M8 Lognormal Models and Life Insurance Applications

Topics in Insurance, Risk, and Finance

Author

Affiliation

Yuyu Chen

Department of Economics, University of Melbourne

Learning outcomes

• Know and derive the properties of lognormal distributions.
• Derive explicit formulas for the distributions and related quantities of $S_n$ and $V_n$ in the lognormal model.
• Define, calculate, and interpret Value-at-Risk.
• Know basic notation for life insurance analysis.
• Compute expected present value of benefit for life insurance and annuity in stochastic models for rates of return.
• Use the equivalence principle to compute premiums of life insurance products.

Review: normal distribution

Lognormal model

• We have studied the moments of cash series.
• To have a more detailed analysis of investment activities (e.g., probability of default), we need to derive distribution functions.
• In particular, distributions can be used to determine the capital requirements for financial institutions (Value-at-Risk for insurance companies and Expected Shortfall for banks).
• The task of finding distributions is however very challenging.
• We will study the case when the accumulation factor $(1+i_t)$ follows a lognormal distribution, in which case we can derive the exact distribution of $S_n$.

Review: normal distribution I

We say a random variable $X$ follows a normal distribution with parameters $\mu$ and ${\sigma }^2$, denoted by $X\sim N(\mu ,{\sigma }^2)$, if the density function of $X$ can be written as $$f(x)=\frac{1}{\sqrt{2\pi }\sigma }\exp (-\frac{(x-\mu {)}^2}{2{\sigma }^2}).$$

• $\mu$ and ${\sigma }^2$ are the mean and variance of $X$.
• If $\mu =0$ and $\sigma =1$, $X$ is usually denoted by $Z$ and is called a standard normal random variable.
• The distribution function and the density function of $Z$ are usually denoted by $\mathrm{\Phi }$ and $\varphi$.

Review: normal distribution II

• Symmetry: $\mathrm{\Phi }(x)+\mathrm{\Phi }(-x)=1$ for $x\in \text{R}$.

• Standalization: For $X\sim N(\mu ,{\sigma }^2)$, we have $$\frac{X-\mu }{\sigma }\sim N(0,1).$$

• Moment generating function: Let $X\sim N(\mu ,{\sigma }^2)$. For $t\in \mathbb{R}$, $$\text{E}[\exp (tX)]=\exp (\mu t+t^2{\sigma }^2/2).$$

• If $X_1\sim N({\mu }_1,{\sigma }_1^2)$ and $X_2\sim N({\mu }_2,{\sigma }_2^2)$ are independent, $$c_1{X}_1+c_2{X}_2\sim N(c_1{\mu }_1+c_2{\mu }_2,c_1^2{\sigma }_1^2+c_2^2{\sigma }_2^2).$$

  Question: If $X_1\sim N({\mu }_1,{\sigma }_1^2)$ and $X_2\sim N({\mu }_2,{\sigma }_2^2)$, does $X_1+X_2$ follow a normal distribution?

Review: normal distribution III

Review: normal distribution example

For $X\sim N(5,5^2)$, what is $\text{p}(X\le 6)$?

We have $$\text{p}(X\le 6)=\text{p}(\frac{X-5}{5}\le \frac{6-5}{5})=\mathrm{\Phi }(0.2)=1-\mathrm{\Phi }(-0.2)=0.57926.$$

Lognormal distribution

Lognormal distribution: definition

Let $Z$ be a standard normal random variable, i.e., $Z\sim N(0,1)$. Let $\mu \in \mathbb{R}$ and $\sigma >0$. A lognormal random variable can be written as $$X=\exp (\mu +\sigma Z).$$

• We write $X\sim LN(\mu ,{\sigma }^2)$.
• Note that $\mu$ and ${\sigma }^2$ here are not the mean and variance of $X$.
• The logarithm of $X$ is normally distributed (with mean $\mu$ and variance ${\sigma }^2$), and hence the name.
• Equivalently, we can write $$X=\exp (Y)$$ where $Y\sim N(\mu ,{\sigma }^2)$.

