frac{k+1}{m}} {}_{\frac{k}{m}|\frac{1}{m}}q_x$$

And the second moment will be

$${}^2A_x^{(m)} = \sum_{k} \nu^{\frac{2k+2}{m}} {}_{\frac{k}{m}|\frac{1}{m}}q_x$$

A recursive formula.

Sometimes, it's hard to determine the actuarial present value for payment expired at the end of the year in a forward manner. Thus, we need to consider a way of getting everything in a backward fashion. One way of doing this is with a life table.

Suppose we have a life table $L_{x}$, then we can find limiting age $\omega$ such that $L_{\omega} =0$ or close to zero.

Then we will assume that $A_{\omega-1} = \nu$.

Then we will consider $A_{x}$ for $x < \omega-1$, we will have

$$A_x = \sum_{k}\nu^{k+1} {}_{k|}q_{x}$$

We also know that ${}_{k|}q_x= {}_{k}p_{x} - {}_{k+1}p_x = {}_k p_x - {}_{k}p_{x} p_{x+k} = {}_k p_x q_{x+k}$

So $$A_x = \sum_{k}{}_k p_x q_{x+k} \nu^{k+1} = q_{x} \nu + \sum_{k=1}^{\infty}{}_k p_x q_{x+k} \nu^{k+1} $$

We also know that ${}_kp_{x} = {}_1p_{x}{}_{k-1}p_{x+1}$

So we will have

$$A_x = q_{x} \nu + p_{x}\sum_{k=1}^{\infty}{}_{k-1}p_{x+1} q_{x+k} \nu^{k+1} \\= q_{x} \nu + p_{x}\nu\sum_{k=0}^{\infty}{}_{k}p_{x+1} q_{x+1+k} \nu^{k+1} \\ = q_x \nu + p_x \nu A_{x+1}$$

We can derive a similar formula for $A_x^{(m)} $. Consider $$A_x^{(m)}= \sum_{k=0}^{\infty} \nu^{\frac{k+1}{m}} {}_{\frac{k}{m}|\frac{1}{m}}q_{x} \\ = \nu^{\frac{1}{m}} {}_{|\frac{1}{m}}q_x + \nu^{\frac{1}{m}}\sum_{k=1}^\infty \nu^{\frac{k}{m}} {}_{\frac{k}{m}}p_{x} {}_{\frac{1}{m}} q_{x+\frac{k}{m}}\\ = \nu^{\frac{1}{m}} {}_{|\frac{1}{m}}q_x + {}_{\frac{1}{m}}p_{x}\nu^{\frac{1}{m}}\sum_{k=1}^\infty \nu^{\frac{k}{m}} {}_{\frac{k-1}{m}}p_{x+\frac{1}{m}}{}_{\frac{1}{m}} q_{x+\frac{k}{m}}\\=\nu^{\frac{1}{m}} {}_{|\frac{1}{m}}q_x + {}_{\frac{1}{m}}p_{x}\nu^{\frac{1}{m}}\sum_{k=0}^\infty \nu^{\frac{k+1}{m}} {}_{\frac{k}{m}}p_{x+\frac{1}{m}}{}_{\frac{1}{m}} q_{x+\frac{1}{m}+\frac{k}{m}}\\= \nu^{\frac{1}{m}} {}_{|\frac{1}{m}}q_x + {}_{\frac{1}{m}}p_{x}\nu^{\frac{1}{m}}\sum_{k=0}^\infty \nu^{\frac{k+1}{m}} {}_{\frac{k}{m}|\frac{1}{m}}q_{x+\frac{1}{m}}$$

So

$$A_x^{(m)} = \nu^{\frac{1}{m}} {}_{|\frac{1}{m}}q_x + {}_{\frac{1}{m}}p_{x}\nu^{\frac{1}{m}}A_{x+\frac{1}{m}}^{(m)}$$

