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AS3429 W2018 Ch.11 Lecture notes

Ch. 11 Interest Rate Risk

11.1 Summary 11.2 The Yield Curve

11.3 Valuation of Insurance & Life Annuities

11.4 Diversifiable and Non-diversifiable risk

AS3429 W2018 Ch.11 Lecture notes

11.2 The yield curve

• section reviews material seen previously (in other courses) o current yields on zero-coupon bonds(risk free) to determine term

structure of interest rates (yield curve), and forward rates can be determined from the yield curve

o use term structure to discount each future payment using the spot interest rate appropriate to the term until that payment is due - this is a replication argument: PV of any cashflow is the cost of purchasing

a portfolio which exactly replicates that cashflow

• text notation; v(t)≡(1+yt)

-t , where yt is the spot rate(current yield on t-year zero coupon bond)

f(t,t+k)) ≡ forward rate, contracted at time zero, effective from t to t+k, expressed as an annual rate • to determine forward rates, use (1+yt)

t+k =(1+yt) t(1+f(t,t+k))

k or

(1+f(t,t+k)) k =(1+yt)

t+k/(1+yt) t = v(t)/v(t+k)

Ex: Given the 10 and 7 year spot rates, determine f (7,10)

AS3429 W2018 Ch.11 Lecture notes

11.3 Valuation of Insurance and Life Annuities

• can redefine PVRV and calculate Premiums and policy reserves using spot rates, For example:

(a) a life annuity due with annual payments issued to x, given a

yield curve {yt}

o PVRV is Y= ∑v(k) (sum from k=0 to Kx) where v(k)=(1+yk) -1

o a&& (x) y = ∑ kpx v(k) (i.e. expected CFs discounted at spot rates) (b) a whole Life Insurance payable at moment of death

o PVRV is Z= v(Tx)

o A (x) y = ∫ v(t) tpx µx+t dt

AS3429 W2018 Ch.11 Lecture notes

Example 1 A survival model follows Makeham’s law with A = 0.001, B = 0.00035 and c =1.075. Calculate the net level annual premium for a 10-year term insurance policy, with sum insured $100,000 payable at the end of the year of death, issued to a life aged 80:

(i) using the spot rates of interest in the table below and, (ii) using a level interest rate of 4.8% per year effective

y1 y2 y3 y4 y5 y6 y7 y8 y9 y10

0.032 0.035 0.038 0.041 0.043 0.045 0.046 0.047 0.048 0.048

AS3429 W2018 Ch.11 Lecture notes

Example 1 A survival model follows Makeham’s law with A=0.001, B=0.00035, and c = 1.075. Calculate the net level annual premium for a 10-year term insurance policy, with sum insured $100,000 payable at the end of the year of death, issued to a life aged 80:

(i) using the spot rates of interest in the table below and, (ii) using a level interest rate of i = 4.8% per year effective

y1 y2 y3 y4 y5 y6 y7 y8 y9 y10

0.032 0.035 0.038 0.041 0.043 0.045 0.046 0.047 0.048 0.048

First, look at (ii) i = 4.8% :

– different i & Makeham parameters than text default model – used SSSM WS to get required values (using given A,B,C and i=.048)

65497.0 1

10 ====

|:[80] A 96885.410 ====|:[80]a

|:[80]|:[80] a 10 1

10 /)A000,100(50.181,13P)ii(forthen &&==

AS3429 W2018 Ch.11 Lecture notes

Example 1: (i) using given Makeham model and yt values

t yt p80+t v(t) tp80 v(t)tp80 v(t)(t-1p80)q80+t-1

0 0.8884470 1 1 1 1 0.032 0.8806070 0.9690 0.88845 0.86089826 0.108093992 2 0.035 0.8722560 0.9335 0.78237 0.73035324 0.099021543 3 0.038 0.8633680 0.8941 0.68243 0.61019067 0.089363899 4 0.041 0.8539130 0.8515 0.58919 0.50170757 0.079397556 5 0.043 0.8438650 0.8102 0.50311 0.40761077 0.069733843 6 0.045 0.8331950 0.7679 0.42456 0.32601864 0.060321165 7 0.046 0.8218750 0.7299 0.35374 0.25820523 0.051692489 8 0.047 0.8098780 0.6925 0.29073 0.20133494 0.043635329 9 0.048 0.7971760 0.6558 0.23546 0.15440475 0.036247113

10 0.048 0.7837440 0.6257 0.18770 0.029882623

5.05072406 0.667389553

ǁ ǁ ä (80:10) A'(80:10)

(i) Using given spot rates can show that P =13,213.75

AS3429 W2018 Ch.11 Lecture notes

Example 1(working page)