Lognormal distribution: density

Proof:

We write $X=\exp (Y)$ where $Y\sim N(\mu ,{\sigma }^2)$. Then for $x>0$, $$\begin{array}{rl}\text{p}(X\le x)& =\text{p}(Y\le \log x)\\ & ={\int }_{-\mathrm{\infty }}^{\log x}\frac{1}{\sigma \sqrt{2\pi }}\exp (-\frac{(y-\mu {)}^2}{2{\sigma }^2})dy\\ & ={\int }_0^x\frac{1}{u\sigma \sqrt{2\pi }}\exp (-\frac{(\log u-\mu {)}^2}{2{\sigma }^2})du.\end{array}$$ Take derivative on both sides to finish.

Lognormal distribution: moments

Proof:

Here, we note that $\log X\sim N(\mu ,{\sigma }^2)$ and the moment generating function of normal distribution gives the desired results.

Lognormal distribution: product

Proof:

For this property, we use the fact that sum of independent normal random variables is still a normal random variable, i.e., we can write $$\begin{array}{r}{X}_1{X}_2=\exp (Y_1)\exp (Y_2)=\exp (Y_1+Y_2),\end{array}$$ where $Y_1\sim N({\mu }_1,{\sigma }_1^2)$ and $Y_2\sim N({\mu }_2,{\sigma }_2^2)$ are independent.

As $Y_1+Y_2\sim N({\mu }_1+{\mu }_2,{\sigma }_1^2+{\sigma }_2^2)$, we have the desired property.

Lognormal distribution: reciprocal

Proof:

Write $X=\exp (\mu +\sigma Z)$. We have $X^{-1}=\exp (-\mu -\sigma Z)$.

Since $-\mu -\sigma Z\sim N(-\mu ,{\sigma }^2)$, we have the desired result.

Lognormal models: $S_n$ and $V_n$

Lognormal model

• Due the properties of lognormal distributions, it is convenient to model the accumulation factors (i.e., $1+i_t$) as lognormal distributions.
• We can derive distributions of $S_n$ and $V_n$ using lognormal models, which turn out to be lognormal distributions.
• However, distributions of $A_n$ and $P_n$ are still not available.
• Even if the accumulation factor is not lognormally distributed, we can still use lognormal distributions to approximate the distribution of $S_n$ and $V_n$.

Lognormal model: $S_n$ in a varying rate model

Suppose that $1+i_t\sim LN(\mu ,{\sigma }^2)$ for $t=1,\dots ,n$. Since $S_n=\prod _{t=1}^n(1+i_t)$, we have $$\log S_n=\sum _{t=1}^n\log (1+i_t).$$

As $\log (1+i_t)\sim N(\mu ,{\sigma }^2)$ and $i_1,\dots ,i_n$ are independent, $\sum _{t=1}^n\log (1+i_t)\sim N(n\mu ,n{\sigma }^2)$. Therefore, $$\begin{array}{rl}\text{p}(S_n\le s)& =\text{p}(\frac{\log S_n-n\mu }{\sqrt{n}\sigma }\le \frac{\log s-n\mu }{\sqrt{n}\sigma })\\ & =\mathrm{\Phi }\left(\frac{\log s-n\mu }{\sqrt{n}\sigma }\right).\end{array}$$

Question: What if the rates are not identically distributed?

Lognormal model: $V_n$ in a varying rate model

The present value of 1 due at the end of year $n$, denoted by $V_n$, is $$V_n=\prod _{t=1}^n(1+i_t{)}^{-1}.$$

Suppose that in a varying rate model $1+i\sim LN(\mu ,{\sigma }^2)$ where $i$ is the rate of return for each period of time. Then $(1+i{)}^{-1}\sim LN(-\mu ,{\sigma }^2)$ $V_n\sim LN(-n\mu ,n{\sigma }^2)$ and $$\begin{array}{rl}\text{p}(V_n\le s)& =\text{p}(\frac{\log V_n+n\mu }{\sqrt{n}\sigma }\le \frac{\log s+n\mu }{\sqrt{n}\sigma })\\ & =\mathrm{\Phi }\left(\frac{\log s+n\mu }{\sqrt{n}\sigma }\right).\end{array}$$

Lognormal model: $S_n$ example

In a varying rate model, suppose the iid returns, $i_k$, are such that $1+i_k\sim LN(0.12,{0.04}^2)$. What is the probability that the accumulated value of 10000 at time 5 is greater than 21000?

