Valuing Equity-Linked Insurance Products...Valuing Equity-Linked Insurance Products Hailiang Yang Department of Statistics and Actuarial Science The University of Hong Kong Hong Kong

Valuing Equity-Linked Insurance Products

Hailiang Yang

Department of Statistics and Actuarial ScienceThe University of Hong Kong

Hong Kong

IMA Workshop Financial and Economic ApplicationsJune 11-15, 2018

Based on joint papers with Hans U. Gerber and Elias S.W. Shiu

Hailiang Yang [email protected]


Motivation

I To value guarantees and options in Variable Annuities

I Variable Annuities= Investment Funds (Mutual Funds)+Rider(s) : Guaranteed Minimum Benefits



Equity-linked death benefit

I x age at issue of policy (time 0)

I Tx time of death

I payment at time Tx

I depends on S(Tx),

I or more generally on S(t), 0 ≤ t ≤ Tx



Goal:

I Calculate E[e−δTx × payment]

I the expectation of the discounted value of the payment

δ valuation force of interest



I Equity-linked products are very popular in the marketnowadays.

I Example: Guaranteed Minimum Death Benefits

I Payoff:

max(S(Tx),K ) = S(Tx) + [K − S(Tx)]+ = K + [S(Tx)− K ]+,

where Tx is the time-until-death random variable for a life agex , S(t) is the price of equity-index at time t, and K is theguaranteed amount.



Mathematical Problem:

I {X (t); t ∈ [0.∞), or t = 0, 1, 2, ...} a random process

I running minimum: m(t) = min0≤s≤t X (s)

I running maximum: M(t) = max0≤s≤t X (s)

I τ a random variable

I we are interested in the distributions of X (τ) and(X (τ),M(τ)) or (X (τ),m(τ)).



Exponential stopping of Brownian motion

I X (t) = µt + σW (t)

I {W (t)}: standard Wiener process

I

fX (τ)(x) =

{κe−αx , if x < 0,κe−βx , if x > 0.

with κ = 2λσ2(β−α)

, α < 0 and β > 0 solutions of the quadratic

equation σ2

2 ξ2 + µξ − λ = 0

I fX (τ),M(τ)(x ,m) = 2λσ2 e

α(m−x)−βm,−∞ < x ≤ m, m ≥ 0



Erlang stopping of Brownian motion

fτ (t) =λntn−1

(n − 1)!e−λt , t > 0,

Let fX (τ)(z) denote the two-sided Laplace transform of fX (τ)(x).

fX (τ)(z) = E[E[e−zX (τ)|τ ]] = E[eDz2τ−µzτ ] = fτ (−Dz2 + µz)

where

fτ (z) =

(λ

z + λ

)n

.



fX (τ)(z) =

(λ

−Dz2 + µz + λ

)n

=

(λ

−D(z + β)(z + α)

)n

= κn(

1

z + β− 1

z + α

)n

,

for −β < z < −α.



LemmaLet a and b be elements with the property that ab = a+ b. Wehave

(a+ b)n = Pn(a) + Pn(b),

where

Pn(x) =n−1∑k=0

(n − 1 + k

k

)xn−k .

If a and b have the property that ab = ν(a+ b) for some numberν = 0, then

(a+ b)n = νnPn(a

ν) + νnPn(

b

ν).



Let

a =−1

α+ z, b =

1

β + z, ν =

1

β − α.

Thus

fX (τ)(z) = κnνn{Pn

(a

ν

)+ Pn

(b

ν

)}

= κnn−1∑k=0

(n − 1 + k

k

)(1

β − α

)k( −1

α+ z

)n−k

+κnn−1∑k=0

(n − 1 + k

k

)(1

β − α

)k( 1

β + z

)n−k

.



Distribution of X (τ)

Note that ( −1α+z )

j is the two-sided Laplace transform of the function(−x)j−1

(j−1)! e−αx I(x<0), and ( 1

β+z )j is that of x j−1

(j−1)!e−βx I(x>0). Hence

fX (τ)(x) =

κne−αx∑n

j=1

(2n−j−1n−j )

(j−1)!(β−α)n−j (−x)j−1, if x < 0,

κne−βx∑n

j=1

(2n−j−1n−j )

(j−1)!(β−α)n−j xj−1, if x > 0,



Reflection principle of Brownian motion

If the drift µ of the Brownian motion X (t) is zero, it follows fromthe reflection principle that

Pr(X (t) ≤ x ,M(t) > y) = Pr(X (t) ≤ x − 2y), y ≥ max(x , 0).



