Hence, premium charged for the year = $89841.34 × 1.25 ÷ 1.04 = $107982.38
Calculation of Capital
The amount of capital necessary is determined such that there is 80% chance that the company will be able to meet all claims at the end of the first year. The aggregate claim data is firstly rounded off to the nearest 1000 and then the frequency of each group was tallied. The distribution of the data is shown in the following graph.
Graph 1 Frequency vs Amount Claimed
The 80th percentile is calculated to give the minimum amount needed to meet 80% of all claims. This amount is the total of premium plus the capital and is equal to $120,643.65.
Thus, the capital required to be contributed by the shareholders at initial startup of the product = 120 643.65 – 107982.38 = $12661.27
Investment Policy
- Asset Model
In the beginning of each year, a portion of the asset is invested into the stock market. Stochastic volatility is utilised to model the random fluctuating movements of the stock prices, providing forecasts for the stock return. In particular the stock price is modelled by the following mathematical form for the (log) stock price P(t):
ln P(t+dt) = ln P(t) + μdt + V(t)Z(t)√dt
In this model, Z(t) are independent normal variables with μ = 0 and σ = 1, and also V(t) are independent random variables whose value is one of 0.2, 0.3 or 0.4 with equal distribution.
The share prices of AMP Limited (AMP) reflect the movement in the stock market, which is relevant in determining the parameter of μ. By comparing the daily real-life data with the expected log-return (ln P(t + dt) – ln P(t)), μ is calculated such that the sum of the squared difference is minimized. The value of μ obtained is -0.06063.
The negative magnitude for μ is explained by the decrease in AMP stock prices. This trend can be associated with the crash in the stock market due to the global financial crisis that has inflicted a significant damage in the stock market.
- Investment Strategy
With the company deciding to use a “constant mix” strategy, the firm has the opportunity to invest a portion of their asset into either stocks or zero coupon bonds. By conducting several experiments of possible asset distribution, the expected utilities are compared to select the most strategic combination.
The expected utility function used in the calculation is
v(w) = -e-0.005w
From that, the optimal investment strategy for all years is to invest 25% into the stock market and 75% into zero coupon bonds.
Profit and Loss Analysis
By using the optimal investment strategy and simulating 7000 possible outcomes, the probability distribution of the profit or loss can be acquired. Negative profits are converted to zero as the company declares bankruptcy. All other positive profits are rounded off into groups of 10000s before the frequencies are recorded and graphed.
The probability distribution can be represented by the following graph.
Graph 2: Profit and Loss Distribution
Statistical information on the profit or loss distribution
Expected Value of Profit or Loss = $72,448.03
Variance of Profit or Loss = 6,319,587,035
Probability of Loss = 26.69%
Expected Shortfall = $25,369.57
Conclusion
In conclusion, by allowing 25% loading on the final premium that should be charged at the beginning of every year is $107982.38. In the initial start up of the insurance policy, the capital contributed by shareholders is $12661.27. The asset at the start of every year will be invested in the share market and zero coupon bonds at a proportion of 25% to 75% respectively until the end of three years.
Recommendations
The recommendation for the company is to allocate a maximum of 25% of the asset in the stock market or implement a new investment policy in order to further minimise the expected loss.
Another suggestion for this company is to purchase insurance from another insurance firm (reinsurance) in order to compensate for the loss in the event of insolvency, as there is a 26.69% chance for the company to make loss.
Furthermore, charging a higher premium can also increase the company’s solvency. However this might reduce the product affordability and hence might attract fewer customers. One strategy that can be adopted is to vary the level of premium based on individual risk for each policyholder. Such factors that could include are their age of property, suburban address and many more. These risks can be evaluated from data obtained through surveys or existing past data.
Additionally, this company can also compare the share prices with other insurance firms like Suncorp (SUN) or QBE Insurance (QBE), this provides a better measure of the stock market environment, hence produce more reliable projections.