We have $\log S_5\sim N(5\times 0.012,5\times {0.04}^2)=N(0.6,0.008)$. Therefore, $$\begin{array}{rl}\text{p}(10000S_5\ge 21000)& =\text{p}(S_5\ge 2.1)\\ & =\text{p}(\frac{\log S_5-0.6}{{0.008}^{0.5}}\ge \frac{\log 2.1-0.6}{{0.008}^{0.5}})\\ & =1-\mathrm{\Phi }\left(\frac{\log 2.1-0.6}{{0.008}^{0.5}}\right)\\ & =1-\mathrm{\Phi }(1.5869)=0.056.\end{array}$$

Lognormal model: $V_n$ example

In a varying rate model, suppose the iid returns, $i_k$, are such that $1+i_k\sim LN(0.08,{0.04}^2)$. What is the probability that the present value of a benefit of 100 at time 10 is less than 40?

We have $\log V_{10}\sim N(-10\times 0.08,10\times {0.04}^2)=N(-0.8,{0.1265}^2)$. Therefore, $$\begin{array}{rl}\text{p}(100V_{10}\le 40)& =\text{p}(V_{10}\le 0.4)\\ & =\text{p}(\frac{\log V_{10}-(-0.8)}{0.1265}\le \frac{\log 0.4-(-0.8)}{0.1265})\\ & =\mathrm{\Phi }\left(\frac{\log 0.4-(-0.8)}{0.1265}\right)\\ & =\mathrm{\Phi }(-0.91936)=0.179.\end{array}$$

Lognormal model: approximation I

In a varying model, even if accumulation factors do not follow a lognormal distribution, one can still use lognormal distributions to approximate $S_n$ and $V_n$.

Lognormal model: approximation II

• Since $\log S_n=\sum _{t=1}^n\log (1+i_t)$, the summation of iid random variables, by the central limit theorem, $\log S_n$ has a normal distribution as its limiting distribution, after normalization.
• Therefore, $S_n$ can be approximated by a lognormal distribution.
• To apply the central limit theorem, technically, you need to verify the moment constraints/assumptions (though it is not required if not asked in this subject).

Lognormal model: $S_n$ and $V_n$ in a fixed rate model

Suppose that in a fixed rate model $1+i\sim LN(\mu ,{\sigma }^2)$ where $i$ is the rate of return for each period of time. Since $S_n=(1+i{)}^n$, we have $$\log S_n=n\log (1+i).$$

Then, $\log S_n\sim N(n\mu ,n^2{\sigma }^2)$. Therefore, $$\text{p}(S_n\le s)=\mathrm{\Phi }\left(\frac{\log s-n\mu }{n\sigma }\right).$$ Similarly, as $\log V_n\sim N(-n\mu ,n^2{\sigma }^2)$, $$\text{p}(V_n\le s)=\mathrm{\Phi }\left(\frac{\log s+n\mu }{n\sigma }\right).$$

Lognormal model: Fixed rate model Example I

In a fixed rate model, the annual accumulation factor follows a lognormal distribution with mean $1.05$ and variance $0.007$. The initial investment is 14000. What is the probability that $14000S_4\ge 20000$?

Suppose that the lognormal distribution has parameters $\mu$ and ${\sigma }^2$. Then we have $$\exp (\mu +0.5{\sigma }^2)=1.05,$$ and $$\exp (2\mu +{\sigma }^2)(\exp ({\sigma }^2)-1)=0.007,$$ from which we obtain $\mu =0.04563$ and ${\sigma }^2=0.006329$.

Lognormal model: Fixed rate model Example II

We have $$S_4\sim LN(4\ast 0.04563,4^2\ast 0.006329)=LN(0.1825,0.1013).$$ Then $$\begin{array}{rl}\text{p}(14000S_4\ge 20000)& =\text{p}(S_4\ge 1.429)\\ & =\text{p}(\log S_4\ge 0.3567)\\ & =\text{p}(Z\ge (0.3567-0.1825)/\sqrt{0.1013})\\ & =\text{p}(Z\ge 0.574)\\ & =1-\mathrm{\Phi }(0.574)=0.2922.\end{array}$$

Lognormal model: a comparison

Let the accumulation factor for one period follow $LN(\mu ,{\sigma }^2)$. Calculate the coefficient of variation of $S_n$ in a varying rate model and fixed rate model respectively.