AS3429 W2018 Ch.11 Lecture notes

Example 2

You are given the following select mortality table

Age x [ ]xq [ ] 1+xq [ ] 2+xq

qx+3 Attained age x+3

70 .0175 .0250 .0315 .0375 73

71 .0190 .0275 .0345 .0425 74

72 .0210 .0300 .0375 .0465 75

and the following interest rates based on yield curve at time t =0:

t Annual forward rate of interest f(t,t+1) 0 0.030 1 0.032 2 0.035

A $250,000 3-year term insurance policy is purchased on [72]. Assume that the death benefit is payable at end of year of death. The policy annual premium (P) is determined using the equivalence principle.

(i) Calculate 2V (ii) Calculate Pr [ L0 > 0] (iii) How would you calculate Var[L0]?

AS3429 W2018 Ch.11 Lecture notes

Example 2 –(working page)

AS3429 W2018 Ch.11 Lecture notes

(iii) Var[L0] calculations : P = 7,066.75

k L0 Pr(K[72] = k) 0 250,000v.03 –P

q[72] = 0.0210

1 250,000v.03v.032 –P[1+v0.03] 1│q[72] = (1-.021)(.03) = 0.02937

2 250,000v.03v.032 v.035 –P[1+v0.03+v0.03v0.032] 2│q[72] =(1-.021)(1-.03)(.0375) =0.03714

≥3 0 – P[1+v0.03 + v0.03 v0.032 ] 3p[72] =0.9140189

P=7,066.75

Var[L0] = E[L0 2 ] – [E[L0]]

2 = E[L0

2 ] (assuming equivalence principle)

AS3429 W2018 Ch.11 Lecture notes

Example 3-Example 1 revisited : Cashflow replication • accumulating CFs at forward rate will exactly fund sums insured, provided

mortality and experience follows assumptions (replication)

• look at 100,000 contracts from Ex.1(DB=100,000) (#’s in 000s) Year Expected Expected claims Net CF Policy k →k+1 Premium Forward rate outgo carried forward Value P k f(k, k+1) C k+1 CF k+1 t V -----------------------------------------------------------------------------------------------------------------------------------------------------------------------

0 →1 1321372 0.0320 1115528 248129 2,792.83 1 →2 1173970 0.0380 1060741 415409 5,309.62 2 →3 1033806 0.0440 999431 513587 7,525.86 3 →4 901744 0.0501 932420 553752 . 4 →5 778537 0.0510 860726 539560 . 5 →6 664803 0.0551 785539 485133 . 6 →7 561004 0.0520 708189 392368 7 →8 467426 0.0540 630102 276143 8 →9 384166 0.0560 552746 144563 6,139.60 9 →10 311127 0.0480 477564 0 0.00

IF CFs are certain AND risk free rates can be locked in, then no interest rate uncertainty(can lock

in forward rates) BUT

(i) mortality is uncertain (although mortality risk can be diversifiable), AND

(ii) if insurer can’t or chooses not to lock in risk free rates then there is interest rate uncertainty

(and interest rate risk is non-diversifiable)

AS3429 W2018 Ch.11 Lecture notes

11.4 Diversifiable and Non-Diversifiable Risk

• a risk is diversifiable if it can be eliminated (relative to its expectation) by increasing number of policies in the portfolio

• Consider a portfolio of N life insurance policies;

- Xi= R.V. for individual risk for the ith policy in portfolio - Assume Xi s identically distributed with mean µ and standard deviation σ - Aggregate risk is ∑Xi and E[∑Xi ] = Nµ and, if Xi s are independent

then Var [∑Xi] = Nσ 2

and

Risk relative to mean value = √Var [∑Xi] = √Nσ = σ/µ (co-efficient of variation) E[∑Xi ] Nµ √N

As N increases, risk relative to mean value tends to zero

• a condition for diversifiability is independence

• A risk is said to be diversifiable if lim (√Var [∑Xi]) = 0 N →∞ N

AS3429 W2018 Ch.11 Lecture notes

11.4.1 Diversifiable Mortality Risk

Assumptions required for diversifiable mortality risk **

(i) the N lives are independent with respect to future mortality (ii) survival model for each of N lives is known (assume each life has

same survival model)

** note: these assumptions are Ch. 11 text default assumptions

Example 1

For 0 ≤ t ≤ t+s, let Dt,s denote the number of deaths between ages x + t and x + t + s from N lives aged x. Show that

lim √Var [Dt,s] = 0 __________________

N→∞ N

AS3429 W2018 Ch.11 Lecture notes

Example 1

Dt,s has binomial distribution with parameters N and t│s qx , or

“ “ “ “ with parameters N and t px(1―s px+t) Then Var[Dt,s]=N[ t px(1 ―spx+t)][1― t px(1 ―spx+t)]