Appendix
① Calculation of μ and σ
In order to find μ and σ we use the fact
E[X] = eμ + ½σ^2 = 20000
ln(20000) = μ + ½σ^2 …………………………………….❶
μ = ln(20000) - ½σ^2 ……………………………………..❸
and Var[X] = e2μ + σ^2 (eσ^2 – 1) = 20000
ln(20000) = 2μ + σ^2 + ln(eσ^2 – 1) …………………….. × ½
½ ln(20000) = μ + ½σ^2 + ½ ln(eσ^2 – 1) ……………………❷
sub ❶ into ❷
½ ln(20000) = ln(20000) + ½ ln(eσ^2 – 1)
½ ln(20000) = -½ ln(eσ^2 – 1)
ln(20000)-1 = ln(eσ^2 – 1)
20000-1 = eσ^2 – 1
1 + 20000-1 = eσ^2
ln(1 + 20000-1) = σ^2
σ = [ln(1 + 20000-1)] ½
σ = 7.070979426 × 10-3 ……………………………………..❹
sub ❹ into ❸
μ = 9.903462553
② Reasoning for expected aggregate claim
Denote the number of claims by N(t) and the loss distribution for the ith claim by X(i,t) for i = 1,2,3,…,N(t). The number of claims that occurs takes the values
N(t) = n(t).
Moreover the total claim payments (aggregate loss) is denoted by S. The aggregate loss is the sum of all claim amounts that occur during the policy year.
Accordingly S = X(1,t) + X(2,t) + X(3,t) + … + X(N(t))
=
The expected claims cost will be the expected value of the aggregate claims such that
E[S] =
To evaluate this probability, conditional probability is utilised. This means the different number of possible claims that will occur and the probability that these numbers of claims occur, so the expected aggregate loss is
E[S] =
With the losses being independent of the number of claims and each loss X(i,t) has the same probability density ƒx (x) then
= E[X| N(t) = n(t)]
= [xi] ƒx (x) dx
= E[Xi]
= n E[X]
Therefore E[S] = n E[X] Pr[N(t) = n(t)]
= E[n(t)] E[X]
Since the average number of claims are 4.5, thus E[n(t)] = 4.5 and E[X] = 20000
∴E[S] = Expected aggregate claim = 4.5 × 20000 = $90 000
③ Claims
This is how the spreadsheet should be set out like producing 7000 simulations each year:
Table 1: Claim Data
-
The number of claims would come from the Random Number Generator using a Poisson distribution with λ = 4.5.
-
The Random Number comes from the function =RAND()
(See appendix 1 for calculation of μ and σ)
-
The Amount claimed comes from the function = LOGINV(B, μ , σ)
-
The Aggregate claimed comes from A × C = D
-
Rounded Aggregate claimed come from the function =ROUND(D,-3)
④ The Calculation of Capital
To calculate the amount of capital such that there is 80% chance to meet all claims in year 1, each aggregate claim data are rounded off to the nearest 1000s and used as bin data. The frequency of each group counted using excel function
COUNTIF(‘Claim Data’! $E$12:$E$7011,A98)
Table 2: Capital + Premium
The 80th percentile of the data is the total amount of premium plus capital that is needed. This value turns out to be$120 643.65. Hence, the amount of capital needed equals to $120,643.65 minus the premium charged which is $12,661.27.
⑤ Calculation of μ using the least squared technique
The value of μ is determined using the least squares method. This involves taking the squared difference between daily actual log returns with the expected log returns, and minimising it using the solver.
For the actual share price P(t) is selected to be the closing AMP share price at time t
Let X = ln P(t + dt) – ln P(t)
The actual log return was calculated by E7= D7 – D6
Table 3: AMP share price data comparison
By using the stochastic volatility model: lnP(t + dt) – lnP(t) = μ dt + V(t) Z(t) √dt
To make calculations simpler, let X = lnP(t + dt) – lnP(t)
So the expected log return: E[lnP(t + dt) – lnP(t)] = E[μ dt + V(t) Z(t) √dt]
E[X] = E[μ dt] + E[V(t) Z(t) √dt]
since μ and dt are constants
E[X] = μ dt + √dt E[V(t)] E[Z(t)]
Since Z(t) are independent N (0,1) random variables, by using the random number generator with Normal distribution, random Z(t) are produced.
