# Insurance Benefits
## Summary
Formulae for the valuation of traditional insurance benefits, for whole life, term and endowment. For a basic review, $T_x$ is the future life time continuous random variable for life $(x)$, and $K_x$ is the curtate future life time. And a new random variable is $K_{x}^{(m)}$ as the benefits which depends on the number of complete period of length $\displaystyle \frac{1} {m}$ years the insured person $(x)$ lived.

## Intro
Assume that the interest rate is a constant. Assume a continuous benefit, like $\$1$ per year would be rather paid in infinitesimal unit of $\$\mathrm{d}t$ in every interval $(t,t + \mathrm{d}t)$.

## Assumption
$Def$

**Basis**: set of assumptions used in life insurance or pension calculations

And in this chapter we use the *Standard Ultimate Survival Model*: **Makeham’s law** with $A = 0.00022$, $B = 2.7 \times 10^{− 6}$, $c = 1.124$, an **ultimate basis**.

We assume a constant interest rate, the mortality model of the *ultimate basis*.

- The **effective annual rate of interst**, $i$, *Compound Interest*
- **annual accumulation factor**, $\newcommand{\ffrac} {\displaystyle \frac} 1+i$
- **discount factor**, $\boxed{ \upsilon = \ffrac{1} {1+i} }$
- **force of interest** (instantameous rate of growth of value), $\boxed{ \delta = \log(1+i) }$
- **nominal annual rate of interest compounded $p$ times per year**, $\boxed{ i^{\,(p)} = p\left( \left( 1+i \right)^{1/p} -1 \right) \Leftrightarrow 1+i = \left( 1+ \ffrac{i^{\,(p)}} {p} \right)^p }$
- **periodic rate**, $\ffrac{i^{\,(p)}} {p}$
- **The effective rate of discount per year**: $\boxed{ d = 1 - \upsilon = i\upsilon = 1 - e^{-\delta} }$
- **the nominal rate of discount compounded $p$ times per year**: $\boxed{ d^{\,(p)} = p\left( 1 - \upsilon^{1/p} \right) \Leftrightarrow \left( 1 - \ffrac{d^{\,(p)}} {p} \right)^p = \upsilon }$

The payment is contingent of the death of insured.

## Valuation of insurance benefits


### Whole life insurance: the continuous case, $\bar{A}_x$
$Def$

**Whole life insurance**: insurance contract that pays benefit IMMEDIATELY upon **death** of the insured, no matter how far in the future.

But the present value of that future payment, which is ONE DOLLAR, depends on payment date, or the date of death, which is *UNKNOWN*, thus it is an $r.v.$.

$$\boxed{ Z = \upsilon^{T_x}=e^{-\delta T_x} = \frac{1} {(1+i)^{T_x}} }$$

**EPV**: the expected present value, or the actuarial present value:

$$\boxed{ \bar{A}_x = E[Z] = E[e^{-\delta T_x}] = \int_{0} ^{\infty} e^{-\delta t} {_{t}}p_x \mu_{x+t} \, \mathrm{d}t }$$

$\dagger$ The bar upon $A$ means that the one dollar benefit is paid right on death. $\ddagger$

$\dagger$ And the EPV is the integral, of the product of the following
- presnet value of $1$ dollar: $1 \cdot e^{-\delta t}$
- probability of payting that at time $t$: ${_{t}}p_{x}\mu_{x+t}\mathrm{d}t$ $\ddagger$
***
$$\boxed{ E[Z^2] = E[e^{-2\delta T_x}]= \int_{0} ^{\infty} e^{-2\delta t} {_{t}}p_x \mu_{x+t} \, \mathrm{d}t = {^2}{\bar{A}_x} }$$

$\odot$ the $2$ on the left upper side of $\bar{A}$ meaning that calculation is at force of interest $2 \delta$, which is equivalent, at a rate of interest $j$, where $1 + j = e^{2\delta} = (1+i)^2$. $\triangleleft$

Then if the death benefit is $S$ then we have

$$E[SZ]=\int_{0} ^{\infty} Se^{-\delta t} {_{t}}p_x \mu_{x+t} \, \mathrm{d}t = S\bar{A}_x \\
Var[SZ]:=V[SZ] = S^2 \left( {^2}{\bar{A}_x} - \left( \bar{A}_x \right)^2 \right)$$

$\odot$ Low values of $Z$ are associated with high values of $T_x$. $\triangleleft$

### Whole life insurance: the annual case, $A_x$
Suppose now that the benefit of $1$ dollar is payable at the end of the year of death of $(x)$, rather than immediately on death. Since we use $K_x$ to denote curtate future life time, the time to the end of the year of death of $(x)$ is $K_x + 1$. So that now

$$\boxed{ Z = \upsilon^{K_x + 1} } \\
\begin{align}
A_x &= E[\upsilon^{K_x + 1}] = \boxed{ \sum_{k = 0}^{\infty} \upsilon^{k+1} \cdot {_{k \mid}}q_{x} } \\
&= \upsilon \cdot q_x + \upsilon^2 \cdot {_{1 \mid}}q_{x} + \upsilon^3 \cdot {_{2 \mid}}q_{x} + \cdots
\end{align}$$

Each term represents the EPV of a death benefit of a dollar paid at time $k$ conditional on the death of $(x)$ in $(k-1,k]$. 

And for the second moment, we have ${^2}A_x = E[\upsilon^{2(K_x + 1)}] = \boxed{ \sum_{k = 0}^{\infty} \upsilon^{2(k+1)} \cdot {_{k \mid}}q_{x} }$. And the variance is $S^2\left( {^2}A_x - \left( A_x \right)^2 \right)$

### Whole life insurance: the $1 /m \text{thly}$ case, $A_x^{(m)}$
$Def$

$K_x^{(m)}$, $r.v.$, where $m > 1$, as the future life time of $(x)$ in years rounded to lower $\ffrac{1} {m} \text{th}$ of a year. Commonly $m = 2,4,12$.