I By changing the probability measure, we can change the drift.If the drift µ is an arbitrary constant, then

Pr(X (t) ≤ x ,M(t) > y) = eyµ/DPr(X (t) ≤ x − 2y),

y ≥ max(x , 0).

The joint density function of X (t) and M(t) is

fX (t),M(t)(x , y) = − ∂2

∂y∂xPr(X (t) ≤ x ,M(t) > y)

= − ∂

∂y[eyµ/D fX (t)(x − 2y)], y ≥ max(x , 0).



Distribution of (X (τ),M(τ))

I We can replace t by τ , then

fX (τ),M(τ)(x , y) = − ∂

∂y[eyµ/D fX (τ)(x − 2y)], y ≥ max(x , 0).

From this we obtain the distribution of (X (τ),M(τ)).



Exponential stopping of Levy process

I moment generating function of X (t) is

E[ezX (t)] = etΨ(z),

where Ψ(z) denotes the Levy exponent of the process

I the mgf of X (τ):

E[ezX (τ)] = E[E[ezX (τ)|τ ]] = E[eτΨ(z)] =λ

λ−Ψ(z).



Suppose that Ψ(z) is a rational function and that the roots of

Ψ(z) = λ

are distinct. Let {αj} and {βk} be the roots with negative andpositive real part, respectively.

λ

λ−Ψ(z)=

∑j

λ

Ψ′(αj)

1

αj − z+

∑k

λ

Ψ′(βk)

1

βk − z,



The pdf of X (τ) is

fX (τ)(x) =

{ ∑j aje

−αjx , if x < 0,∑k bke

−βkx , if x ≥ 0,

where

aj =−λ

Ψ′(αj)

and

bk =λ

Ψ′(βk).



Exponential stopping of jump diffusion

Let {X (t); t ≥ 0} be a Brownian motion (with drift and diffusionparameters µ and σ) extended by independent jumps in bothdirections. The downward jumps form an independent compoundPoisson process; the frequency of these jumps is ν. Similarly, theupward jumps forms another independent compound Poissonprocess with Poisson parameter ω.



Assume that the pdf of each downward jump is

m∑j=1

Ajvje−vjx , x > 0,

with∑m

j=1 Aj = 1 and 0 < v1 < v2 < ... < vm, and that the pdf ofeach upward jump is

n∑k=1

Bkwke−wkx , x > 0,

with∑n

k=1 Bk = 1 and 0 < w1 < w2 < ... < wn.



Then

Ψ(z) = Dz2 + µz − ν

m∑j=1

Ajz

vj + z+ ω

n∑k=1

Bkz

wk − z,

where

D = σ2/2.

Under the assumption that the m+ n+ 2 solutions of the equationΨ(z) = λ are distinct, the density function of X (τ) is given above.In the case of mixtures (all A′

i s and B ′i s positive), the solutions are

distinct and real as we have the following interlacing relationship:

−∞ < αm+1 < −vm < ... < −v1 < α1 < 0

< β1 < w1 < ... < wn < βn+1 < ∞.



For Levy process, we have

M(τ) and [M(τ)− X (τ)] are independent

[X (τ)−M(τ)] and m(τ) have the same distribution

Therefore

E[ezX (τ)] = E[ezM(τ)]E[ezm(τ)],

a version of the celebrated Wiener-Hopf factorization.



MX (τ)(z) =λ

λ−Ψ(z);

MX (τ)(z) is the mgf E[ezX (τ)] when the expectation exists. Thezeros of the denominator are the poles of MX (τ)(z).

Because 0 ≤ M(τ) < ∞, the mgf E[ezM(τ)] is an analytic functionof z with negative real part and it has no negative zeros. Similarly,the mgf E[ezm(τ)] is an analytic function of z with positive realpart and it has no positive zeros.



Thus

E[ezm(τ)] ∝[ m∏j=1

(z + vj)

][m+1∏j=1

1

z − αj

],

E[ezM(τ)] ∝[ n∏k=1

(z − wk)

][n+1∏k=1

1

z − βk

].

As each mgf takes the value 1 when z = 0, we have

E[ezm(τ)] =

[ m∏j=1

z + vjvj

][m+1∏j=1

−αj

z − αj

],

E[ezM(τ)] =

[ n∏k=1

wk − z

wk

][n+1∏k=1

βkβk − z

].



Hence, the pdf of M(τ) is

fM(τ)(x) =n+1∑k=1

b∗ke−βkx , x > 0.

where

b∗k =

[ n∏i=1

wi − βkwi

][ n+1∏i=1,i =k

βiβi − βk

]βk .