====

]D[Var s,t

2 txsxttxsxt

))p1(p1)(p1(pN ++++++++

−−−−−−−−−−−−

]D[Var s,tlim

N ====

∞∞∞∞→→→→

AS3429 W2018 Ch.11 Lecture notes

11.4.2 Non-diversifiable Risk

Example2 (Non-diversifiable Interest rate risk)

An insurer issues a whole life insurance policy to (40), with level premiums payable continuously throughout the term of the policy, and with sum insured $50,000 payable immediately on death. Assume Makeham’s survival model with A=0.0001, B=0.00035 and c=1.075.

(a) can show assuming i=5% that P=$1,010.36

(b) Now suppose that the effective annual interest rate is modelled as a random variable, denoted i, with the following distribution: i=4% with probability 0.25, i=5% with probability 0.5, i=6% with probability 0.25.

Calculate the (i) expected value and (ii) the standard deviation of the present value of the future loss on the contract. Assume that the future lifetime is independent of the interest rate

Appendix A-Conditional Expectations and Conditional Variance

E[Y]= E[ E[Y│X] ] and V[Y] = E [V[Y│X] ] + V[ E[Y│X] ]

E[L0]= E[ E[L0│i] ] and V[L0] = E[V[L0│i]] + V[ E[L0│i] ]

AS3429 W2018 Ch.11 Lecture notes

Example 2 working page

AS3429 W2018 Ch.11 Lecture notes

Example 2 PART (b) working page/recap continued

(i) E[L0]=E[E[L0│i] ] where E[L0│i]=50,000 A 40 —1,010.36 a 40│i i E[L0│i] Probability .04 $1,587.43 .25 .05 0 .50 .06 -1,071.49 .25

E[L0]=128.99 = E[E[L0│i] ] =.25(1587.43)+0(.50) —1071.49(.25)

(ii) V[L0] = E[V[L0│i]] + V [E[L0│i]] (a) + (b)

(a): L0│i =50,000vi T

—1,010.36 i|T a

(T=T40)

=(50,000+1010.36/δi) vi T

— 1,010.36/ δi

i 2

i0 ])A([)/36.1010000,50(]iL[V −−−−δδδδ++++====

2 A

V[L0│i]= 14,675 2 with prob 0.25

= 14,014 2 with prob 0.5

= 13,316 2 with prob .25

E[V[L0│i]] = (.25)14,675 2 + (.50)14,014

2 + (.25)13,316

2 =196,364,672

AS3429 W2018 Ch.11 Lecture notes

Example 2 recap;

(ii) V[L0] = E[V[L0│i]] + V [E[L0│i]] (continued) (a) + (b)

(b): V[E[L0│i]]=E[E[L0│i] 2 ]-[E[E[L0│i]]]

2 =E[E[L0│i]

2 ]- (128.99)

i E[L0│i] Probability .04 $1,587.43 .25 .05 0 .50 .06 -1,071.49 .25

E[E[L0│i]

2 ]= 917,006.21 = .25(1587.43)

2 + .5(0)

2 + .25(-1,071.49)

then V[E[L0│i]] = 917,006.21 - 128.99 2

= 900,368

So, V[L0] = E[V[L0│i]] + V[E[L0│i]]=197,265,130 and √V[L0]=14,045

V[L0] = 196,364,762 + 900,368 (rx due to future (rx due to interest

lifetime uncertainty) rate uncertainty)

AS3429 W2018 Ch.11 Lecture notes

V[L0] =E[V[L0│i]] + V[E[L0│i]] =197,265,130

= 196,364,762 + 900,368 (rx due to future (rx due to interest

lifetime uncertainty) rate uncertainty)

Now, for a portfolio of N i.i.d. contracts (L0 ≡ L 0, j) :

V[L]= E [V[∑∑∑∑ ====

1j L0│i] ] + V[ E[∑∑∑∑

====

1j L0│i] ]

= E[N (V[L0│i])] + V[N (E[L0│i])]

V[L]=N E[V[L0│i]] + N 2 V[ E[L0│i]

=N196,364,762 + N 2 900,368 (interest rate uncertainty more important

for large N (and is non-diversifiable)

Note: mortality risk is diversifiable only if future survival model is known and the separate lives are independent with respect to mortality

AS3429 W2018 Ch.11 Lecture notes

Example 3 (Non-diversifiable Mortality risk example) A portfolio consists of 10 identical one-year term insurance policies issued simultaneously to lives aged x with independent lifetimes. Each policy has a sum insured of $1,000 payable at end of year of death and each policy was purchased with a single premium of $90. The insurer used i = 4% for all calculations but is unsure about the mortality of this group of policyholders.