$$\boxed{ K_x^{(m)} = \frac{1} {m} \left \lfloor mT_x \right \rfloor }$$

For example:

$$\begin{array}{ccccc} \hline
T_x & K_x & K_x^{(2)} & K_x^{(4)} & K_x^{(12)} \\ \hline
23.675 & 23 & 23.5 & 23.5& {23\frac{8}{12}} \\
\upsilon^{T_x} & \upsilon^{K_{x}+1} & \upsilon^{K_{x}^{(2)} \, +1/2} & \upsilon^{K_{x}^{(4)} \, +1/4} & \upsilon^{K_{x}^{(12)} \, +1/12} \\ \hline
\end{array}$$
And for the probability:

$$Pr[K_x^{(m)} = k] = Pr[k \leq T_x < k + \frac{1} {m}] = {_{k \mid \frac{1} {m}}}q_x$$

For the expected value:

$$E[Z] = \boxed{ A_x ^{(m)} = \sum_{k=0}^{\infty} \upsilon^{\frac{k+1} {m}} {_{\frac{k} {m} \mid \frac{1} {m}}}q_x} = \upsilon^{1/m} {_{\frac{1} {m}}}q_x + \upsilon^{2/m} {_{{\frac{1} {m}} \mid {\frac{1} {m}}}}q_x + \upsilon^{3/m} {_{{\frac{2} {m}} \mid {\frac{1} {m}}}}q_x + \cdots $$

### Recursions
Although paying the death benefit at the year end, but sometimes we have to deal with the annual life tables. We can use **backwards recursion** to find the $A_x$.

Let the highest age in the table be $\omega$, so that we can assume that all lives expire by age $\omega$, expecially that $q_{\omega - 1} = 1$. Or if there's not a highest age, just choose one so that $q_{\omega - 1}$ is close enough to $1$.

And in that case, $K_{\omega - 1} = 0$

$$\boxed{ A_{\omega - 1} = E\left[ \upsilon ^{1+K_{\omega - 1}} \;\right] = \upsilon }$$

So that 

$$\begin{align}
A_x &= \sum_{k=1}^{\omega -x-1} \upsilon ^{k+1} \cdot {_{k}}p_x \cdot q_{x+k} \\
&= \upsilon \cdot q_x + \upsilon^2 \cdot p_x \cdot q_{x+1} + \upsilon^3 \cdot {_{2}}p_{x} \cdot q_{x+2} + \upsilon^4 \cdot {_{3}}p_{x} \cdot q_{x+3} + \cdots \\
&= \upsilon \cdot q_x + \upsilon \cdot p_x \left( \upsilon \cdot q_{x+1} + \upsilon^2 \cdot p_{x+1} \cdot q_{x+2} + \upsilon^3 \cdot {_{2}}p_{x+1} \cdot q_{x+3} + \cdots \right) \\
&= \boxed{\upsilon \cdot q_x + \upsilon \cdot p_x \cdot A_{x+1}}
\end{align}$$

And with this formula we can calculate $A_{x_{_0}}$ of the minimum age, backwards from $A_{\omega - 1} = \upsilon$.

And more terms backward are
0. $q_{\omega - 1} = 1$
1. $A_{\omega}$ not exists
2. $A_{\omega - 1} = v$
3. $A_{\omega - 2} = v q_{\omega - 2} + v \cdot p_{\omega - 2} \cdot v$
4. $A_{\omega - 3} = v q_{\omega - 3} + v^2 \cdot p_{\omega - 3} q_{\omega - 2} + v^3 p_{\omega - 3} p_{\omega - 2}$
5. $A_{\omega - i} = v \cdot q_{w-i} + \displaystyle \sum_{j=2}^i \left( v^j \left( \prod_{k=i-j+2}^{i} p_{\omega - k} \right) q_{\omega - (i-j+1)}\right) $
***
And for monthly benefits, the similar formula can be derived like:

$$\begin{align}
A_{x}^{(m)} &= \upsilon^{1/m}\cdot{_{\frac{1} {m}}}q_x + \upsilon^{2/m} \cdot {_{\frac{1} {m}}}p_x \cdot {_{\frac{1} {m}}} q_{x+\frac{1} {m}} + \upsilon^{3/m} \cdot {_{\frac{2} {m}}}p_x \cdot {_{\frac{1} {m}}} q_{x+\frac{2} {m}} + \cdots \\
&= \upsilon^{1/m} \cdot {_{\frac{1} {m}}}q_x + \upsilon^{1/m} \cdot {_{\frac{1} {m}}}p_x \left( \upsilon^{1/m} \cdot {_{\frac{1} {m}}}q_{x+\frac{1} {m}} + \upsilon^{2/m} \cdot {_{\frac{1} {m}}}p_{x+\frac{1} {m}} \cdot {_{\frac{1} {m}}}q_{x+\frac{2} {m}} + \cdots \right) \\
&= \boxed{\upsilon^{1/m} \cdot {_{\frac{1} {m}}}q_x + \upsilon^{1/m} \cdot {_{\frac{1} {m}}}p_x \cdot A^{(m)}_{x + \frac{1} {m}}}
\end{align}$$

>**e.g.**
>
> Using the Standard Ultimate Survival Model, and an interest rate of $5\%$ per year effective, calculate the mean and standard deviation of the present value of a benefit of $100000$ dollar payable $(\text{a})$ immediately on death, $(\text{b})$ at the end of the month of death, $(\text{c})$ at the end of the year of death for lives aged $20$, $40$, $60$, $80$ and $100$, and comment on the results.
>
> $(\text{a})$ Mean: $100000\bar{A}_{x}$; Standard Deviation: $100000 \sqrt{\sideset{^2}{_x}{\bar{A}} - \left( \bar{A}_{x} \right)^2}$; for $x = 20, 40, 60, 80$, with the effective rate of interest $j = (1+i)^2 - 1 = 10.25\%$
>
> $(\text{b})$ and  $(\text{c})$ just replace each $\bar{A}$ to $A^{(12)}$ and $A_{x}$.

