Notation Used Throughout

Symbol	Meaning
\(l_x\)	Number of lives age \(x\) per starting cohort (radix usually 10,000 or 100,000).
\(q_x\)	\((l_x - l_{x+1})/l_x\); probability of dying in year of age \(x\).
\(p_x\)	\(1 - q_x\); probability of surviving from \(x\) to \(x+1\).
\(i, v, \delta\)	Effective annual rate, discount factor \(v = 1/(1+i)\), force of interest \(\delta = \ln(1+i)\).

List of Practical Experiments

1. Fitting Gompertz's law of mortality.
2. Construction of a life table from \(q_x\).
3. Computation of \(A_x\) by recursion.
4. Computation of \(\bar A_x\) via the UDD factor.
5. Computation of \(\ddot a_x\).
6. Net annual premium.
7. Reserve \({}_tV_x\) computation.
8. Thiele's recursion for reserves.
9. Joint-life and last-survivor APVs.
10. Gross-premium reserve with expenses.

Experiment 1 — Fitting Gompertz's Law

1. Problem

The force of mortality is \(\mu_{40} = 0.003,\; \mu_{50} = 0.008,\; \mu_{60} = 0.020,\; \mu_{70} = 0.052\). Fit Gompertz's law \(\mu_x = Bc^{x}\).

2. Aim

To estimate the Gompertz parameters \(B\) and \(c\) by linearising \(\ln\mu_x\) against \(x\).

3. Formula

Applying it:

1. Take logs: \(\ln\mu_x = \ln B + x\ln c\) — a straight line in \(x\).
2. The slope is \(\ln c\) (estimate from the average change per year); the intercept is \(\ln B\).
3. Recover \(c = e^{\ln c}\) and \(B = e^{\ln B}\).

4. Calculation

Blank working table:

\(x\)	\(\mu_x\)	\(\ln\mu_x\)
40	0.003
50	0.008
60	0.020
70	0.052

\(x\)	\(\mu_x\)	\(\ln\mu_x\)
40	0.003	−5.809
50	0.008	−4.828
60	0.020	−3.912
70	0.052	−2.957

The increments are ≈ +0.95 per 10-year step, so \(\ln c \approx 0.095\) and \(c = e^{0.095} = 1.10\). Then \(\ln B = -5.809 - 40(0.095) = -9.609\), so \(B = e^{-9.609} = 0.0000671\).

5. Result

The fitted law is \(\mu_x \approx 0.0000671\times 1.10^{x}\) (a least-squares slope gives the same \(c \approx 1.099\)).

Experiment 2 — Construction of a Life Table from \(q_x\)

1. Problem

Given \(l_{30} = 9700\) and \(q_{30} = 0.0020,\; q_{31} = 0.0022,\; q_{32} = 0.0024\), construct the \(l_x\) and \(d_x\) columns to age 33.

2. Aim

To build the survivorship (\(l_x\)) and death (\(d_x\)) columns of a life table recursively.

3. Formula

Applying it:

1. Apply \(l_{x+1} = l_x(1 - q_x)\) recursively from the radix.
2. Compute \(d_x = l_x - l_{x+1}\).

4. Calculation

Blank working table:

\(x\)	\(q_x\)	\(l_x\)	\(d_x\)
30	0.0020	9700
31	0.0022
32	0.0024
33	—		—

\(x\)	\(q_x\)	\(l_x\)	\(d_x\)
30	0.0020	9700.0	19.4
31	0.0022	9680.6	21.3
32	0.0024	9659.3	23.2
33	—	9636.1	—

E.g. \(l_{31} = 9700\times0.998 = 9680.6\); \(d_{30} = 9700 - 9680.6 = 19.4\).

5. Result

The life table for ages 30–33 is \(l_x = 9700.0,\; 9680.6,\; 9659.3,\; 9636.1\) with deaths \(d_x = 19.4,\; 21.3,\; 23.2\).

Experiment 3 — Computation of \(A_x\) by Recursion

1. Problem

With limiting age \(\omega = 100\), \(i = 0.05\) (so \(v = 0.9524\)) and \(q_{97} = 0.40,\; q_{98} = 0.50,\; q_{99} = 0.60\), compute \(A_{97}, A_{98}, A_{99}\) by recursion from \(A_{100} = v\).

2. Aim

To compute the whole-life assurance \(A_x\) by backward recursion.

3. Formula

Applying it:

1. Start at the limiting age with \(A_\omega = v\) (certain death within the year).
2. Recurse downward with \(A_x = v\,q_x + v\,p_x\,A_{x+1}\).