Similarly, we obtain the pdf of m(τ),

fm(τ)(x) =m+1∑j=1

a∗j e−αjx , x < 0,

where

a∗j =

[ m∏i=1

αj + vivi

][ m+1∏i=1,i =j

−αi

αj − αi

](−αj).



Now

fX (τ),M(τ)(x , y) = fM(τ),X (τ)−M(τ)(y , x − y)

= fM(τ)(y)fX (τ)−M(τ)(x − y)

= fM(τ)(y)fm(τ)(x − y)

=

[n+1∑k=1

b∗ke−βky

][m+1∑j=1

a∗j e−αj (x−y)

]

=m+1∑j=1

n+1∑k=1

a∗j b∗ke

−(βk−αj )y−αjx .



Link to ruin theory

For t ≥ 0, let the surplus of a company at time t be

U(t) = u − X (t),

where u is a positive number representing the initial surplus. Let

T = inf{t : U(t) ≤ 0}

be the time of ruin.



Link to ruin theory

Then,

Pr(M(τ) ≥ u) = Pr(τ > T ) = E[e−λT ].

Thus, knowing the Laplace transform, with respect to λ, of thetime of ruin random variable is equivalent to knowing thedistribution of the M(τ).



Link to ruin theory

In particular, if {X (t)} is a Levy process with an upward jump pdfof the linear combination of exponential, we can use the results inSection 9 of Albrecher, Gerber and Yang (2010), with w(x) ≡ 1and w0 = 1, to obtain a closed-form expression for Pr(M(τ) ≥ u).



Geometric stopping of a random walk

Random walk

I X (t) = X1 + · · ·+ Xt , X (0) = 0

I X1,X2, ... are i.i.d. r.v.’s, with Pr{Xt = 1} = p1,Pr{Xt = 0} = p0, Pr{Xt = −1} = p−1, p1 + p0 + p−1 = 1.



Geometric distribution

Let τ be an independent r.v. with a geometric distribution (sayparameter π), such that

Pr{τ = t} = (1− π)πt , t = 0, 1, 2, ... .

Its probability generating function (pgf) is

E [zτ ] =1− π

1− πz.



The pgf of X (τ)

E[zX (τ)

]= E

[E[zX (τ)|τ

]]= E

[E[zX (1)

]τ]= E

[(p1z + p0 + p−1z

−1)τ]

=1− π

1− π(p1z + p0 + p−1z−1).

This is a rational function of z , which we can expand by partialfractions. Let

0 < α < 1 < β < ∞denote the the solutions of the quadratic equation

πp1z2 − (1− πp0)z + πp−1 = 0.



The pgf of X (τ)

E[zX (τ)

]= C

α

z − α− C

β

z − β,

with

C =1− π

π

1

p(β − α)=

(1− α)(β − 1)

β − α.

To identify the distribution of X (τ), we note that

E[zX (τ)

]= C

α/z

1− α/z+ C

1

1− z/β.



The distribution of X (τ)

Pr{X (τ) = j} = Cα−j , j = −1,−2, ...,

Pr{X (τ) = j} = Cβ−j , j = 0, 1, 2, ... .

Thus X (τ) has a two-sided geometric distribution.



The record highs and lows of the random walk

I

M(t) = max{0,X (1), ...,X (t)}

denote the running maximum and, similarly, m(t) the runningminimum after t steps.



Joint distribution in the trinomial tree model

Suppose X (t) = X1 + ...+ Xt ,

where Xi takes three values: −1, 0, 1

and P(Xi = 1) = P(Xi = −1) = p/2, P(Xi = 0) = q withp + q = 1.

We assume that X1,X2, ... is an i.i.d. sequence. Since the randomwalk X (t) is symmetric, the reflection principle is true (the proof isthe same as that for simple symmetric random walk).

P({X (t) = j ,M(t) ≥ k}) = P(X (t) = 2k − j)




Now we assume that the random walk is not symmetric,

In this case, we have

P(X (t) = j ,M(t) ≥ k) = (p−1/p1)k−jP(X (t) = 2k − j).




Since τ is independent of X (t), we have

P(X (τ) = j ,M(τ) ≥ k) = (p−1/p1)k−jP(X (τ) = 2k − j).

From this we can obtain the joint probability function of X (τ) andM(τ).



Distributions of M(τ) and m(τ)

Both M(τ) and m(τ) have geometric distributions:

Pr{M(τ) ≥ k} = β−k , k = 0, 1, 2, ...,

Pr{m(τ) ≤ k} = α−k , k = 0,−1,−2, ...,

or,

Pr{M(τ) = k} = (β − 1)β−k−1, k = 0, 1, 2, ...,

Pr{m(τ) = k} = (1− α)α−k , k = 0,−1,−2, ... .