### Term insurance
#### The continuous case $\bar {A}_{{x:\overline {n}|}}^{1}$

Now the death benefit is payable only if the policyholder dies *within* a fixed term of $n$ years. So now the present value, of a benefit of $1$ dollar, is

$$Z = \begin{cases}
\upsilon^{T_x} = e^{-\delta T_x} & \text{ if } T_x \leq n \\ 
0 & \text{ if } T_x > n 
\end{cases}$$

And here the EPV is

$$\boxed{ E[Z] = \bar {A}_{{x:\overline {n}|}}^{1} = \int_{0} ^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \,}$$

And similarly, the expected value of the square of the present value is

$$\boxed{ E[Z^2] = \sideset{^2}{_{{x:\overline {n}|}}^{1}}{\bar{A}} = \int_{0} ^{n} e^{-2\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \,}$$

#### The annual case ${A}_{{x:\overline {n}|}}^{1}$

Now the death benefit is payable at the *end* of the year of death, provided that the policyholder dies *within* a fixed term of $n$ years. So now the present value, of a benefit of $1$ dollar, is

$$Z = \begin{cases}
\upsilon^{K_x + 1} & \text{ if } K_x \leq n - 1 \\ 
0 & \text{ if } K_x \geq n 
\end{cases}$$

And here the EPV is

$$\boxed{ E[Z] = {A}_{{x:\overline {n}|}}^{1} = \sum_{k = 0} ^{n-1} \upsilon^{k + 1} {_{k \mid}}q_{x} }$$

And similarly, the expected value of the square of the present value is

$$\boxed{E[Z^2] = {^2}{A}_{{x:\overline {n}|}}^{1} = \sum_{k = 0} ^{n-1} \upsilon^{2( k + 1)} {_{k \mid}}q_{x} }$$

#### The $1 /m \text{thly}$ case, ${A^{(m)}}{_{x:\overline {n}|}^{1}}$
Now the death benefit is payable at the *end* of the $1 /m \text{thly}$ year of death, provided that the policyholder dies *within* a fixed term of $n$ years. So now the present value, of a benefit of $1$ dollar, is

$$Z = \begin{cases}
\upsilon^{K_x^{(m)} \; +  \frac{1} {m}} & \text{ if } K_x^{(m)} \leq n - \frac{1} {m} \\ 
0 & \text{ if } K_x^{(m)} \geq n 
\end{cases}$$

And here the EPV is

$$\boxed{ E[Z] = {A^{(m)}}{_{x:\overline {n}|}^{1}} = \sum_{k = 0} ^{mn-1} \upsilon^{(k + 1)/m} {_{\frac{k} {m} \mid \frac{1} {m}}}q_{x} }$$

And similarly, the expected value of the square of the present value is

$$\boxed{ E[Z^2] = {^2}{A^{(m)}}{_{{x:\overline {n}|}}^{1}} = \sum_{k = 0} ^{n-1} \upsilon^{2( k + 1 )/m} {_{k \mid}}q_{x} }$$

### Pure Endowment
$\odot$ "Pure Endowment" is a term. And "Endowment" normally means "Endowment Insurance". $\triangleleft$

$Def$

A $n\text{-year}$ pure endowment with sum insured $100000$ dollars, issued to $(x)$, will pay $100000$ dollars in $n$ years if $(x)$ is still alive at that time

So now the present value, of a benefit of $1$ dollar, is

$$Z = \begin{cases}
0 & \text{ if } T_x < n & P = 1 - {_{n}}p_{x}\\ 
\upsilon^{T_x} & \text{ if } T_x \geq n & P = {_{n}}p_{x}
\end{cases}$$

So that the EPV is

$${_n}E_{x}:= \boxed{ E[Z] = {A}{_{{x:\overline {n}|}}^{\;\;1}} = \upsilon^n \cdot {_{n}}p_{x}  }$$

### Endowment insurance
It's a combination of a *term insurance* and a *pure endowment*. The sum insured is payable on the death of $(x)$ should $(x)$ die within a fixed term, say $n$ years, but if $(x)$ survives for $n$ years, the sum insured is payable at the end of the $n$th year.

So the present value is

$$\begin{align}
Z &= \begin{cases}
\upsilon^{T_x} = e^{-\delta T_x} & \text{ if } T_x < n \\ 
\upsilon^n & \text{ if } T_x \geq n 
\end{cases} \\
&= \upsilon^{\min{(T_x \, , \, n)}} \\
&= e^{-\delta \min{(T_x \, , \, n)}}
\end{align}
$$

Then the EPV should be

$$\begin{align}
E[Z] &= \int_{0} ^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t &+ &\int_{n} ^{\infty} e^{-\delta n} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \int_{0} ^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t &+ &e^{-\delta n} {_{n}}p_{x} \\
&= \bar {A}_{{x:\overline {n}|}}^{1} &+ &{A}{_{{x:\overline {n}|}}^{\;\;1}}
\end{align}$$

In complete actuarial notation, it would be like

$$\boxed{ \bar {A}_{{x:\overline {n}|}} = \bar{A}_{{x:\overline {n}|}}^{1} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \, }$$

And similarly, the expected value of the squared present value of the benefit is 

$$\boxed{ {^{2}}\bar{A}_{{x:\overline {n}|}} = \int_{0} ^{n} e^{-2\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t + e^{-2\delta n} {_{n}}p_{x} = {^{2}}\bar{A}_{{x:\overline {n}|}}^{1} + {^{2}}{A}{_{{x:\overline {n}|}}^{\;\;1}} \, }$$

***

And if the benefit is paid at the end of the year of death, then the present value would be like

$$\begin{align}
Z &= \begin{cases}
\upsilon^{K_x + 1} & \text{ if } K_x \leq n - 1 \\ 
\upsilon^n & \text{ if } K_x \geq n 
\end{cases} \\
&= \upsilon^{\min{(K_x + 1,n)}}
\end{align}$$

So that the EPV is like

$$\boxed{ \sum_{k=0} ^{n-1} \upsilon^{k+1} {_{k\mid}}q_{x} + \upsilon^n  P[K_x \geq n] = {A}_{{x:\overline {n}|}}^{1} + \upsilon^n {_{n}}p_{x} \, }$$

And in complete actuarial notation, it will be like

$$\boxed{ {A}_{{x:\overline {n}|}} = {A}_{{x:\overline {n}|}}^{1} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \, }$$

Similarly, the expected value of the squared present value of the benefit is

$$\boxed{ {^{2}}{A}_{{x:\overline {n}|}} = \sum_{k=0} ^{n-1} \upsilon^{2(k+1)} {_{k\mid}}q_{x} + \upsilon^{2n} {_{n}}p_{x} = {^{2}}{A}_{{x:\overline {n}|}}^{1} + {^{2}}{A}{_{{x:\overline {n}|}}^{\;\;1}} \, }$$