Let's consider a more generic case and here we will introduce some actuarial notations that are non standard. Let pmf be

$$f(k; x, n, m) = \begin{cases} {}_{\frac{k}{m}| \frac{1}{m}}q_x & k \le nm-1 \\ p_{x+n} & k \ge nm\end{cases}$$

$$g(\alpha, \beta, n, m, \nu, t,x) = \alpha \sum_{i=0}^{nm-1} \nu^{t(k+1)/m} {}_{\frac{i}{m}| \frac{1}{m}}q_x + \beta\nu^{tn} {}_np_{x}$$

Then we will have

$$g(\alpha, \beta, n, m, \nu, t, x) - \beta\nu^{tn} {}_np_{x} = \alpha \sum_{i=0}^{nm-1} \nu^{t(i+1)/m} {}_{\frac{i}{m}| \frac{1}{m}}q_x$$

$$g(\alpha, \beta, n, m, \nu, t, x) - \beta\nu^{tn} {}_np_{x} = \alpha \left(\nu^{t/m} {}_{| \frac{1}{m}}q_x - {}_{\frac{1}{m}}p_x\nu^{t/m}\nu^{tn} {}_{n-\frac{1}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}+ {}_{\frac{1}{m}}p_x\nu^{t/m}\sum_{i=0}^{nm-1} \nu^{t(i+1)/m} {}_{\frac{i}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}\right)$$

$$=\alpha \left(\nu^{t/m} {}_{| \frac{1}{m}}q_x - {}_{\frac{1}{m}}p_x\nu^{t/m}\nu^{tn} {}_{n-\frac{1}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}\right)+ {}_{\frac{1}{m}}p_x\nu^{t/m}\alpha\sum_{i=0}^{nm-1} \nu^{t(i+1)/m} {}_{\frac{i}{m}| \frac{1}{m}}q_{x+\frac{1}{m}} $$

We also know that

$$g(\alpha, \beta, n, m, \nu, t, x +\frac{1}{m}) - \beta\nu^{tn} {}_np_{x+\frac{1}{m}} = \alpha \sum_{i=0}^{nm-1} \nu^{t(i+1)/m} {}_{\frac{i}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}$$

So we will have

$$g(\alpha, \beta, n, m, \nu, t, x) - \beta\nu^{tn} {}_np_{x} = \alpha \left(\nu^{t/m} {}_{ \frac{1}{m}}q_x - {}_{\frac{1}{m}}p_x\nu^{t/m}\nu^{tn} {}_{n-\frac{1}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}\right)+ {}_{\frac{1}{m}}p_x\nu^{t/m}\left(g(\alpha, \beta, n, m, \nu, t, x +\frac{1}{m}) - \beta\nu^{tn} {}_np_{x+\frac{1}{m}} \right)$$

$$g(\alpha, \beta, n,m, \nu , t,x) - \beta\nu^{tn} {}_np_{x}= \alpha \left(\nu^{t/m} {}_{| \frac{1}{m}}q_x - {}_{\frac{1}{m}}p_x\nu^{t/m}\nu^{tn} {}_{n-\frac{1}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}\right)+ {}_{\frac{1}{m}}p_x\nu^{t/m}\left(g(\alpha, \beta, n, m, \nu, t, x +\frac{1}{m}) - \beta\nu^{tn} {}_np_{x+\frac{1}{m}} \right)$$

Reducing unnecessary hyper-parameters, we will have

$$h(x)= \alpha\nu^{t/m} {}_{ \frac{1}{m}}q_x - \alpha{}_{\frac{1}{m}}p_x\nu^{t/m}\nu^{tn} {}_{n-\frac{1}{m}| \frac{1}{m}}q_{x+\frac{1}{m}}+ {}_{\frac{1}{m}}p_x\nu^{t/m}h(x+\frac{1}{m}) - \nu^{t/m}\beta\nu^{tn} {}_np_{x} + \beta\nu^{tn} {}_np_{x}$$

$$h(x) = \alpha\nu^{t/m} {}_{ \frac{1}{m}}q_x - \alpha\nu^{tn+t/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{t/m}h(x+\frac{1}{m}) - \nu^{t/m}\beta\nu^{tn} {}_np_{x} + \beta\nu^{tn} {}_np_{x}$$

