List of Practical Experiments (Official Syllabus)

1. Risk computation for different utility models.
2. Discrete and continuous risk calculations.
3. Calculation of aggregate claims for collective risks.
4. Calculation of aggregate claim for individual risks.
5. Computing ruin probabilities and aggregate losses.
6. Annuity and present value of contract.
7. Computing premium for different insurance schemes.
8. Practical based on life models and tables.

Experiment 1 — Risk Computation for Different Utility Models

1. Problem

An individual with wealth ₹2,00,000 faces a loss of ₹50,000 with probability 0.2 (else 0). Find the maximum premium \(G^*\) the individual would pay for full insurance under linear, logarithmic and exponential utility.

2. Aim

To compute the maximum acceptable premium under three utility functions and compare the implied risk aversion.

3. Formula

Applying it:

1. The maximum premium solves \(u(w - G^*) = E[u(w - X)]\), where \(X\) is the loss.
2. Linear \(u(w) = w\): \(G^* = E(X)\) (risk-neutral).
3. Log \(u(w) = \ln w\): \(w - G^* = \exp\!\big(E[\ln(w - X)]\big)\).
4. Exponential \(u(w) = -e^{-aw}\): solve \(-e^{-a(w-G^*)} = E[-e^{-a(w-X)}]\) for \(G^*\).

4. Calculation

Blank working table (\(a = 0.00002\)):

Utility	Max premium \(G^*\)	Risk premium
Linear (risk-neutral)
Logarithmic
Exponential

Linear: \(G^* = E(X) = 0.2\times 50000 = \text{₹}10{,}000\).

Logarithmic: \(E[\ln(w-X)] = 0.8\ln(200000) + 0.2\ln(150000) = 9.7649 + 2.3837 = 12.1485\); \(w - G^* = e^{12.1485} = 1{,}88{,}817\), so \(G^* = \text{₹}11{,}183\).

Exponential: RHS \(= -0.8e^{-4} - 0.2e^{-3} = -0.02461\), so \(0.00002(w - G^*) = -\ln(0.02461) = 3.7046\), giving \(w - G^* = 1{,}85{,}230\) and \(G^* = \text{₹}14{,}770\).

Utility	Max premium \(G^*\)	Risk premium
Linear (risk-neutral)	₹10,000	₹0
Logarithmic	₹11,183	₹1,183
Exponential	₹14,770	₹4,770

5. Result

The maximum premium rises with risk aversion: ₹10,000 (linear, pure premium) < ₹11,183 (log, mild aversion) < ₹14,770 (exponential, strong aversion).

Experiment 2 — Discrete and Continuous Risk Calculations

1. Problem

(a) Annual claim count \(N \sim\) Poisson(5) with each claim a fixed ₹10,000 — find \(E(S)\) and \(\text{Var}(S)\). (b) Claim size \(X \sim\) Exponential with mean ₹20,000 over 10 independent policies — find \(E(S)\), \(\text{Var}(S)\) and \(P(S > \text{₹}2{,}50{,}000)\).

2. Aim

To compute the mean, variance and a tail probability of aggregate claims in a discrete and a continuous setting.

3. Formula

Applying it:

1. Discrete: \(S = 10000\,N\), so \(E(S) = 10000\,E(N)\), \(\text{Var}(S) = 10000^2\,\text{Var}(N)\).
2. Continuous: sum of 10 iid Exp(mean 20000) is Gamma(10); \(E(S) = 10\mu\), \(\text{Var}(S) = 10\mu^2\).
3. Tail probability by the CLT: \(P(S > s) \approx P\!\big(Z > (s - E(S))/\sigma_S\big)\).

4. Calculation

Blank working table:

Quantity	Discrete	Continuous
\(E(S)\)
\(\text{Var}(S)\)
\(\sigma_S\)
Tail probability	—

Quantity	Discrete	Continuous
\(E(S)\)	₹50,000	₹2,00,000
\(\text{Var}(S)\)	\(5\times10^8\)	\(4\times10^9\)
\(\sigma_S\)	₹22,361	₹63,246
Tail probability	—	\(P(Z > 0.79) = 0.215\)

5. Result

Discrete: \(E(S) = \text{₹}50{,}000,\; \sigma_S = \text{₹}22{,}361\). Continuous: \(E(S) = \text{₹}2{,}00{,}000,\; \sigma_S = \text{₹}63{,}246\), and \(P(S > \text{₹}2{,}50{,}000) \approx 0.215\).

