1. Revision 2. Revision pv 3. - note that there are other equivalent formulae! 1 pv A x A 1 x:n A 1

1. Revision 2. Revision 3. pv

1 − pv - note that there are other equivalent formulae! 16.5 4. • Ax • A1 x:n • A_x:n1 • ax • ax:n • _n|ax

5. Kx= int[Tx] - or, as an approximation: Tx≈ Kx+1₂

6. Male: 0.993, 0.937 Female: 0.994, 0.9617 7. Male: 0.007, 0.063 Female: 0.006, 0.0383 8. lx+n− lx+n+m lx or (npx)(mqx+n)

9. Note for this question the interest rates should have been 0%, 5% and 10.25% as noted in the tutorial

A₁₀₀= 1; 0.894; 0.803 (males), variance = 0.00429 A₁₀₀= 1; 0.881; 0.781 (females), variance = 0.005634

Tutorial 2

1. Proof - see next page 2. Proof

3. 0.0109

4. Proof - write down a formula for ex and ex+1, then multiply the second by px and

add px 5. µx= −_l1_xdxdlx ∴ lxµx= −_dxdlx R∞ 0 lx+tµx+tdt = − R∞ 0 dlx+t dx+tdt = −[lx+t]∞0 = −[l∞− lx] = lx 6. Proof 7. 0.8389; 0.0251; 0.9926 8. 0.996939017; 0.9969408

if your first answer was 0.9969353 note the UDD (sqx≈ sqx) only applies for integer

xand s ∈ (0, 1], you first need to derive another approx is needed for non-integer x as we did in the tutorial

9. 0.656338028; 0.678626725 10. lx= k(e_x)Ax

11. $61,391.33 12. $61,391.33 13. £3,737.28

f_x(t) = d dtFx(t) = d dt(P[Tx≤ t]) = lim h→0+ 1 h(P[Tx≤ t + h] − P[Tx≤ t]) = lim h→0+ 1 h(P[T ≤ x + t + h|T > x] − P[T ≤ x + t|T > x]) = lim h→0+ 1 h( P[T ≤ x + t + h] − P[T ≤ x] P[T > x] − [T ≤ x + t] − P[T ≤ x] P[T > x] )) = lim h→0+ 1 h 1 S(x)(P[T ≤ x + t + h] − P[T ≤ x + t]) = lim h→0+ 1 h S(x + t) S(x) ( P[T ≤ x + t + h] − P[T ≤ x + t] S(x + t) ) = Sx(t) lim h→0+ 1 hP[T ≤ x + t + h|T > x + t] = Sx(t)µx+t =tpxµx+t

Tutorial 3

1. 2A_x:n − (Ax:n)2

where2A_x:n denotes Ax:n calculated at i∗= (1 + i)2− 1

2. Proof - do by splitting A1_x:n into Ax - n|Ax. Needs to be done from first principles

-so define all three of these as the expected value of random variables and split the values by Kx

cov[X ,Y ] =2A1_x:n− (Ax)(A1x:n)

Var[J] =2_n|A_x− (_n|Ax)2 where J is the random variable denoting the present value of the deferred assurance as defined in the first half of the question.

Note that it is easier here to calculate Var[J] from first principles but that was not what the question asked...

3. ¨ax:n = ∑n−1_{j=0 j}pxvj Var[ ¨a_min[K x+1,n]] = 1 d2[2Ax:n− (Ax:n)2] ¨ ax:n = ax:n+ 1 − vnnpx Var[ ¨a_min[K x+1,n]] = 1 d2[2Ax:n− (Ax:n)2]

4. Denote the present value of the payments under the deferred annuity due by Y . Y = 0 if Kx≤ n

Y = vna_K

x−n+1 if Kx> n

Therefore, in one term, Y = vna¨_{max[Kx−n+1,0]} ...

Var[vna¨

max[Kx−n+1,0]] = 1

d2[E[(vmax[Kx+1,n])2] − (E[vmax[Kx+1,n])2]

...

Var[vna¨_max[K

x−n+1,0]] = v2n

d2[nqx+npx2Ax+n− (nqx+npxAx+n)2]

For the deferred annuity; it is not possible to simplify much past 1

i2Var[v max[Kx,n]_] 5. R_∞ 0 vttpxdt 1 δ2[ 2_A¯ x− ( ¯Ax)2]

6. 3,4,5 year select periods: l46−l50 l_[41]+2 6 year select: l[41]+5_l −l50 [41]+2 7. Expected values: 4%: £45,640 6%: £32,692 4% select: £45,510 6% select: £32,533

4% payable immediately on death: two approximations possible; one is claims ac-celaration; the other you end up multiplying by i

£33,658 or £33,663

Variances: 4%: 289,290,000 6%: 341,033,136 8. ¨a_x:n−1₂(1 − vnnpx)