We know that $S_n\sim LN(n\mu ,n{\sigma }^2)$ in the varying rate model and $S_n\sim LN(n\mu ,n^2{\sigma }^2)$ in the fixed rate model. Then, in the varying rate model, $$cv(S_n)=\frac{\sqrt{\text{var}(S_n)}}{\text{E}(S_n)}=\sqrt{\exp (n{\sigma }^2)-1}.$$ In the fixed rate model, $$cv(S_n)=\frac{\sqrt{\text{var}(S_n)}}{\text{E}(S_n)}=\sqrt{\exp (n^2{\sigma }^2)-1}.$$ Hence, $S_n$ is more spread-out in the fixed rate model, which is intuitive (why?).

Question: Why not use variance to compare the spreadness?

Value-at-Risk

With the distribution of $S_n$, more risk metrics of $S_n$ can be calculated. For a random variable $X$, Value-at-Risk (VaR) at level $p\in (0,1)$ is defined as $${\text{VaR}}_p(X)=F^{-1}(p)=inf\{x:\text{p}(X\le x)\ge p\}.$$

• Here, $X$ is regarded as losses.
• In general, $p$ is close to 1 (e.g., $p=0.99$).
• VaR is used as a regulatory risk measure in the realm of bank and insurance.
• VaR: the event that the loss is greater than this level has a probability less than $1-p$ (what is special about VaR?).
• For a continuous random variable $X\sim F$, ${\text{VaR}}_p(X)$ is the inverse of $F$.

Value-at-Risk : example A

Suppose that a loss $X$ has the distribution that $P(X=-7)=0.04$ and $P(X=5.5)=0.96$. What is ${\text{VaR}}_{0.95}(X)$?

Ans: ${\text{VaR}}_{0.95}(X)=5.5$.

Value-at-Risk : example B I

In a varying rate model, the annual accumulation factor follows a lognormal distribution with parameters $\mu =0.075$ and ${\sigma }^2={0.025}^2$. Let the initial investment be 1000 and the the accumulated value at the end of 5 years be $X$. What are ${\text{VaR}}_{0.25}(X)$ and ${\text{VaR}}_{0.75}(X)$?

Since rates are independent and they follow the lognormal distribution, $S_5\sim LN(5\ast \mu ,5\ast {\sigma }^2)=LN(5\ast 0.075,5\ast {0.025}^2)$. We have $$\begin{array}{rl}\text{p}(S_5\le s/1000)& =\text{p}(\frac{\log (S_5)-5\ast 0.075}{\sqrt{5}\ast 0.025}\le \frac{\log (s/1000)-5\ast 0.075}{\sqrt{5}\ast 0.025})\\ & =\mathrm{\Phi }\left(\frac{\log (s/1000)-5\ast 0.075}{\sqrt{5}\ast 0.025}\right).\end{array}$$

Value-at-Risk : example B II

By letting the probability function equal to $0.25$ and $0.75$, we get $${\text{VaR}}_{0.25}(X)=1000\exp (\sqrt{5}\ast 0.025{\mathrm{\Phi }}^{-1}(0.25)+5\ast 0.075)=1401$$ and $${\text{VaR}}_{0.75}(X)=1000\exp (\sqrt{5}\ast 0.025{\mathrm{\Phi }}^{-1}(0.75)+5\ast 0.075)=1511.$$

Life insurance applications

Life insurance applications

We will apply the stochastic model for rate of returns in pricing life insurance products.

• Life insurance will be studied in greater details in another subject. We will cover some simple examples in this subject.
• The pricing of life insurance products usually depends on whether a person survives after or dies before a specified time.

For simplicity, the following assumptions are imposed.

• The rate of returns are iid (we use $i$ as a generic copy of $i_1,\dots ,i_n$).
• The event of death/survival is independent of the rates of return.

Traditional life insurance products

• Term life insurance: a lump sum is paid on the death of the insured if the death occurs before the end of a specified period.
• Whole life insurance: a lump sum is paid on the death of the insured whenever it occurs.
• Term life annuity: a series of payments is made to the insured up to a specified period if he/she is alive.
• Whole life annuity: a series of payments is made to the insured as long as he/she is alive.