We note that M(τ) ≥ k is the event that X (n) reaches level kbefore or at time τ . Similarly, m(τ) ≤ k is the event that X (n)falls to level k before or at time τ .



The distributions of M(τ) and X (τ)−m(τ) are the same

We note that for each t, the r.v.’s M(t) and X (t)−m(t) have thesame distribution. This follows from

M(t) = max{0,X1,X1 + X2, ... ,X1 + X2 + ...+ Xt},X (t)−m(t) = max{0,Xt ,Xt + Xt−1, ... ,Xt + Xt−1 + ...+ X1}.

Hence, the distributions of M(τ) and X (τ)−m(τ) are the same.Similarly, the distributions of m(τ) and X (τ)−M(τ) are the same.



The r.v.’s M(τ) and X (τ)−M(τ) are independent

Because the conditional distribution of X (τ)−M(τ), given M(τ),is the conditional distribution of X (τ), given X (n) ≤ 0 forn = 1, ..., τ , and hence the same for all values of M(τ). To seethis, consider the first time t when X (t) = M(τ); thus τ ≥ t andX (n)− X (t) ≤ 0 for n = t, ..., τ . Then observe that theconditional distribution of τ − t does not depend on t.



The joint distribution of X (τ) and M(τ)

Pr{X (τ) = j ,M(τ) = h}= Pr{M(τ)− X (τ) = h − j , M(τ) = h}= Pr{M(τ)− X (τ) = h − j}Pr{M(τ) = h}= Pr{m(τ) = −(h − j)}Pr{M(τ) = h}= (1− α)αh−j(β − 1)β−h−1.

Similarly, for h = 0,−1,−2, ... and j ≥ h, we have

Pr{X (τ) = j ,m(τ) = h}= Pr{X (τ)−m(τ) = j − h, m(τ) = h}= Pr{X (τ)−m(τ) = j − h}Pr{m(τ) = h}= Pr{M(τ) = j − h}Pr{m(τ) = h}= (β − 1)β−(j−h)−1(1− α)α−h.



For k = 0, 1, 2, ... and j ≤ k, we find that

Pr{X (τ) = j ,M(τ) ≥ k} = Cα−j(α

β)k .

Of course for j ≥ k ,

Pr{X (τ) = j ,M(τ) ≥ k} = Pr{X (τ) = j} = Cβ−j .

Similarly, for k = 0,−1,−2, ... and j ≥ k one shows that

Pr{X (τ) = j ,m(τ) ≤ k} = Cβ−j(β

α)k .

Of course

Pr{X (τ) = j ,m(τ) ≤ k} = Pr{X (τ) = j} = Cα−j

for j ≤ k.



Remark

The proofs that M(τ) and X (τ)−M(τ) are independent, and thatX (τ)−m(τ) has the same distribution as m(τ), are valid for ageneral random walk. It follows that

PX (τ)(z) = E[zX (τ)] = E[zM(τ)+X (τ)−M(τ)]

= E[zM(τ)]× E[zX (τ)−M(τ)]

= E[zM(τ)]× E[zm(τ)] = PM(τ)(z)Pm(τ)(z).

This formula is a version of the Wiener-Hopf factorization.



Remark

If X1 takes integer values from −n to +m, then

PX (τ)(z) =1− π

1− πPX1(z)]=

1− π

1− π∑m

j=−n pjzj

=(1− π)zn

g(z),

where g(z) =

(1− π

∑mj=−n pjz

j

)zn is a polynomial of degree

m + n.



RemarkBecause 1 > π = π

∑mj=−n pj , we have

|zn| > |g(z)− zn|, for |z | = 1.

Then by Rouche’s Theorem, g(z) has the same number of zerosinside the complex disk of radius 1 as the function zn. Denotethese n zeros of g(z) as α1, ..., αn. Denote the other zeros of g(z),those with absolute value greater than 1, as β1, ..., βm. Then, thepgf PX (τ)(z) is proportional to

zn(∏nj=1(z − αj)

)(∏mj=1(z − βj)

)=

1(∏nj=1(1− αj/z)

)(∏mj=1(z − βj)

) .



Remark

Note that PM(τ)(1) = 1 and Pm(τ)(1) = 1. BecauseM(τ) ≥ 0, PM(τ)(z) exists for each z with |z | < 1. Similarly,Pm(τ)(z) exists for each z with |z | > 1. Therefore

PM(τ)(z) =

∏mj=1(βj − 1)∏mj=1(βj − z)

, Pm(τ)(z) =

∏nj=1(1− αj)∏m

j=1(1− αj/z).