4. Calculation

Blank working table:

\(x\)	\(q_x\)	\(A_x\)
100	1.00	0.9524
99	0.60
98	0.50
97	0.40

\(x\)	\(q_x\)	\(A_x\)
100	1.00	0.9524
99	0.60	0.9343
98	0.50	0.9212
97	0.40	0.9075

\(A_{99} = 0.9524(0.60) + 0.9524(0.40)(0.9524) = 0.5714 + 0.3629 = 0.9343\), and similarly for \(A_{98}, A_{97}\).

5. Result

\(A_{97} = 0.9075,\; A_{98} = 0.9212,\; A_{99} = 0.9343\) — the assurance value rises toward \(v\) as age approaches the limiting age.

Experiment 4 — \(\bar A_x\) via the UDD Factor

1. Problem

Given \(A_{55} = 0.2750\) and \(i = 0.06\), compute the continuous assurance \(\bar A_{55}\) under the uniform-distribution-of-deaths (UDD) assumption.

2. Aim

To convert a discrete assurance to its continuous counterpart using \(\bar A_x = (i/\delta)A_x\).

3. Formula

Applying it:

1. Compute \(\delta = \ln(1 + i)\).
2. Multiply \(A_x\) by \(i/\delta\).

4. Calculation

Blank working table:

Quantity	Value
\(\delta = \ln 1.06\)
\(i/\delta\)
\(\bar A_{55}\)

Quantity	Value
\(\delta = \ln 1.06\)	0.05827
\(i/\delta = 0.06/0.05827\)	1.02971
\(\bar A_{55} = 1.02971\times0.2750\)	0.28317

5. Result

\(\bar A_{55} = 0.28317\) — slightly above the discrete \(A_{55} = 0.2750\), since continuous payment discounts benefits for a shorter time on average.

Experiment 5 — Computation of \(\ddot a_x\)

1. Problem

Given \(A_{55} = 0.2750\) and \(i = 0.06\), compute the whole-life annuity-due \(\ddot a_{55}\) (and \(a_{55}\), \(\bar a_{55}\)).

2. Aim

To obtain the life annuity from the assurance using \(\ddot a_x = (1 - A_x)/d\).

3. Formula

Applying it:

1. Compute \(d = i/(1+i)\).
2. Apply \(\ddot a_x = (1 - A_x)/d\); then \(a_x = \ddot a_x - 1\) and \(\bar a_x \approx \ddot a_x - 0.5\).

4. Calculation

Blank working table:

Quantity	Value
\(d = 0.06/1.06\)
\(\ddot a_{55}\)
\(a_{55},\; \bar a_{55}\)

Quantity	Value
\(d = 0.06/1.06\)	0.05660
\(\ddot a_{55} = (1 - 0.2750)/0.05660\)	12.81
\(a_{55} = 12.81 - 1\)	11.81
\(\bar a_{55} \approx 12.81 - 0.5\)	12.31

5. Result

\(\ddot a_{55} = 12.81\) (annuity-due), \(a_{55} = 11.81\) (immediate), \(\bar a_{55} \approx 12.31\) (continuous).

Experiment 6 — Net Annual Premium

1. Problem

Compute the net annual premium for a whole-life policy on (55) with sum assured ₹10,00,000, using \(A_{55} = 0.2750\) and \(\ddot a_{55} = 12.81\) (from Experiment 5).

2. Aim

To compute the level net annual premium \(P_x = A_x/\ddot a_x\) and the money premium.

3. Formula

Applying it:

1. Compute \(P_{55} = A_{55}/\ddot a_{55}\) (premium per ₹1 of sum assured).
2. Multiply by the sum assured.

4. Calculation

Blank working table:

Quantity	Value
\(P_{55}\) (per ₹1)
Annual premium (₹)

Quantity	Value
\(P_{55} = 0.2750/12.81\)	0.02147
Premium \(= 10{,}00{,}000\times0.02147\)	₹21,470

5. Result

The net annual premium is \(P_{55} = 0.02147\) per ₹1, i.e. ₹21,470 for a ₹10 lakh whole-life policy.

Experiment 7 — Reserve \({}_tV_x\) Computation

1. Problem

For the whole-life policy on (55) with \(P_{55} = 0.02147\), compute the duration-5 net premium reserve using \(A_{60} = 0.3300,\; \ddot a_{60} = 11.83\).

2. Aim

To compute the prospective net premium reserve \({}_tV_x = A_{x+t} - P_x\,\ddot a_{x+t}\).

3. Formula

Applying it:

1. Take \(A_{x+t}\) and \(\ddot a_{x+t}\) at the valuation age.
2. Apply the prospective reserve formula; scale by the sum assured.