Experiment 3 — Collective Risk Model (Aggregate Claims)

1. Problem

In the collective risk model \(S = X_1 + \cdots + X_N\) (with \(N\) random and \(X_i\) iid, independent of \(N\)), take \(N \sim\) Poisson(100) and \(X \sim\) Exponential with mean 5000. Find \(E(S),\; \text{Var}(S)\), a 20 %-loaded premium, and the probability the premium is exceeded.

2. Aim

To apply the compound-Poisson formulas for the mean and variance of aggregate claims and evaluate a loaded premium.

3. Formula

Applying it:

1. General: \(E(S) = E(N)E(X)\), \(\text{Var}(S) = E(N)\text{Var}(X) + \text{Var}(N)[E(X)]^2\).
2. Compound Poisson (\(E(N) = \text{Var}(N) = \lambda\)): \(E(S) = \lambda E(X)\), \(\text{Var}(S) = \lambda E(X^2)\).
3. For Exp(mean \(\mu\)): \(E(X) = \mu\), \(E(X^2) = 2\mu^2\).
4. Premium \(P = 1.2\,E(S)\); tail probability by the CLT.

4. Calculation

Blank working table:

Quantity	Value
\(E(S)\)
\(\text{Var}(S)\)
\(\sigma_S\)
Premium \(P = 1.2E(S)\)
\(P(S > P)\)

\(E(X) = 5000,\; E(X^2) = 2\times 5000^2 = 5\times 10^7\).

Quantity	Value
\(E(S) = 100\times 5000\)	₹5,00,000
\(\text{Var}(S) = 100\times 5\times10^7\)	\(5\times10^9\)
\(\sigma_S\)	₹70,711
Premium \(P = 1.2\times 5{,}00{,}000\)	₹6,00,000
\(P(S > P) = P(Z > 1.41)\)	0.079

5. Result

\(E(S) = \text{₹}5{,}00{,}000,\; \sigma_S = \text{₹}70{,}711\). With a 20 % loading the premium is ₹6,00,000, and the probability that claims exceed it is only about 7.9 %.

Experiment 4 — Individual Risk Model (Aggregate Claims)

1. Problem

A portfolio has 200 policies; each pays ₹1,00,000 on a claim with probability 0.01. Find \(E(S)\), \(\text{Var}(S)\) and a 95 % upper premium.

2. Aim

To compute aggregate-claim moments in the individual risk model and set a premium at the 95 % confidence level.

3. Formula

Applying it:

1. Each \(X_i = 100000\,I_i\) with \(I_i \sim\) Bernoulli(0.01): \(E(X_i) = 100000q\), \(\text{Var}(X_i) = 100000^2 q(1-q)\).
2. Sum over 200 independent policies: \(E(S) = 200E(X_i)\), \(\text{Var}(S) = 200\,\text{Var}(X_i)\).
3. 95 % premium \(= E(S) + 1.645\,\sigma_S\).

4. Calculation

Blank working table:

Quantity	Value
\(E(X_i)\) / \(\text{Var}(X_i)\)
\(E(S)\)
\(\text{Var}(S)\) / \(\sigma_S\)
95 % premium

Quantity	Value
\(E(X_i)\) / \(\text{Var}(X_i)\)	1000 / \(9.9\times10^7\)
\(E(S) = 200\times 1000\)	₹2,00,000
\(\text{Var}(S) = 1.98\times10^{10}\) / \(\sigma_S\)	₹1,40,712
95 % premium \(= 200000 + 1.645\times 140712\)	₹4,31,472

5. Result

\(E(S) = \text{₹}2{,}00{,}000,\; \sigma_S = \text{₹}1{,}40{,}712\); a premium of about ₹4,31,472 covers aggregate claims with 95 % confidence.