***

Finally if the death benefit is payable at the end of the $1 /m \text{th}$ year of death, then the present value of the benefit is

$$\begin{align}
Z &= \begin{cases}
\upsilon^{K_x^{(m)} \; +  \frac{1} {m}} & \text{ if } K_x^{(m)} \leq n - \frac{1} {m} \\ 
\upsilon^n & \text{ if } K_x^{(m)} \geq n 
\end{cases}\\
&= \upsilon^{\min{(K_x^{(m)} \; +  \frac{1} {m}\:,\: n)}} 
\end{align}$$

Here the EPV of the benefit is

$$\sum_{k = 0} ^{mn-1} \upsilon^{(k + 1)/m} {_{\frac{k} {m} \mid \frac{1} {m}}}q_{x} + \upsilon^n P[K_x^{(m)} \geq n] = {A^{(m)}}{_{x:\overline {n}|}^{1}} + \upsilon^n {_{n}}p_{x}$$

And in complete actuarial notation we have

$$\boxed{ A^{(m)}_{x:\overline {n}|} = {A^{(m)}}{_{x:\overline {n}|}^{1}} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \, }$$

### Deferred insurance benefits
Deferred insurance refers to insurance which does not begin to offer death benefit cover until the end of a deferred period. 

For example, a death benefit will be paid immediately on the death of $(x)$, provided that $(x)$ dies between ages $x+u$ and $x + u + n$. So in this condition, the present value random variable ($p.v.\;r.v.$ later) is

$$Z = \begin{cases}
0 & \text{ if } T_x < u \text{ or } T_x \geq u+n \\ 
e^{-\delta T_x} & \text{ if } u \leq T_x < u+n
\end{cases}$$

And the actuarial notation for the EPV of the deferred term insurance benefit is

$$\begin{align}
{_{u\mid}}\bar{A}_{{x:\overline {n}|}}^{1} &= \int_{u} ^{u+n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&\stackrel{s = t - u}= \int_{0} ^{n} e^{-\delta(s+u)} {_{s+u}}p_{x} \mu_{x+s+u} \, \mathrm{d}s \\
&= \boxed{ e^{-\delta u} {_{u}}p_{x} \int_{0} ^{n} e^{-\delta s} {_{s}}p_{x+u} \mu_{x+s+u} \, \mathrm{d}s \, }\\
&= \boxed{ e^{-\delta u} {_{u}}p_{x} \bar{A}_{x+u:\overline {n}|}^{1} \,} \\
&= \boxed{ {_{u}}E_{x}\bar{A}_{x+u:\overline {n}|}^{1} \,}
\end{align}$$

Another fomula is

$$\begin{align}
{_{u\mid}}\bar{A}_{{x:\overline {n}|}}^{1} &= \int_{u} ^{u+n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \int_{0} ^{u+n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t - \int_{0} ^{u} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \boxed{ \bar{A}_{{x:\overline {u+n}\displaystyle{|}}}^{1} - \bar{A}_{{x:\overline {u}|}}^{1}}
\end{align}$$

$\odot$ Thus, the EPV of a deferred term insurance benefit can be found by differencing the EPVs of term insurance benefits for terms $u + n$ and $u$. $\triangleleft$

Then we can use deffered term insurance polices to decompose the $n\text{-year}$ term insurance

$$\begin{align}
\bar {A}_{{x:\overline {n}|}}^{1} &= \int_{0}^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \sum_{r=0} ^{n-1} \int_{r}^{r+1} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \boxed{ \sum_{r=0} ^{n-1} {_{r\mid}}\bar{A}_{{x:\overline {1}|}}^{1}}
\end{align}$$

and similarly for whole life insurance policy.

$$\begin{align}
\bar {A}_{x} &= \int_{0}^{\infty} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \sum_{r=0} ^{\infty} \int_{r}^{r+1} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\
&= \sum_{r=0} ^{\infty} {_{r\mid}}\bar{A}_{{x:\overline {1}\displaystyle{|}}}^{1}
\end{align}$$

In particular, we should note that

$$\begin{align} 
A_x &= \boxed{{A}_{{x:\overline {n}|}}^{1} + {_{n\mid}}A_x }  \\
{A}_{{x:\overline {n}|}}^{1} &= A_x - {_{n\mid}}A_x  \\
&= A_x - \upsilon^n {_{n}}p_{x} A_{x+n}
\end{align}$$

And we can also get that

$$\begin{align} 
{_{n\mid}}A_x &= A_x - {A}_{{x:\overline {n}|}}^{1}\\
&= \sum_{k=0}^{\infty} \upsilon^{k+1} {_{k\mid}}q_{x} - \sum_{k=0}^{n-1} \upsilon^{k+1} {_{k\mid}}q_{x} \\
&= \sum_{k=n}^{\infty} \upsilon^{k+1} {_{k\mid}}q_{x} \\
&= \upsilon^{n} \cdot {_{n}}p_{x} \cdot A_{x+n}
\end{align}$$

>**e.g.** Given $F_0(t) = 1 - \left( 1 - \ffrac{t} {10} \right)^{(1/2)}$, $t \in [0,10]$, find the $A_x$ using recursion, using $i=0.07$.
> 
> First we need to find $\upsilon = \ffrac{1} {1+0.07} = 0.935$; then 
>
>$${_{t}}p_{x}= S_x(t)= \frac{S_0(x+t)} {S_0(x)}=\frac{\left( 1-(x+t)/10 \right)^{1/2} } {\left( 1 - x/10 \right)^{1/2}}= \left( \frac{10-(t+x)} {10-x} \right)^{1/2}$$
>
>$$p_{x} = \left( \frac{9-x} {10-x} \right)^{1/2}, q_{x} = 1-p_{x}$$
> 
>Then using the regression: $A_x = \upsilon q_{x} + \upsilon p_{x} A_{x+1} $, from the initial condition: $q_{\omega-1}=1,A_{\omega-1}=\upsilon$
>
>$$\begin{array}{cccc}\hline
x & p_x & q_x & A_x \\\hline
0 & 0.948683 & 0.051317 & 0.630980951 \\
1 & 0.942809 & 0.057191 & 0.657577631 \\
2 & 0.935414 & 0.064586 & 0.685628879 \\
3 & 0.925820 & 0.074180 & 0.715230902 \\
4 & 0.912871 & 0.087129 & 0.746491857 \\
5 & 0.894427 & 0.105573 & 0.779537603 \\
6 & 0.866025 & 0.133975 & 0.814524014 \\
7 & 0.816497 & 0.183503 & 0.851667972 \\
8 & 0.707107 & 0.292893 & 0.891346428 \\
9 & 0        & 1        & 0.934579439 \\\hline
\end{array}$$
>
>Or use the definition: $A_0 = \displaystyle \sum_{k=0}^{9} \upsilon^{k+1} \cdot {_{k}}p_{0} \cdot q_{k},\dots, \boxed{ A_x = \sum_{k=0}^{\omega-1-x} \upsilon^{k+1} \cdot {_{k}}p_{x} \cdot q_{x+k}\,},A_9 = \sum_{k=0}^{0} \upsilon^{k+1} \cdot {_{k}}p_{9} \cdot q_{9+k}$