For this formula, we can easily get the recursive formula for whole life insurance actuarial present value $A_x^{(m)}$ We can do it by simply setting $\alpha =1$ and $\beta =0$ and sending $n \to \infty$, specifically, we have

$$\left|\nu^{tn+t/m} {}_{n| \frac{1}{m}}q_{x}\right| \le |\nu|^{tn+t/m}$$

as $n$ goes to infinity, $\lim_{n\to \infty}|\nu|^{tn+t/m} = 0$

And we know that $A_x^{(m)}$ is the first moment so $t=1$ and therefore we will have

$$A_x^{(m)} = h(x) = \nu^{1/m} {}_{ \frac{1}{m}}q_x + {}_{\frac{1}{m}}p_x\nu^{1/m}h\left(x+\frac{1}{m}\right) = \nu^{1/m} {}_{ \frac{1}{m}}q_x + {}_{\frac{1}{m}}p_x\nu^{1/m}A_{x+\frac{1}{m}}^{(m)}$$

We can also get the formula for actuarial present value for term insurance $A_{x:\angl{n}}^{(m)}$ by setting $\alpha=1$ and $\beta=0$ and $t=1$ and not sending $n \to \infty$ rather think $n$ as a hyper-parameter, so we will get

$$A^{(m)}{}_{x:\angl{n}}^{1}=h(x) = \nu^{t/m} {}_{ \frac{1}{m}}q_x - \nu^{tn+t/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{t/m}h(x+\frac{1}{m}) = \nu^{1/m} {}_{ \frac{1}{m}}q_x - \nu^{n+1/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{1}$$

We can also get the recursive formula for pure endowment by setting $\alpha=0$ and $\beta=1$ and $t=1$

$$A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1} = h(x) = {}_{\frac{1}{m}}p_x\nu^{1/m}h(x+\frac{1}{m}) - \nu^{1/m}\nu^{n} {}_np_{x} + \nu^{n} {}_np_{x} = \nu^{n} {}_np_{x} = A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{\quad\quad1}$$

We can also get synthetic endowment (endowment insurance) recursive formula by setting $\alpha =\beta =t=1$,

We will have $$A^{(m)}{}_{x:\angl{n}}= h(x) = \nu^{1/m} {}_{ \frac{1}{m}}q_x - \nu^{n+1/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{1/m}h(x+\frac{1}{m}) - \nu^{n+1/m} {}_np_{x} + \nu^{n} {}_np_{x}=\nu^{1/m} {}_{ \frac{1}{m}}q_x - \nu^{n+1/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}} - \nu^{n+1/m} {}_np_{x} + \nu^{n} {}_np_{x}$$

One can also see that we have decomposition $A^{(m)}{}_{x:\angl{n}} =A^{(m)}{}_{x:\angl{n}}^{1} + A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1}$

Indeed, we have

$$A^{(m)}{}_{x:\angl{n}} = \nu^{1/m} {}_{ \frac{1}{m}}q_x - \nu^{n+1/m} {}_{n| \frac{1}{m}}q_{x}+ {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}} - {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{\quad\quad 1} + A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1}$$

and we know $$A^{(m)}{}_{x:\angl{n}}^{1}- {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{1}= \nu^{1/m} {}_{ \frac{1}{m}}q_x - \nu^{n+1/m} {}_{n| \frac{1}{m}}q_{x}$$

So, we will have

$$A^{(m)}{}_{x:\angl{n}} = A^{(m)}{}_{x:\angl{n}}^{1}- {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{1}+ {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}} - {}_{\frac{1}{m}}p_x\nu^{1/m}A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{\quad\quad 1} + A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1}$$

Rearranging a little bit, we will have

$$A^{(m)}{}_{x:\angl{n}} - A^{(m)}{}_{x:\angl{n}}^{1}- A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1}= {}_{\frac{1}{m}}p_x\nu^{1/m}\left(A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}- A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{1} - A^{(m)}{}_{x+\frac{1}{m}:\angl{n}}^{\quad\quad 1}\right) $$

Since for $x$ big enough, we will have $A^{(m)}{}_{x:\angl{n}} - A^{(m)}{}_{x:\angl{n}}^{1}- A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1} = 0$ (base case) and so $A^{(m)}{}_{x:\angl{n}} - A^{(m)}{}_{x:\angl{n}}^{1}- A^{(m)}{}_{x:\angl{n}}^{\,\,\,\,\,1} =0$ for every $x$.