Notation: I

• $(x)$: a life aged $x$
• $K_x$: the remaining integer number of years that $(x)$ lives. So $K_x$ is a discrete random variable taking non-negative integer amounts.
• $q_x$: the probability that the insured $(x)$ dies within one year (dies in the period $(x,x+1)$, i.e., $$q_x=\text{p}(K_x=0).$$
• $p_x$: the probability that the insured $(x)$ survives the first year, i.e, $$p_x=\text{p}(K_x\ge 1).$$

Notation: II

• ${}_k{q}_x$: the probability that the insured $(x)$ dies within $k$ year (dies in the period $(x,x+k)$, i.e., $${}_k{q}_x=\text{p}(K_x<k).$$
• ${}_k{p}_x$: the probability that the insured $(x)$ survives the first $k$ year, i.e, $${}_k{p}_x=\text{p}(K_x\ge k).$$
• ${}_{k|}{q}_x$: the probability that the insured $(x)$ dies after $k$ year and before $k+1$ years (dies in the period $[x+k,x+k+1)$, i.e., $${}_{k|}{q}_x=P(K_x=k)={}_k{p}_x-{}_{k+1}{p}_x.$$

Notation: III

Some important properties:

• ${}_k{p}_x=\prod _{t=1}^k{p}_{x+t-1}$. However, ${}_k{q}_x\ne \prod _{t=1}^k{q}_{x+t-1}$.
• ${}_{t+u}{p}_x={{}_t{p}_x}_u{p}_{x+t}$.
• ${}_k{p}_x+{}_k{q}_x=1$.
• ${}_{k|}{q}_x=P(K_x=k)={}_k{p}_x{q}_{x+k}={}_k{p}_x-{}_{k+1}{p}_x.$

Example A

Suppose that $q_{30}=0.001332$, $q_{31}=0.001409$, and $q_{32}=0.001508$. What is the probability that $(30)$ dies at the age 32?

We have $$\begin{array}{rl}\text{p}(K_{30}=2)& ={}_{2|}{q}_{30}\\ & ={}_2{p}_{30}{q}_{32}\\ & =p_{30}{p}_{31}{q}_{32}\\ & =(1-q_{30})(1-q_{31})q_{32}=0.001504.\end{array}$$

Example B

Let ${}_k{q}_0=1-(1-k/120{)}^{1/6}$ for $k=0,1,\dots ,120$. Find the probability that

• $(0)$ survives beyond 30,
• $(30)$ dies before 50.

We have $${}_{30}{p}_0=1-{}_{30}{q}_0=(1-30/120{)}^{1/6}=0.9532,$$ and $${}_{20}{q}_{30}=1{-}_{20}{p}_{30}=1-\frac{{}_{50}{p}_0}{{}_{30}{p}_0}=1-(7/9{)}^{1/6}=0.0410$$

Life insurance I

• Consider a whole life insurance which pays $1$ at the year end of the death of $(x)$ (if $K_x=k$, the benefit is paid at $k+1$). The present value of this benefit is denoted as ${\tilde{A}}_x$.
• Given that $K_x=k$, by independence, the expectation of ${\tilde{A}}_x$ is

$$\begin{array}{rl}\text{E}[{\tilde{A}}_x|K_x=k]& =\text{E}[\prod _{t=1}^{k+1}(1+i_t{)}^{-1}|K_x=k]\\ & =\text{E}[\prod _{t=1}^{k+1}(1+i_t{)}^{-1}]\\ & =\prod _{t=1}^{k+1}\text{E}[(1+i_t{)}^{-1}]\\ & =\text{E}{[(1+i{)}^{-1}]}^{k+1}.\end{array}$$

Life insurance II

Then we have $$\begin{array}{rl}\text{E}[{\tilde{A}}_x]& =\sum _{k=0}^{\mathrm{\infty }}\text{E}[{\tilde{A}}_x|K_x=k]\text{p}(K_x=k)\\ & =\sum _{k=0}^{\mathrm{\infty }}\text{E}[{\tilde{A}}_x|K_x=k]{}_{k|}{q}_x\\ & =\sum _{k=0}^{\mathrm{\infty }}\text{E}{[(1+i{)}^{-1}]}^{k+1}{}_{k|}{q}_x.\end{array}$$