The value of qx is the same for every policyholder. With probability 0.3 qx = 0 for every policyholder and with probability 0.7 qx = 0.2 for every policyholder. Calculate the variance of the present value of future losses at issue random variable for the entire portfolio.

L= ∑Li,j where Li,j are i.i.d. (Li,j ≡ L0 )

Calculate:

V[L]= 10E[V[L0│qx]] + (10 2 )V[E [L0│qx]]

first, calculate V[L0]= E[V[L0│qx]] + V[E [L0│qx]]

AS3429 W2018 Ch.11 Lecture notes

Example 3 calculations:

Calculations for V[L0]= E[V[L0│qx]] + V[E [L0│qx]] :

Lo│qx = 1000v-90 with prob. qx = - 90 with prob. (1- qx) then E[Lo│qx ] = 1000vqx — 90

q E[L0│q] Probability 0.2 $102.308 .7 0.0 -90.00 .3

E[L0]=44.6156 = E[E[L0│q] ]=(0.7)(102.308) +(0.3)(-90)

V[E[L0│q]]=E[E[L0│q] 2 ]- E[E[L0│q]]

2 =E[E[L0│q]

2 ]- (44.6156)

= ((0.7)(102.308) 2 +(0.3)(-90)

2 ) – (44.6156)

2 = 7,766.30

V[L0│q]]=E[(L0│q) 2 ]-[E[L0│q]]

2 =[(1000v-90)

2 (qx)+(-90)

2 (1-qx)]– (1000vqx -90)

V[L0│q]= 147,928.99 with prob. 0.7 (qx=.2) = 0 with prob. 0.3 (qx = 0)

E[V[L0│q]] = (.7)(147,928.99) + (0.3)(0)=103,550.29

V[L0]= E[V[L0│qx]] + V[E [L0│qx]] = 103,550.29 +7,766.30

AS3429 W2018 Ch.11 Lecture notes

Example 3 continued

V[L0]= E[V[L0│qx]] + V[E [L0│qx]] = 103,550.29 +7,766.30

then V[L]= 10E[V[L0│qx]] + (10 2 )V[E [L0│qx]]

(10)103,550.29 + (100)7,766.30 = 1,813,133

= 1,035,503 + 776,630 (rx due to future (rx due to survival

lifetime uncertainty) model uncertainty)

57% 43% If N=100 then 88% of variance is due to survival model uncertainty:

V[L] = 100(103,550.29) + (100)

2 (7,766.30)

= 10,355,030 + 77,663,000 = 88,018,029 (12%) (88%)

AS3429 W2018 Ch.11 Lecture notes

Bernoulli shortcut: • recall, if X is Bernoulli then E(X)=q and Var(X)=q(1-q) • If Y is a r.v. that can only have values x1 and x2 with probabilities of

(1-q) and q respectively, then Y=x1+(x2-x1)X and E[Y]=(x2-x1)q+x1 and Var[Y]= (x2-x1)

2 q(1-q)

Bernoulli shortcut can optionally be used in above problem (a few different ways):

(i) V[Lo│qx ]: Lo│qx = 1000v-90 with prob. qx = - 90 with prob. (1- qx)

then using Bernoulli, V[Lo│qx ]=(1000v) 2 qx(1-qx) (x1 = -90,x2 =1000v-90)

and use this to calc. E[V[Lo│qx]]=(0.7)(1000v) 2 (.2)(.8)+(.3)0=103,550.29

(ii) V[E[Lo│qx ]] =(-90-102.308) 2 (.3)(.7)= 7,766.30 (see below)

q E[L0│q] Probability 0.2 $102.308 .7 0.0 -90.000 .3

Used x1=102.308,x2=–90, q=.3 in Bernoulli shortcut formulae

Also, could calculate E[Lo ] =102.308 + (-90-102.308)(.3)= 44.6156

AS3429 W2018 Ch.11 Lecture notes

Extra Practice problem:

The following data is provided for this problem

δ 50A 50 2 A 50a

.04 0.33 0.14 16.75

.06 0.21 0.08 13.17

A life insurance company issues a 100 i.i.d. policies of $50,000 on individuals aged 50. The Death benefit is payable at moment of death and the premiums P are payable continuously, where P = $878.38. You are also given that δ =4% with probability of 50% and δ = 6% with probability 50%.

Calculate ]L[Var . (answer: 189,566)