Experiment 5 — Ruin Probabilities and Aggregate Losses

1. Problem

For a compound-Poisson surplus \(U(t) = u + ct - S(t)\) with \(\lambda = 100\), \(X \sim\) Exp(mean 5000) and loading \(\theta = 0.2\), find the ultimate ruin probability \(\psi(u)\) for \(u = 30{,}000\) and \(u = 1{,}00{,}000\).

2. Aim

To use the closed-form ruin probability for exponential claims and interpret the effect of the initial surplus.

3. Formula

Applying it:

1. Premium rate \(c = (1+\theta)\lambda E(X)\).
2. For exponential claims, \(\psi(u) = \dfrac{1}{1+\theta}\, \exp\!\Big(-\dfrac{\theta u}{(1+\theta)E(X)}\Big)\).
3. Evaluate at each \(u\) and interpret (Lundberg bound \(\psi(u) \le e^{-Ru}\)).

4. Calculation

Blank working table:

Initial surplus \(u\)	\(\psi(u)\)
30,000
1,00,000

\(c = 1.2\times 100\times 5000 = \text{₹}6{,}00{,}000\) per unit time; with \(\theta = 0.2,\; E(X) = 5000\), \(\psi(u) = (1/1.2)\,e^{-u/30000} = 0.833\,e^{-u/30000}\).

Initial surplus \(u\)	\(\psi(u)\)
30,000	\(0.833\,e^{-1} = 0.306\)
1,00,000	\(0.833\,e^{-3.33} = 0.030\)

5. Result

With ₹30,000 initial surplus the long-run ruin probability is about 31 %; raising the surplus to ₹1,00,000 cuts it to about 3 % — larger reserves greatly improve solvency.

Experiment 6 — Annuity and Present Value of Contract

1. Problem

(a) An annuity-certain pays ₹1,00,000 at each year-end for 20 years at \(i = 6\%\). (b) A whole-life annuity-due for \((x=60)\) with \(\ddot a_{60} = 12.5\) at \(i = 5\%\) pays ₹50,000/year. Find the present values, including the monthly and continuous approximations.

2. Aim

To compute present values of an annuity-certain and a life annuity, and the standard monthly and continuous adjustments.

3. Formula

Applying it:

1. Annuity-certain: \(a_{\overline{n}|} = (1 - v^n)/i\) with \(v = 1/(1+i)\).
2. Life annuity: APV \(= \text{(annual payment)}\times \ddot a_x\).
3. Monthly: \(\ddot a_x^{(12)} \approx \ddot a_x - 11/24\); continuous: \(\bar a_x \approx \ddot a_x - 1/2\).

4. Calculation

Blank working table:

Contract	Annuity factor	Present value
Annuity-certain (20 yr, 6 %)
Whole-life annuity-due	12.5
Monthly approximation
Continuous approximation

\(v = 1/1.06 = 0.9434,\; v^{20} = 0.3118\), so \(a_{\overline{20}|} = (1 - 0.3118)/0.06 = 11.4699\).

Contract	Annuity factor	Present value
Annuity-certain (20 yr, 6 %)	11.4699	₹11,47,000
Whole-life annuity-due	12.5	₹6,25,000
Monthly approximation	12.5 − 0.458 = 12.042	₹6,02,100
Continuous approximation	12.5 − 0.5 = 12.0	₹6,00,000

5. Result

The 20-year annuity-certain is worth ₹11,47,000; the whole-life pension of ₹50,000/year is worth ₹6,25,000 (annual), ₹6,02,100 (monthly) or ₹6,00,000 (continuous).

Experiment 7 — Premium for Different Insurance Schemes

1. Problem

From a life table at \(i = 5\%,\; x = 30\): \(A_{30} = 0.105,\; \ddot a_{30} = 18.85,\; A^{1}_{30:\overline{20}|} = 0.022,\; \ddot a_{30:\overline{20}|} = 12.93,\; {}_{20}E_{30} = 0.355,\; \ddot a_{30:\overline{10}|} = 7.87\). For a face value of ₹10,00,000 compute the net annual premium of whole-life, 20-year term, 20-year endowment and 10-year limited-payment whole-life policies.

2. Aim

To compute net single and net annual premiums for the four standard insurance schemes.

3. Formula

Applying it:

1. Net single premium (NSP) \(= \text{face}\times A\).
2. Net annual premium \(= \text{NSP}/\ddot a\) (the relevant annuity).
3. Endowment \(A_{x:\overline{n}|} = A^{1}_{x:\overline{n}|} + {}_{n}E_x\).