>**e.g.** Exercise 4.1
>
>Given $i = 6\%$ and the following table
>
>$$\begin{array}{ccc} \hline
x & l_x & A_x \\ \hline
35 & 100000.00 & 0.151375 \\ 
36 &  99737.15 & 0.158245 \\ 
37 &  99455.91 & 0.165386 \\ 
38 &  99154.72 & 0.172804 \\ 
39 &  98831.91 & 0.180505 \\ 
40 &  98485.68 & 0.188492 \\ \hline
\end{array}$$
>
>($\text{a}$) ${_{5}}E_{35}$
> 
>>First we cauculate $\upsilon = \ffrac{1} {1+0.06} = 0.943396$

>>${_5}p_{35} = \ffrac{l_{35+5}} {l_{35}} = 0.984857$

>>${_5}E_{35} = \upsilon^5 \cdot {_5}p_{35} = 0.7359423$

> ($\text{b}$) ${A}_{{35:\overline {5}\displaystyle{|}}}^{1}$

>> $$\begin{align}
{A}_{{35:\overline {5}|}}^{1} &= \sum_{k=0}^{5-1} \upsilon^{k+1} \cdot {_{k\mid}}p_{35} \\
&= \upsilon^1 \cdot \frac{l_{35} - l_{36}} {l_{35}} + \upsilon^2 \cdot \frac{l_{36} - l_{37}} {l_{35}} + \upsilon^3 \cdot \frac{l_{35} - l_{35}} {l_{35}} + \upsilon^4 \cdot \frac{l_{35} - l_{35}} {l_{35}} + \upsilon^5 \cdot \frac{l_{35} - l_{35}} {l_{35}} \\
&= 0.012656 
\end{align}$$

> ($\text{c}$) ${_{5\mid}}A_{35}$

>>$$\begin{align}
{_{5\mid}}A_{35} &= \upsilon^5 \cdot {_{5}}p_{35} \cdot A_{40} \\
&= A_{35} - {A}_{{35:\overline {5}\displaystyle{|}}}^{1} \\
&= 0.151375 - 0.012656 = 0.138719
\end{align}$$

>**e.g.** Exercise 4.5
>
>Show that
>
>$${A}_{{x:\overline {n}|}} = \sum_{k=0}^{n-2} \upsilon^{k+1} {_{k}}q_{x} + \upsilon^n \cdot {_{n-1}}p_{x}$$
>
>$$\begin{align}
{A}_{{x:\overline {n}|}} &= {A}{_{{x:\overline {n}|}}^{1}} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \\
&= \left( \sum_{k=0}^{n-1} \upsilon^{k+1} \cdot {_{k\mid}}q_{x}\right) + \upsilon^n \cdot {_{n}}p_{x} \\
&= \sum_{k=0}^{n-2} \upsilon^{k+1} {_{k}}q_{x} + \upsilon^{n-1+1} {_{n-1\mid}}q_{x} + \upsilon^n \cdot {_{n}}p_{x} \\
&= \sum_{k=0}^{n-2} \upsilon^{k+1} {_{k}}q_{x} + \upsilon^n \cdot {_{n-1}}p_{x}
\end{align}$$

>**e.g.** Exercise 4.11
>
>- Under an endowment insurance issued to a life aged $x$, let $X$ denote the present value of a unit sum insured, payable at the moment of death or at the end of the $n\text{-year}$ term.
>- Under a term insurance issued to a life aged $x$, let $Y$ denote the present value of a unit sum insured, payable at the moment of death within the $n\text{-year}$ term.
>
>Given that $V [ X ] = 0.0052, \upsilon^n = 0.3, {_{n}}p_{x} = 0.8, E [ Y ] = 0.04$, calculate $V [ Y ]$.
>
>>So far we already know that
>>
>>$$\begin{array}{cc}
X = \begin{cases}
\upsilon^{T_{x}} & \text{ if } T_{x} \leq n \\ 
\upsilon^{n} & \text{ if } T_{x} > n
\end{cases} & Y = \begin{cases}
\upsilon^{T_{x}} & \text{ if } T_{x} \leq n \\ 
0 & \text{ if } T_{x} > n
\end{cases}
\end{array}$$
>>
>>Obviously we need to construct a Intermediate variable to connect them
>>
>>$$Z = X-Y = \begin{cases}
0 & \text{ if } T_{x} \leq n \\ 
\upsilon^{n} & \text{ if } T_{x} > n
\end{cases}$$
>>
>>A pure endowment! So that now
>>
>>$$\begin{align}
V[Y] &= V[X] + V[Z] - 2Cov[X,Z] \\
&= V[X] + E[Z^2] - \left( E[Z] \right)^2 - 2\left( E[XZ] - E[X] \cdot E[Z] \right)\\
&= V[X] + \upsilon^{2n} \cdot {_{n}}p_{x} - \left( \upsilon^{n} \cdot {_{n}}p_{x} \right)^2 - 2\left( E[Z^2] - \left( E[Y] + \upsilon^n \cdot {_{n}}p_{x} \right) \cdot E[Z] \right)\\
&= 0.0052 + 0.3^2 \times 0.8 -\left( 0.3 \times 0.8 \right)^2 \\
&\;\;\;\;-2 \times \left( 0.3^2 \times 0.8 - \left( 0.04 + 0.3 \times 0.8 \right) \times 0.3 \times 0.8 \right)\\
&= 0.01
\end{align}$$