4. Calculation

Blank working table:

Quantity	Value
\(P_{55}\,\ddot a_{60}\)
\({}_5V_{55}\) (per ₹1)
Reserve for ₹10 lakh

Quantity	Value
\(P_{55}\,\ddot a_{60} = 0.02147\times11.83\)	0.2540
\({}_5V_{55} = 0.3300 - 0.2540\)	0.0760
Reserve for ₹10 lakh	₹76,000

5. Result

The duration-5 reserve is \({}_5V_{55} = 0.0760\) per ₹1, i.e. ₹76,000 for a ₹10 lakh policy.

Experiment 8 — Thiele's Recursion (Reserve Build-up)

1. Problem

For the whole-life policy on (55) with \(P_{55} = 0.02147,\; i = 0.06,\; q_{55} = 0.008,\; b = 1,\; {}_0V = 0\), compute the year-1 reserve \({}_1V\) by Thiele's recursion.

2. Aim

To verify the reserve build-up recursively via Thiele's equation.

3. Formula

Applying it:

1. Accumulate the opening reserve plus premium at interest.
2. Subtract the expected death cost, then divide by the survival probability.

4. Calculation

Blank working table:

Quantity	Value
\(({}_0V + P)(1+i)\)
\(q_{55}\,b\)
\({}_1V\)

Quantity	Value
\((0 + 0.02147)(1.06)\)	0.02276
\(q_{55}\,b = 0.008\times1\)	0.008
\({}_1V = (0.02276 - 0.008)/0.992\)	0.01488

5. Result

\({}_1V = 0.01488\). Iterating the recursion year by year reproduces the entire reserve schedule.

Experiment 9 — Joint-Life and Last-Survivor APVs

1. Problem

For a couple aged (65, 60) with independent lives, \(A_{65} = 0.40,\; A_{60} = 0.28,\; A_{65:60} = 0.46\), \(\ddot a_{65:60} = 10.9,\; \ddot a_{60} = 14.8\). Find the last-survivor assurance and the reversionary annuity to the wife.

2. Aim

To compute last-survivor and reversionary values from single-life and joint-life quantities.

3. Formula

Applying it:

1. Last-survivor assurance: \(A_{\overline{65:60}} = A_{65} + A_{60} - A_{65:60}\).
2. Reversionary annuity (wife after husband's death): \(\ddot a_{65|60} = \ddot a_{60} - \ddot a_{65:60}\).

4. Calculation

Blank working table:

Quantity	Value
Last-survivor \(A_{\overline{65:60}}\)
Reversionary annuity \(\ddot a_{65|60}\)

Quantity	Value
\(A_{\overline{65:60}} = 0.40 + 0.28 - 0.46\)	0.22
\(\ddot a_{65|60} = 14.8 - 10.9\)	3.9

5. Result

The last-survivor assurance is \(A_{\overline{65:60}} = 0.22\); the reversionary annuity to the wife is \(\ddot a_{65|60} = 3.9\).

Experiment 10 — Gross-Premium Reserve with Expenses

1. Problem

A 20-year endowment on (40), sum assured 1, has \(A_{40:\overline{20}|} = 0.336,\; \ddot a_{40:\overline{20}|} = 11.45\) and net premium \(P = 0.02934\). Expenses: initial \(E_0 = 0.04\), renewal \(E_m = 0.005\)/year, premium-related \(e_p = 5\%\) of each premium. Find the gross premium \(G\).

2. Aim

To compute the gross (office) premium by the equivalence principle, allowing for expenses.

3. Formula

Applying it:

1. Equate the EPV of gross premiums (net of the 5 % premium expense) to the EPV of benefits plus expenses.
2. Solve \(G(1 - e_p)\,\ddot a = A + E_0 + E_m\,\ddot a\) for \(G\).

4. Calculation

Blank working table:

Quantity	Value
LHS coefficient \(0.95\times11.45\)
RHS \(0.336 + 0.04 + 0.005\times11.45\)
Gross premium \(G\)

Quantity	Value
LHS coefficient \(0.95\times11.45\)	10.8775
RHS \(0.336 + 0.04 + 0.05725\)	0.43325
\(G = 0.43325/10.8775\)	0.03983

5. Result

The gross premium is \(G = 0.03983\) — about 36 % above the net premium (0.02934), the loading covering the 4 % initial expense, 0.5 %/year renewal expense and 5 %-of-premium commission.

Lab Record Format (to be followed for every experiment)

1. 1. Problem — the data/rates given and what is to be computed.
2. 2. Aim — the actuarial quantity the experiment produces.
3. 3. Formula — the formula or recursion, then the numbered steps that apply it.
4. 4. Calculation — the filled table with the arithmetic worked through.
5. 5. Result — the final value with its interpretation.