For the continuous case, we can sending $m\to \infty$ and get similar result:

$$\bar{A}{}_{x:\angl{n}} = \bar{A}{}_{x:\angl{n}}^{1}+ \bar{A}{}_{x:\angl{n}}^{\,\,\,\,\,1}$$

One thing is interesting about pure endowment is that its variance is associated with the probability of an individual's death more specifically, we have $A_{x:\angl{n}}^{\,\,\,\,1} = \nu^n{}_np_x = {}_xE_n$

and $${}^2A_{x:\angl{n}}^{\,\,\,\,1} = \nu^{2n}{}_np_x = \nu^n{}_xE_n = {}^2_xE_n$$

And so the variance$$\operatorname{Var}[Z] = {}^2_xE_n - ({}_xE_n)^2 = \nu^n{}_xE_n - \nu^n{}_np_x{}_xE_n = {}_xE_n\nu^n(1-{}_np_x{})={}_xE_n\nu^n{}_nq_x$$ as we can see that if $n$ is small and $x$ is small, then the variance is small. That is because when individuals are young, it's unlikely for them to die and so the payout is mostly determined (it will be paid). However, as people aging, the probability becomes larger and so it's hard to say exactly if the payout will come.

Deferred insurance.

Deferred insurance has payout at specific time frame $u \le T_x \le u +n$.

More specifically the present value $$Z = \begin{cases}0 & T_x< u \land u+n < T_x \\ \exp(-\delta T_x) &u \le T_x \le u+n \end{cases}$$

and so the expected present value will be

$${}_u |\bar{A}_{x:\angl{n}}^1=\int_u^{u+n}\exp(-\delta t){}_tp_x \mu_{x+t}\, dt \\= \int_0^n \exp(-\delta t -\delta u){}_{t+u}p_x \mu_{x+u+t}\, dt \\= \exp(-\delta u){}_{u}p_{x}\int_0^n \exp(-\delta t){}_{t}p_{x+u} \mu_{x+u+t}\, dt\\ = {}_u E_x A_{x+u:\angl{n}}^{1}$$

In particular, we have $${}_u |\bar{A}_{x:\angl{n}}^1 = \int_0^{u+n}\exp(-\delta t){}_tp_x \mu_{x+t}\, dt - \int_0^{u}\exp(-\delta t){}_tp_x \mu_{x+t}\, dt = \bar{A}_{x:\angl{u+n}}^1-\bar{A}_{x:\angl{u}}^1$$

which is quite analogous to ${}_{a|b}q_{x} = {}_a p_{x} - {}_{a+b} p_{x}$ but with reversing subscripts.

Decomposition of term insurance into deferred term insurance.

Consider $$\bar{A}_{x:\angl{n}}^1 = \int_0^n \exp(-\delta t) {}_tp_x\mu_{x+t} \, dt = \sum_{i=0}^{n-1} \int_{i}^{i+1}\exp(-\delta t) {}_tp_x\mu_{x+t} \, dt = \sum_{k=0}^{n-1}{}_k |\bar{A}_{x:\angl{1}}^1 $$

If we send $n\to \infty$, we will have

$\bar{A}_{x} = \sum_{k=0}^{\infty}{}_k |\bar{A}_{x:\angl{1}}^1$

We will also have $A_x = A_{x:\angl{n}}^1 + {}_n| A_{x}$

We want to establish relations among $A_x$ and $\bar{A}_x$ and $A^{(m)}_x$ we start by assuming the UDD assumption. First notice that under UDD, we have $$\frac{d}{ds}{}_sq_x = \frac{d}{ds}sq_x = q_x$$

for $0\le s <1$

and

$$\frac{\frac{d}{ds}{}_sq_x}{{}_sp_x} = \mu_{x+s} \implies q_x = {}_sp_x\mu_{x+s}.$$