Tutorial 4 1. Nx+1−Nx+n+1 Dx , Nx+m+1 Dx , Nx+m+1−Nx+m+n+1 Dx , Nx Dx, Mx−Mx+n Dx , Mx−Mx+n+Dx+n Dx , Mx+m Dx 2. ... see Tut 3! 3. Proofs - except: ¨a(m)x ≈ ¨ax−m−1_2m 4. Proof 5. tVx:n = 1 − ¨ a_x+t:n−t ¨ ax:n

6. TYPO in question: should be x :n in the last subscript

Answer: because the reserve required for an endowment assurance is the reserve required for an endowment, plus the reserve required for an assurance; as an endow-ment assurance is both an endowendow-ment and an assurance.... (if the policyholder dies, we need to have the reserve for an assurance, if he survives, we need to have the reserve for a pure endowment).

1. Note for this question that ’look up in the tables’ means that you should use available values and simplifying formulae (such as ¨a_x= ax+ 1) to arrive at your answer, not

that the values are necessarily published somewhere in your yellow books. Not all variances are given, some more should be solvable.

a_50:15 = 10.75750214 commutation functions

a_50:15 = 10.75711239 from published figs in the tables Variance: 1.74302 5|a50= 12.03046224 5|a50= 12.03032535 Variance: 34.4759 5|a50:15 = 8.482324877 5|a50:15 = 8.482106580 ¨ a₅₀= 17.44419 ¨ a₅₀= 17.444 A1 50:15 = 0.062848947 A1 50:15 = 0.062854454 A_50:15 = 0.567184493 A_50:15 = 0.56719 5|A50= 0.315049648 5|A50= 0.3150549 2. 4%: £45,640, variance 289,290,400 6%: £32,692, variance 341,033,136 select 4%: £45,510, variance 283,539,900 select 6%: £32,533, variance 332,303,911 payable immediately on death: 4%: £46,544 payable immediately on death: 6%: £33,658 3. UDD: 0.988232 CFM: 0.988234994 4. A_71:3 = 0.892063339 (4%) A_71:3 = 0.843984265 (6%) ¨ a_63:20 = 11.7938 (4%) ¨ a_63:20 = 10.3288 (6%)

5. 0.065088624 - note that the question should have said ’over each of the next two years’ to avoid any potential confusion

6. A1_x:n = 0.067637358 Variance: 0.039672262

Tutorial 5

1. Proof

2. 0.000341799

Reserves - see spreadsheet printout 3. 0.101686157

Reserves - see spreadsheet printout 4. Proof

5. (DA)1_x:n = (n + 1)A1_x:n− (IA)1 x:n 6. 1000(13₂₄a¨x+11₂₄ax(_{1+ j}1 )) 7. £2,813.58 8. ( ¯Ia)¯ x= R∞ 0 tvttpxdt (I ¯a)x= ∑∞t=0 t|a¯x

Page 1

prospective retrospective 0.000345763

annuity assurance total annuity assurance total

1 6.15 0.002155 0.000028 1.05 0 0.000028 2 5.39 0.001928 0.000066 2.14 0 0.000066 3 4.58 0.001672 0.000087 3.28 0 0.000087 4 3.75 0.001403 0.000108 4.47 0 0.000108 5 2.87 0.001112 0.000119 5.72 0 0.000119 6 1.96 0.000777 0.000101 7.03 0 0.000101 7 1 0.000417 0.000071 8.39 0 0.000071 8

net premium reserve: (question 3) premium:

prospective retrospective 0.101835550

annuity pure end total annuity pure end total

1 6.15 0.732932 0.106454 1.05 0 0.106454 2 5.39 0.766162 0.217733 2.14 0 0.217733 3 4.58 0.800915 0.334064 3.28 0 0.334064 4 3.75 0.837245 0.455672 4.47 0 0.455672 5 2.87 0.875231 0.582802 5.72 0 0.582802 6 1.96 0.914969 0.715722 7.03 0 0.715722 7 1 0.956521 0.854685 8.39 0 0.854685 8

Tutorial 6

1. tVx:n = v[qx+t+ px+t(t+1Vx:n)] − Px:n

Premiums, retrospective and prospective reserves all need to be calculated on the same basis.

2. Expected Death Strain: £1,380,000 Mortality Profit: -£12,920,000 Revised EDS: £1,378,620

Revised Mortality Profit: -£12,907,080 3. Proof

4. EDS first year: -0.0694349 Mortality profit: -0.0694349 EDS 10 years in: -0.1692801

EDS year beginning age 69: -0.2305948 ADS year beginning age 69: -10.375

5. Full solution to this is available here:

http://www.actuaries.org.uk/__data/assets/pdf_file/0007/141973/FandI_CT5_200809_Report.pdf