Similarly V(t) is made using the function: =RANDBETWEEN(2,4)/10
The initial guess was chosen to be μ = –0.05. This choice is based on the observation of the downfall period of the share prices.
P(t) (starting from H7) =EXP(LN(H6) + G4*I4 + F6*G6*sqrt(I4))
E[X]: J7 =I7 – I6
Table 4: Least square - mu
By using the Solver Function, set the target cell as the sum of squared difference (K1018) and the changing cell G4, this minimises the μ.
Hence after using the solver the minimised μ is –0.060626045
⑥ Table of Investment Return and Expected Utility
Table 5: Optimal Investment and Utility
The above table is constructed for each year 1, year 2 and year 3 for the purpose of profit or loss and the return on investment calculation.
The first two columns of this table are the same procedure from the Table 1: Claims Table for column A and D.
Stock(1) column gives the return on the α% of asset invested in the stock market. The formula used to achieve this is;
($H$7)*$D$7*EXP(‘Least square – mu’!$G$4 + RANDBETWEEN(2,4)/10)*NORMINV((RAND()),0,1))
The above formula basically computes the following calculation:
ln P(t+dt) = ln P(t) + μdt + V(t)Z(t)
ln P(t+dt) – ln P(t) = μdt + V(t)Z(t)
ln = μdt + V(t)Z(t)
= Return on investment
Similarly, the ZCB(1) column gives the return on investment from zero coupon bond given that the interest rate is 4% p.a. The return of (100-α)% of asset invested on zero coupon bond at the end of year 1 is:
In excel, this is calculated by the formula: (1-$H$7)*$D$7*(1.04)
The sum of corresponding Stock(1) and ZCB(1) minus the random aggregate claim gives the ending asset value at the end of year 1. These values of wealth are then used in the calculation of U(w) = E[v(w)]. It is assumed that the utility function used is the exponential utility. v(w) = -e
The last column gives the value of v(w=x)*P(w=x),with x equally distributed with probability of . Hence, the sum of this column gives U(w) of a specific value of α.
The level of α at which U(w) is highest is the most optimum allocation of investment. From trial and error, it is found that the optimum α stays the same throughout the 3 years period at 25%.
⑦ Profit or Loss Probability Distribution
At the end of the 3rd year, the amount of profit or loss is calculated. Thus, in addition to the table in appendix (6), the table for year 3 has one extra column, which gives the values for expected profit. This column is the amount of asset at the end of year 3 subtracted by the expected loss and the capital.
Table 6: Profit or Loss
If the company makes a negative profit, the asset at the end of the year will be written off to zero. However, positive assets are rounded off to the nearest 10,000s for the purpose of the evaluation of profit or loss distribution.
Excel function for column A(3): if(stock(3)+ZCB(3)-AC>=0, stock(3)+ZCB(3)-AC,0)
These groups are tabulated and tallied before the distribution graph is plotted,
Refer to graph 2.
The variant of the profit or loss distribution can be calculated using the variant formula.
Var[X] = E[(X – E[X])2]
To do this by excel, one extra column is created for the calculation of (X – E[X])2(1/7000) where 1/7000 is the probability for each value of x. In this case x is the value of profits and E[X] is the average of the profits.
The sum of this column will give the variance of the profits, which equals to 6,319,587,035. The formula used is SUM().
See appendix for the calculation of μ and σ.
See appendix for reasoning of the expected aggregate claim
See appendix on the sample excel format
See calculation for Capital
Calculation of μ using the least squared technique
Investment Return and Expected Utility
Details on excel functions, calculation on expected value and variance.
Pg 196 – 197, Michael Sherris, (2009), Principles of Actuarial Science, McGraw Hill