$$\bar{A}_x = \int_{0}^{\infty}\exp(-\delta t) {}_tp_x\mu_{x+t}dt\\ = \sum_{k=0}^{\infty}\int_k^{k+1}\exp(-\delta t) {}_tp_x\mu_{x+t}dt \\ = \sum_{k=0}^{\infty}{}_kp_{x}\exp(-\delta k)\int_0^{1}\exp(-\delta t ) {}_{t}p_{x+k}\mu_{x+k+t}dt \\= \sum_{k=0}^{\infty}{}_kp_{x}\exp(-\delta k)q_{x+k}\int_0^{1}\exp(-\delta t ) dt \\=\frac{1-\exp(-\delta)}{\delta} \sum_{k=0}^{\infty}\nu^{k} {}_kp_{x}q_{x+k} \\=\frac{1-\nu}{\delta \nu} \sum_{k=0}^{\infty}\nu^{k+1} {}_{k|}q_{x}\\ = \frac{1-\frac{1}{1+i}}{\delta \frac{1}{1+i}} A_x = \frac{i}{\delta}A_x.$$

Now consider $$A_x^{(m)}= \sum_{k=0}^{\infty}\nu^{\frac{k+1}{m}} {}_{\frac{k}{m}|\frac{1}{m}}q_x = \sum_{k=0}^{\infty}\nu^{\frac{k+1}{m}} \left({}_{\frac{k}{m}}p_x-{}_{\frac{k+1}{m}}p_x\right) $$

We know ${}_sp_{x} = 1-{}_s q_{x} = 1-s q_x$ for $0\le s \le1$ and so we will have

$${}_{k+s}p_{x}= {}_{k}p_{x} {}_sp_{x+k} = {}_{k}p_{x}(1-sq_{x+k})$$

and so $${}_{\frac{k}{m}}p_x-{}_{\frac{k+1}{m}}p_x = (1-\frac{k}{m} q_x) -(1-\frac{k+1}{m}q_x)=\frac{1}{m}q_x$$

if $\frac{k+1}{m} \le 1$. And so we will have

We also know that $\sum_{i=0}^{m-1} \nu^{\frac{i+1}{m}} = \nu^{1/m}\frac{1-\nu}{1-\nu^{\frac{1}{m}}}$ and so we will have

$$A_{x}^{(m)} = \frac{1-\nu}{1-\nu^{\frac{1}{m}}} \frac{1}{\nu}\sum_{k=0}^{\infty}\nu^{k+1}{}_{k|}p_x\\ =\frac{i}{1-\left(\frac{1}{1+i}\right)^{1/m}}\cdot (\frac{1}{1+i})^{1/m}A_x = \frac{i}{(1+i)^{1/m}-1} A_x = \frac{i}{i^{(m)}}A_x.$$ Yes I have finished this exercise.

We can approximate the following for term insurance with payment upon death:

$$\bar{A}_{x:\angl{n}} \approx \frac{i}{\delta} A_{x:\angl{n}}^1 + {}_nE_x.$$

Claim acceleration approach.

Let's assume that the insured individual's death occurs uniformly across the year, then for $A_{x}^{(m)}$ the payout most likely happen at the middle of the year and so if $m=2k+1$, then $k+1=\frac{m+1}{2}$ will be the time when the claim is paid. So we will have the following approximation:

$$A_x^{(m)} \approx \nu^{\frac{m+1}{2m}} q_x +\nu^{1+\frac{m+1}{2m}} {}_{1|}q_x+\nu^{2+\frac{m+1}{2m}} {}_{2|}q_x+... \\ =\sum_{k=0}^{\infty}\nu^{\frac{m+1}{2m}+k}{}_{k|}q_x\\=\nu^{\frac{-m+1}{2m}}\sum_{k=0}^{\infty}\nu^{k+1}{}_{k|}q_x\\\approx(1+i)^{\frac{m-1}{2m}}A_x$$

If we sent $m\to \infty$, we will have

$\bar{A}_{x}=(1+i)^{1/2}A_x$

And we will also have the same with term insurance payable upon $1/m$-th within the year.

$$\bar{A}{}_{x:\angl{n}}^1 \approx (1+i)^{\frac{1}{2}}A{}_{x:\angl{n}}^1 + {}_n E_x$$