## Relating $\bar{A}_x$, $A_x$, and $A_{x}^{(m)}$
### UDD
$${_{s}}q_{y} \overset{\mathrm{UDD}}{ = } s\cdot q_{y} \Rightarrow \frac{\mathrm{d}} {\mathrm{d}s}\left({_{s}}q_{y}\right) = q_{y} \\
f_{y}(s) = {_{s}}p_{y} \cdot \mu_{y+s}$$
$$\Rightarrow {_{s}}p_{y} \cdot \mu_{y+s} \overset{\mathrm{UDD}}{ = } \:q_y$$

Then we have

$$\begin{align}
\bar{A}_x &= \int_{0}^{\infty} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \; \mathrm{d}t = \sum_{k=0} ^{\infty} \int_{k}^{k+1} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \; \mathrm{d}t \\
&\overset{t=k+s}{ = } \sum_{k=0} ^{\infty} \int_{0}^{1} e^{\delta}e^{-\delta s}e^{-\delta}e^{-\delta k} {_{k+s}}p_{x} \mu_{x+k+s} \; \mathrm{d}(k+s) \\
&= \sum_{k=0} ^{\infty} \int_{0}^{1}e^{(1-s)\delta}e^{-(1+k)\delta} {_{k}}p_{x}\cdot {_{s}}p_{x+k} \mu_{x+k+s} \mathrm{d}s \\
&= \sum_{k=0} ^{\infty} {_{k}}p_{x} \upsilon^{k+1} \int_{0}^{1}e^{(1-s)\delta} {_{s}}p_{x+k} \mu_{x+k+s} \mathrm{d}s \\
&\overset{\mathrm{UDD}}{ = }  \sum_{k=0} ^{\infty} {_{k}}p_{x} \cdot q_{x+k} \upsilon^{k+1} \int_{0}^{1}e^{(1-s)\delta} \mathrm{d}s =A_x \frac{e^{\delta} - 1} {\delta}
\end{align}$$

So finally since $e^{\delta} = 1 + i$, so that

$$\boxed{\bar{A}_x \approx \frac{i} {\delta}A_x}$$

Similarly, we have a table

$$\begin{array}{ccc} \hline
&\text{UDD Method} & \text{Similar idea} \\ \hline
\text{Whole life} & \bar{A}_x \approx \frac{i} {\delta}A_x & A_x^{(m)} = \frac{i} {i^{(m)}} \; A_x \\
n\text{-year term} &  \bar {A}_{{x:\overline {n}|}}^{1} \approx \frac{i} {\delta}{A}_{{x:\overline {n}|}}^{1} & {A^{(m)}}{_{x:\overline {n}|}^{1}} = \frac{i} {i^{(m)}} \; {A}_{{x:\overline {n}|}}^{1} \\
\text{Endowment} & \bar {A}_{{x:\overline {n}|}}\approx \frac{i} {\delta} \bar {A}_{{x:\overline {n}|}}^{1} + \upsilon^n {_{n}}p_{x} & A^{(m)}_{x:\overline {n}|} = \frac{i} {i^{(m)}} \; {A}_{{x:\overline {n}|}}^{1} + \upsilon^n {_{n}}p_{x} \\ \hline
\end{array}$$

And the deferred one: ${_{n|}}\bar{A}_x \approx \ffrac{i} {\delta} {_{n|}}{A}_x$

$\dagger$ The $i^{(m)}$ is defined by

$$1+i = \left( 1+ \frac{i^{(m)}} {m} \right)^m$$

### Using the claims acceleration approach
>**e.g.** From the relation of $A_x$ and $A_x^{(4)}$
>
>From year $K_x$ to $K_x + 1$, four different payment may happen for $A_x^{(4)}$, which are $K_x + 1/4$, $K_x + 2/4$, $K_x + 3/4$, $K_x + 4/4$. So that if we assume that the deaths occur evenly over the year, then the average payment time is

>$$\frac{1} {4} \left( K_x + 1/4 + K_x + 2/4 + K_x + 3/4 + K_x + 4/4\right) = K_x + \frac{5} {8}$$
>
>NOT $\ffrac{2} {4}$

$Conclusion$

For $K_{x}^{(m)}$, we have the average is 

$$\frac{1} {m} \left( K_x + 1/m + K_x + 2/m + \cdots + K_x + m/m\right) = K_x + \frac{m+1} {2m}$$

So that in general, for an $m-\text{thly}$ death benefit, assuming deaths are uniformly distributed over the year of age, the average time of payment of the death benefit is $( m + 1 )/ 2m$ in the year of death. So we have the approximation

$$\begin{align}
A_x^{(m)} &\approx q_x \cdot \upsilon^{\frac{m+1} {2m}} + {_{1|}}q_x \cdot \upsilon^{1+ \frac{m+1} {2m}} + {_{2|}}q_x \cdot \upsilon^{2+ \frac{m+1} {2m}} + \cdots \\
&= \sum_{k=0}^{\infty} {_{k|}}q_x \cdot \upsilon^{k+ \frac{m+1} {2m}} \\
&= \left( 1+i \right)^{\frac{m-1} {2m}} \sum_{k=0}^{\infty} {_{k|}}q_x \cdot \upsilon^{k+1}
\end{align}$$

That is 

$$\boxed{ A_x^{(m)} \approx \left( 1+i \right)^{\frac{m-1} {2m}} A_x }$$

$\dagger$ Note that

$$A_x^{(m)} = \sum_{k=0}^{\infty} {_{\frac{k} {m}|\frac{1} {m}}}q_x \cdot \upsilon^{(k+1)/m} \neq \sum_{k=0}^{\infty} {_{k|}}q_x \cdot \upsilon^{k+ \frac{m+1} {2m}}$$ $\ddagger$

And for the continuous benefit EPV, $\bar{A}_x$, we can just let $m \to \infty$ so that

$$\boxed{ \bar{A}_x \approx \left( 1+i \right)^{1/2} A_x }$$

This implies that assume that lives die uniformly through the year, the benefit paid immediately on death on average is half a year ealier than the benefit valued by $A_x$.