Remark

In the special case of a simple random walk, we have

PX (τ)(z) = C(β − α)z

(z − α)(β − z)=

β − 1

β − z× (1− α)z

z − α.



Applications to Valuing Equity-linked Insurance Products

I To value guarantees and options in Variable Annuities

I Variable Annuities= Investment Funds (Mutual Funds)+Rider(s) : Guaranteed Minimum Benefits



Some Examples

I Guaranteed Minimum Death Benefits

I Payoff: max{S(Tx),K}

where S(t) denotes the price of a stock or stock index at timet, Tx is the future life time of policyholder aged x .

I Note that max{S(Tx),K} = (K − S(Tx))+ + S(Tx).



Some Examples

I High water mark method or low water mark method

I Payoff: max{MS(Tx),K}

where MS(t) denotes the running maximum of the the priceof a stock or stock index up to time t,

I Note that max{MS(Tx),K} = (MS(Tx)− K )+ + K .



Mathematical problem

I How to calculate

E [e−δTxb(S(Tx),MS(Tx))]. (orE [v (Kx+1)b(S(Kx),MS(Kx))].)

where b(., .) is a benefit function, (v is the discount factor perunit time, Kx denotes the curtate future life time r.v.).



Index process

I S(t) price of one unit of a fund at time t

I S(t) = S(0)eX (t), t ≥ 0 X (t) Brownian motion, a Levyprocess (in the discrete time case, a random walk)

I Tx (or Kx) is assumed to be independent of S(t)



The reduced problem

I Idea: the pdf of Tx can be approximated by

n∑i=1

Aiλie−λi t , t > 0

(the Ai ’s can be negative, as long as the sum is pdf)



The reduced problem

E[e−δTxb(S(Tx))] = E[E[e−δTxb(S(Tx))|Tx ]]

=

∫ ∞

0e−δt fTx (t)E[b(S(t))|Tx = t]dt

=

∫ ∞

0e−δt fTx (t)E[b(S(t))]dt

≃∫ ∞

0e−δt

n∑i=1

Aiλie−λi tE[b(S(t))]dt

=n∑

i=1

Ai

∫ ∞

0e−δtλie

−λi tE[b(S(t))]dt.



The reduced problem

So, it suffices that we know how to calculate

E[e−δτb(S(τ))] =

∫ ∞

0e−δtλe−λtE[b(S(t))]dt



The reduced problem

δ can be eliminated

∫ ∞

0e−δtλe−λtE[b(S(t))]dt

=λ

λ+ δ

∫ ∞

0(λ+ δ)e−(λ+δ)tE[b(S(t))]dt

rule: do the calculation without discountingbut replace λ by λ+ δ multiply by λ

λ+δ



I Want to calculate

E[b(S(τ),MS(τ))] = E[b(S(0)eX (τ),S(0)eM(τ))]

where M(t) is the running maximum of X (s) up to time tand τ is an exponential random variable.

I so we need

fX (τ),M(τ)(x , y) the joint pdf of (X (τ),M(τ))



Examples

In the following we assume that X (t) is a Brownian motion.

E[e−δτb(S(τ))] = E[b(S(τ∗))]

= κ

∫ 0

−∞b(S(0)ex)e−αxdx + κ

∫ ∞

0b(S(0)ex)e−βxdx .



Examples

I (1) b(s) = (K − s)+, K < S(0) out-of-the-money put option

E[e−δτ (K − S(τ))+] =κK

−α(1− α)

[K

S(0)

]−α

I (2) b(s) = (s − K )+, K > S(0) out-of-the-money call option

E[e−δτ (S(τ)− K )+] =κK

β(β − 1)

[S(0)

K

]β



Examples

Lookback options

E[e−δτ [ max0≤t≤τ

S(t)− K ]+]

= E[e−δτ [S(0)eM(τ) − K ]+]

out-of-the-money

=λ

λ+ δ

K

β − 1

[S(0)

K

]β.



Examples

Barrier options

Let L denote the barrier and ℓ = ln[L/S(0)]. Consider theup-and-in option (S(0) < L), the value is given by

Pr(M(τ) ≥ ℓ)Eb(L) =[S(0)

L

]βEb(L),

where

Eb(s) = E[b(S(τ))|S(0) = s]



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Documents

Valuing Equity-Linked Insurance Products...Valuing Equity-Linked Insurance Products Hailiang Yang Department of Statistics and Actuarial Science The University of Hong Kong Hong Kong