Life annuity I

• Consider unit payments at the beginning of every year given that $(x)$ is alive (the benefit will be paid out at $k$ if $K_x\ge k$). The present value of this benefit is denoted as ${\tilde{a}}_x$.
• Given that $K_x=k$, by independence, the expectation of ${\tilde{a}}_x$ is $$\begin{array}{rl}\text{E}[{\tilde{a}}_x|K_x=k]& =\text{E}[1+\sum _{t=1}^k\prod _{m=1}^t(1+i_m{)}^{-1}|K_x=k]\\ & =\text{E}[1+\sum _{t=1}^k\prod _{m=1}^t(1+i_m{)}^{-1}]\\ & =1+\sum _{t=1}^k\prod _{m=1}^t\text{E}[(1+i_m{)}^{-1}]\\ & =\sum _{t=0}^k\text{E}{[(1+i{)}^{-1}]}^t=\frac{1-\text{E}{[(1+i{)}^{-1}]}^{k+1}}{1-\text{E}[(1+i{)}^{-1}]}.\end{array}$$

Life annuity II

Then we have $$\begin{array}{rl}\text{E}[{\tilde{a}}_x]& =\sum _{k=0}^{\mathrm{\infty }}\text{E}[{\tilde{a}}_x|K_x=k]\text{p}(K_x=k)\\ & =\sum _{k=0}^{\mathrm{\infty }}\text{E}[{\tilde{a}}_x|K_x=k]{}_{k|}{q}_x\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{1-\text{E}{[(1+i{)}^{-1}]}^{k+1}}{1-\text{E}[(1+i{)}^{-1}]}{}_{k|}{q}_x\\ & =\frac{1-\text{E}[{\tilde{A}}_x]}{1-\text{E}[(1+i{)}^{-1}]}.\end{array}$$

Life annuity in arrears

• If we wish to calculate the present value of annuities in arrears (units are paid at the end of each period), we can follow a similar procedure.
• However, we notice that the only difference between annuity in advance and annuity in arrears is that the former one pays 1 at the inception. The other payments are the same as long as the insured lives.
• Thus, the difference between the two annuities is simply that the annuity in advance is 1 greater.

Premium

• Policyholders need to pay premiums in exchange for the benefits provided by the insurance products.
• Premiums can either by paid in a lump sum or in installments (like annuity).
• A premium rate, denoted by $P$, is the a fixed amount paid each time.
• A fair premium rate is determined by the equivalence principle: $$\text{EPV of premiums=EPV of benefits.}$$

Here, EPV stands for expected present value.

Premium : life insurance

Suppose $(x)$ needs to buy a life insurance with benefit $S$.

• If the premium is paid in a lump sum at the start of the contract, then the fair premium rate is simply $$P=S\text{E}[{\tilde{A}}_x].$$
• If the premiums are paid at the start of each period given that $(x)$ is alive, then the fair premium rate is determined by $$P\text{E}[{\tilde{a}}_x]=S\text{E}[{\tilde{A}}_x].$$

Premium : example I

Suppose $(40)$ has the following probabilities: $$q_{40}=0.7,q_{41}=0.8,q_{42}=1.$$ The person purchases a life insurance with a benefit at the end of the year of death of 200000. Interest rates are iid with $i_k\sim U(0.05,0.07)$. What is the fair premium rate paid at the inception of each year?

Premium : example II

We first compute the mean of the discounting factor $$\begin{array}{rl}\text{E}[(1+i_k{)}^{-1}]& ={\int }_{0.05}^{0.07}(1+x{)}^{-1}\frac{1}{0.07-0.05}dx\\ & =50(\log 1.07-\log 1.05)=0.94342.\end{array}$$ Since the fair premium is determined by $$P\text{E}[{\tilde{a}}_{40}]=200000\text{E}[{\tilde{A}}_{40}],$$ We need to find $\text{E}[{\tilde{A}}_{40}]$ and $\text{E}[{\tilde{a}}_{40}]$.

Premium : example III

We have $$\begin{array}{rl}\text{E}[{\tilde{A}}_{40}]& =\sum _{k=0}^{\mathrm{\infty }}\text{E}{[(1+i{)}^{-1}]}^{k+1}{}_{k|}{q}_{40}\\ & =0.94342q_{40}+{0.94342}^2{p}_{40}{q}_{41}+{0.94342}^3{p}_{40}{p}_{41}{q}_{42}\\ & =0.66040+0.21361+0.05038=0.92439,\end{array}$$ and that $$\text{E}[{\tilde{a}}_{40}]=\frac{1-\text{E}[{\tilde{A}}_{40}]}{1-\text{E}[(1+i{)}^{-1}]}=1.33643.$$ Therefore, $P=138337.$