Similarly for the endowment

$$\boxed{ \bar{A}_{x:\overline{n}|} \approx (1+i)^{1/2} A^1_{x:\overline{n}|} + \upsilon^n \cdot {_{n}}p_x}$$

$$\boxed{ {A}_{x:\overline{n}|}^{(m)} \approx (1+i)^{1/2} {A^{(m)}}{_{x:\overline {n}|}^{1}} + \upsilon^n \cdot {_{n}}p_x}$$

## Variable insurance benefits
Consider an event $E$, then the indicator random variables is defined as

$$I(E) = \begin{cases}
1 & \text{ if } E \text{ is true} \\ 
0 & \text{ if } E \text{ is not true} 
\end{cases}$$

Now we can see that we use the expectation of this variable to replace the probability of the event.

$$\newcommand{\EE}{\mathbb{E}}\EE[I(E)] = \Pr[E]$$

However in continuous condition, the indicator variable is defined as $I(t < T_x \leq t + \mathrm{d}t)$, so that 

$$\begin{align}
\EE[I(t < T_x \leq t + \mathrm{d}t)] &= \Pr[t < T_x \leq t + \mathrm{d}t] \\
&= \Pr[t < T_x] \cdot \Pr[T_x \leq t + \mathrm{d}t | t < T_x] \\
&\approx {_{t}}p_{x} \cdot \mu_{x+t} \mathrm{d} t
\end{align}$$

>**e.g.**
>
>For a discrete case, PV: $Z = 1000 \cdot I\left( T_x \leq 10 \right) \upsilon^{10} + 2000 \cdot I\left( 10 < T_x \leq 20 \right) \upsilon^{20}$. Then we have the EPV
>
>$$1000 \cdot {_{10}}q_{x} \upsilon^{10} + 2000 \cdot {_{10|10}}q_{x} \upsilon^{20}$$
>***
>An **increasing insurance** policy with a death benefit of $T_x$ payable at the moment of death. That is, the benefit is the number of years lived by an insured life from age $x$ to his or her death. So it's not hard to find the EPV
>
>$$\int_{0}^{\infty} t e^{-\delta t}{_{t}}p_{x} \cdot \mu_{x+t} \; \mathrm{d} t:=\left( \bar{I}\bar{A}\right)_x$$
>
>Here's how to use indicator to find this value
>
>$$\begin{align}
\left( \bar{I}\bar{A}\right)_x &= \EE\left[\lim_{\mathrm{d}t \to 0} \sum_{i=1}^{\infty} t_i\upsilon^{t_i} I\left( t_i < T_x \leq t_i + \mathrm{d}t \right)\right] \\
&= \lim_{\mathrm{d}t \to 0} \sum_{i=1}^{\infty} t_i\upsilon^{t_i} \EE\left[ I\left( t_i < T_x \leq t_i + \mathrm{d}t \right) \right] \\
&=\int_{0}^{\infty} t \cdot \upsilon^{t} \cdot {_{t}}p_{x} \cdot \mu_{x+t} \mathrm{d} t \\
\end{align}$$
>
>First we define the present value, PV: $Z = T_x e^{-\delta T_x}$
>
>$$\EE[Z^k] = \int_{0}^{\infty} \left( t e^{-\delta t} \right)^k {_{t}}p_{x} \cdot \mu_{x+t} \; \mathrm{d} t$$
>
>And if the policy ceases after $n$ years, we have
>
>$$\left( \bar{I}\bar{A}\right)^{1}_{x:\overline{n}|} = \int_{0}^{n} t e^{-\delta t}{_{t}}p_{x} \cdot \mu_{x+t} \; \mathrm{d} t$$
>
>Another common one policy is with geometrically increasing benefit. One form is to return the profits on the premium back to the insurer. Another is like
>
>>Benefit would be $(1+j)^k$ if death happens in $(x+k,x+k+1]$. So that the EPV would be
>>
>>$$\sum_{k=0}^{n-1} (1+j)^k \upsilon^{k+1} {_{k|}}q_{x} = \frac{1} {1+j} \sum_{k=0}^{n-1} \left(\frac{1+j} {1+i} \right)^{k+1} {_{k}}p_{x} q_{x+k}$$
>>
>>So that we can define $v^* = \ffrac{1+j} {1+i}$, and $i^* = \ffrac{1} {v^*} - 1 = \ffrac{i-j} {1+j}$, and finally the EPV would be, under the assumption $j<i$
>>
>>$$\frac{1} {1+j} \sum_{k=0}^{n-1} \left(v^* \right)^{k+1} {_{k|}}p_{x} q_{x+k}:= A^{1}_{x:\overline{n}|i^*}$$

>**e.g.** Example 4.6
>
>$n\text{-year}$ term insurance policy issued to $(x)$, under which the death benefit is $k+1$ if death occurs between ages $x+k$ and $x+k+1$, for $k = 1, 2, \dots, n-1$
>
>We can directly write out the EPV of this death benefit:
>
>$$\left( IA \right)^{1}_{x:\overline{n}|}:=\sum_{k=0}^{n-1} \upsilon^{k+1} \cdot (k+1) {_{k|}}q_{x}$$
>
>And by the way if the term $n$ is infinite, we can denote that as
>
>$$\left( IA \right)_{x}:=\sum_{k=0}^{\infty} \upsilon^{k+1} \cdot (k+1) {_{k|}}q_{x}$$

>However to find the variance
>
>$$Z = \begin{cases}
\left( K_x + 1\right) \upsilon^{K_x + 1} & \text{ if } K_x \leq n \\ 
0 & \text{ if } K_x > n 
\end{cases}$$
>
>$$E[Z^2] = \sum_{k=0}^{n-1} \upsilon^{2(k+1)} (l+k)^2 {_{k|}}q_{x}$$
>
>So that the variance is just $V[Z] = \displaystyle \sum_{k=0}^{n-1} \upsilon^{2(k+1)} (l+k)^2 {_{k|}}q_{x} - \left( \left( IA \right)^{1}_{x:\overline{n}|} \right)^2$



<center> **Summary Table** </center>

Before everything...