4. Calculation

Blank working table:

Scheme	\(A\)	Annuity	Net annual premium
Whole-life	0.105	18.85
20-yr term	0.022	12.93
20-yr endowment		12.93
10-pay whole-life	0.105	7.87

Scheme	\(A\)	Annuity	Net annual premium
Whole-life	0.105	18.85	10,00,000×0.105/18.85 = ₹5,570
20-yr term	0.022	12.93	22,000/12.93 = ₹1,701
20-yr endowment	0.022 + 0.355 = 0.377	12.93	3,77,000/12.93 = ₹29,157
10-pay whole-life	0.105	7.87	1,05,000/7.87 = ₹13,342

5. Result

Whole-life ₹5,570; term ₹1,701 (cheapest — no survival benefit); endowment ₹29,157 (dearest — guaranteed maturity payout); 10-pay whole-life ₹13,342 (about 2.4× ordinary whole-life, since premiums are paid over only 10 years).

Experiment 8 — Life Models & Tables

1. Problem

Using the life-table extract below, compute survival/death probabilities, the force of mortality at age 60, and fit a Gompertz law to \(\mu(50) = 0.005,\; \mu(70) = 0.030\).

Age x	lx	dx	qx	px
30	97500	140	0.00144	0.99856
31	97360	146	0.00150	0.99850
40	95800	240	0.00250	0.99750
50	92300	520	0.00563	0.99437
60	85000	1200	0.01412	0.98588

2. Aim

To read survival probabilities from a life table, approximate the force of mortality, and fit a Gompertz mortality law.

3. Formula

Applying it:

1. Survival: \({}_{n}p_x = l_{x+n}/l_x\); death: \({}_{n}q_x = 1 - l_{x+n}/l_x\).
2. Force of mortality (constant-force within the year): \(\mu_x \approx -\ln p_x\).
3. Gompertz \(\mu(x) = Bc^x\): from two ages, \(c^{x_2-x_1} = \mu(x_2)/\mu(x_1)\), then \(B = \mu(x_1)/c^{x_1}\).

4. Calculation

Blank working table:

Quantity	Value
\({}_{10}p_{50} = l_{60}/l_{50}\)
\({}_{20}q_{30} = 1 - l_{50}/l_{30}\)
\(\mu_{60} \approx -\ln p_{60}\)
Gompertz \(c = (\mu_{70}/\mu_{50})^{1/20}\)
Gompertz \(B = \mu_{50}/c^{50}\)

Quantity	Value
\({}_{10}p_{50} = 85000/92300\)	0.9209
\({}_{20}q_{30} = 1 - 92300/97500\)	0.0533
\(\mu_{60} \approx -\ln(0.98588)\)	0.01422
\(c = (0.030/0.005)^{1/20} = 6^{1/20}\)	1.0937
\(c^{50} = 6^{2.5} = 88.18,\; B = 0.005/88.18\)	\(5.67\times10^{-5}\)

So the fitted law is \(\mu(x) = 5.67\times 10^{-5}\times 1.0937^{x}\). Adding a constant term \(A\) gives the Makeham law \(\mu(x) = A + Bc^x\) when accidental (age-independent) risk matters. For \(\bar A_x\) under Gompertz there is no closed form; evaluate \(\bar A_x = \int_0^\infty v^t\,{}_tp_x\,\mu(x+t)\,dt\) numerically (Simpson's rule, spreadsheet or R).

5. Result

\({}_{10}p_{50} = 0.9209\) and \({}_{20}q_{30} = 0.0533\); the force of mortality at 60 is about 0.0142/year, and the Gompertz fit is \(\mu(x) = 5.67\times 10^{-5}\times 1.0937^{x}\).

Lab Record Format (to be followed for every experiment)

1. 1. Problem — the data/model and what is to be computed.
2. 2. Aim — the quantity (premium, probability, present value) the experiment produces.
3. 3. Formula — the formula, then the numbered steps that apply it.
4. 4. Calculation — the filled table with the arithmetic worked through.
5. 5. Result — the final figure with units and interpretation.