- The **effective annual rate of interst**, $i$, *Compound Interest*
- **discount factor**, $\upsilon = \ffrac{1} {1+i}$
- **force of interest**, $\delta = \log(1+i)$
- **The effective rate of discount per year**, $d = 1 - \upsilon = i\upsilon = 1 - e^{-\delta}$

<center> Insurance </center>

$$\begin{array}{ccc} \hline
& \text{Whole life} & \text{Term} \\ \hline
\text{Continuous} & \bar{A}_x = \displaystyle \int_{0} ^{\infty} e^{-\delta t} {_{t}}p_x \mu_{x+t} \, \mathrm{d}t & \bar {A}_{{x:\overline {n}|}}^{1} = \displaystyle \int_{0} ^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t \\[1em]
\text{Annual} & A_x = \displaystyle \sum_{k = 0}^{\infty} \upsilon^{k+1} \cdot {_{k \mid}}q_{x} & {A}_{{x:\overline {n}|}}^{1} = \displaystyle \sum_{k = 0} ^{n-1} \upsilon^{k + 1} {_{k \mid}}q_{x} \\[1em]
m\text{thly} & A_x ^{(m)} = \displaystyle \sum_{k=0}^{\infty} \upsilon^{(k + 1)/m} {_{\frac{k} {m} \mid \frac{1} {m}}}q_x & {A^{(m)}}{_{x:\overline {n}|}^{1}} = \displaystyle \sum_{k = 0} ^{mn-1} \upsilon^{(k + 1)/m} {_{\frac{k} {m} \mid \frac{1} {m}}}q_{x} \\[1em] \hline
\end{array}
$$
***

<center> Recursion </center>

$$\boxed{ A_x = \sum_{k=0}^{\omega-1-x} \upsilon^{k+1} \cdot {_{k}}p_{x} \cdot q_{x+k} = \upsilon \cdot q_{x} + \upsilon \cdot p_{x} \cdot A_{x+1}}$$

$$A_{x}^{(m)} = \upsilon^{1/m} \cdot {_{\frac{1} {m}}}q_{x} + \upsilon^{1/m} \cdot {_{\frac{1} {m}}}p_{x} \cdot A_{x + \frac{1} {m}}^{(m)} $$
***

<center> Endowment </center>

$$\begin{array}{cc} \hline
\text{Pure Endowment} & \text{Endowment} \\ \hline
{_n}E_{x} = {A}{_{{x:\overline {n}|}}^{\;\;1}} = \upsilon^n \cdot {_{n}}p_{x}  & 
\begin{array}{rc}
\text{Continuous} & \bar {A}_{{x:\overline {n}|}} = \bar{A}_{{x:\overline {n}|}}^{1} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \\[1em]
\text{Annual} & {A}_{{x:\overline {n}|}} = {A}_{{x:\overline {n}|}}^{1}+{A}{_{{x:\overline {n}|}}^{\;\;1}} \\[1em]
m\text{thly} & A^{(m)}_{x:\overline {n}|} = {A^{(m)}}{_{x:\overline {n}|}^{1}} + {A}{_{{x:\overline {n}|}}^{\;\;1}} \\[1em]
\end{array} \\ \hline
\end{array}
$$
***

<center> Deferred Insurance </center>

$$\begin{align}
{_{u\mid}}\bar{A}_{{x:\overline {n}|}}^{1} &= {_{u}}E_{x} \cdot \bar{A}_{x+u:\overline {n}|}^{1} \\
&= {A}{_{{x:\overline {n}|}}^{\;\;1}} \cdot \bar{A}_{x+u:\overline {n}|}^{1} \\
&= \boxed{ \bar{A}_{{x:\overline {u+n}\displaystyle{|}}}^{1} - \bar{A}_{{x:\overline {u}|}}^{1}} 
\end{align}$$

- Decomposition of $n\text{-year}$ term insurance
$$
\bar{A}_{{x:\overline {n}|}}^{1} = \int_{0}^{n} e^{-\delta t} {_{t}}p_{x} \mu_{x+t} \, \mathrm{d}t = \sum_{r=0} ^{n-1} {_{r\mid}}\bar{A}_{{x:\overline {1}\displaystyle{|}}}^{1}
$$
- Decomposition of whole life insurance
$$\bar{A}_{x} = \sum_{r=0} ^{\infty} {_{r\mid}}\bar{A}_{{x:\overline {1}\displaystyle{|}}}^{1}$$
- One last formula!

$$\begin{align}
{_{n|}}A_x &= {_{n}}E_x \cdot A_{x+n} = \boxed{ \upsilon^n \cdot {_{n}}p_{x} A_{x+n} } \\
&= A_x - A^1_{x:\overline{n}|}
\end{align}$$

$$\boxed{A_x = {A}_{{x:\overline {n}|}}^{1} + {_{n\mid}}A_x }$$

## Questions from quizs

- The **expected present value** of a policy issued to a life aged $x$ with a $1$ *dollar* death benefit payable *at the end of the* **month** ($m=12$) of death (regardless of when that death occurs).

$$A^{(12)}_x = \sum_{k=0}^{\infty} \upsilon^{(k + 1)/12} {_{\frac{k} {12} \mid \frac{1} {12}}}q_x$$

- The **expected present value** of a policy issued to a life aged $x$ a $1$ *dollar* death benefit payable *right after death* given the insurer doesn't suvive $n$ term year.

$$\bar{A}^1_{x:\overline{n}|}$$

- ${_{4}}E_{35}$ comes first, then ${_{4|}}A_{35} = {_{4}}E_{35} A_{35+4}$ comes next, then the $A^{1}_{35:\overline{4}\displaystyle{|}} = \sum\limits_{k=0}^{3} \upsilon^{k+1} {_{k|}}q_{35} = A_{35} - {_{4|}}A_{35}$. And finally $A_{35:\overline{4}\displaystyle{|}} = A^{1}_{35:\overline{4}\displaystyle{|}} + A^{\;\;\;1}_{35:\overline{4}\displaystyle{|}} = A^{1}_{35:\overline{4}\displaystyle{|}} + {_{4}}E_{35}$

41ce6723a04013019c0a39c7fa6fe4b1582a663f
ipynb
Jupyter Notebook
FinMath/Actuarial Mathematics/AcM_note_Chap04.ipynb
XavierOwen/Notes
d262a9103b29ee043aa198b475654aabd7a2818d
MIT
2018-11-27T10:31:08.000Z
2019-01-20T03:11:58.000Z
FinMath/Actuarial Mathematics/AcM_note_Chap04.ipynb
XavierOwen/Notes
d262a9103b29ee043aa198b475654aabd7a2818d
MIT
FinMath/Actuarial Mathematics/AcM_note_Chap04.ipynb
XavierOwen/Notes
d262a9103b29ee043aa198b475654aabd7a2818d
MIT
2020-07-14T19:57:23.000Z
2020-07-14T19:57:23.000Z
Qwen/Qwen-72B
1. YES
2. YES