stock loan valuation under brownian-motion based and ...norberg r, the markov chain market, astin...

Stock Loan Valuation Under Brownian-MotionBased and Markov Chain Stock Models

David Prager1

1Associate Professor of MathematicsAnderson University (SC)

Based on joint work with Professor Qing Zhang, University of Georgia

IMA Workshop: Financial and Economic ApplicationsJune 11, 2018

1 What Is a Stock Loan?

2 History and Background

3 A Markov Chain Model

4 Specific Examples

5 Conclusion and Directions for Further Study

What is a Stock Loan?

• Client (borrower) owns share of stock.• Use as collateral to obtain, for a fee, loan from bank

(lender).• Upon loan maturity (or before, for American maturity),

client may either:• Repay the loan (principal and interest).• Default (surrender the stock).

Stock Loan Problem

• For a given stock, maturity, principal, and loan interest rate,what is the fair value of the fee charged by bank?

• Notations:• q = Loan Principal• γ = Loan Interest Rate• r = Risk-Free Rate• c = Bank Service Fee• Amount borrower gets = q − c

Stock Loan Problem

• For a given stock, maturity, principal, and loan interest rate,what is the fair value of the fee charged by bank?

• Notations:• q = Loan Principal• γ = Loan Interest Rate• r = Risk-Free Rate• c = Bank Service Fee• Amount borrower gets = q − c

Two Examples: Borrower’s Perspective

Stock ClosingPrice6/5/17

RepaymentAmount6/5/18*

Apple(AAPL)

$ 155.99 $172.40

SouthernCo.(SO)

$ 50.81 $ 56.15

*Assumes γ = 0.1 and q = share price.

Two Examples: Borrower’s Perspective

Stock ClosingPrice6/5/17

RepaymentAmount6/5/18*

ClosingPrice6/5/18

Borrower’sDecision

Apple(AAPL)

$ 155.99 $ 172.40 $ 191.83 Repay

SouthernCo.(SO)

$ 50.81 $ 56.15 $ 43.93 Default

*Assumes γ = 0.1 and q = share price.

Two Examples: Lender’s Perspective

Stock ClosingPrice6/5/17

CashPaidOut*

Borrower’sDecision

Lender’sNominalProfit

Apple(AAPL)

$ 155.99 $ 153.99 Repay 172.40-153.99=18.41

SouthernCo.(SO)

$ 50.81 $ 48.81 Default 43.93-48.81=(4.88)

*Assumes c = $2 and q = share price.

Perpetual Stock Loans (Xia and Zhou, 2007)

• Stock obeys Geometric Brownian Motion:

St = x exp((

r − δ − (σ2/2))

t + σWt

)where δ is the dividend yield and x = S0.

• Bank collects dividends during loan period.• The loan is perpetual.

Evaluating the Stock Loan: Preliminaries

• Let V (x) ≡

supτ∈T0

E[e−rτ

(x exp

((r − δ − σ2

2

)τ + σWτ

)− qeγτ

)+

].

• Assumption: V (x) = x − q + c > 0

Evaluating the Stock Loan: Preliminaries

For a ∈ R+, let:• τa ≡ inf

[t ≥ 0 : e−γtSt = a

]• g(a) ≡ E

[e−rτa (Sτa − qeγτa)+

]= (a− q)E

[e(γ−r)τa Iτa<∞

].

Solving for the Value Function: Case 1

Case 1: If δ = 0 and γ − r ≤ σ2

2, then

1 g(a) =(a− q)x

aand

2 V (x) = x .

Solving for the Value Function: Case 2

Let a0 ≡

(q

[√(σ2 −

γ−r+δσ

)2+ 2δ + σ

2 + γ−r+δσ

])(√(

σ2 −

γ−r+δσ

)2+ 2δ − σ

2 + γ−r+δσ

) .

Case 2: If δ > 0, or δ = 0 and γ− r >σ2

2, and q < a0 ≤ x , then

1 g(a) attains its maximum at a = x and2 V (x) = x − q.

Solving for the Value Function: Case 3

Let a0 ≡

(q

[√(σ2 −

γ−r+δσ

)2+ 2δ + σ

2 + γ−r+δσ

])(√(

σ2 −

γ−r+δσ

)2+ 2δ − σ

2 + γ−r+δσ

) .

Case 3: If δ > 0, or δ = 0 and γ − r >σ2

2, and a0 > x , then

1 g(a) attains its maximum on [q ∨ x ,∞) at a = a0 and2 V (x) = g(a0).

Finite Maturity Stock Loans, Mean-Reverting Model

• Assume the stock loan matures at time T <∞ andmaturity is European.

• Assume the stock price obeys the mean-reverting model.

St = eXt

dXt = a(L− Xt )dt + σdWt

where a > 0 is the rate of reversion and L is the equilibriumlevel.

Finite Maturity Stock Loans, Mean-Reverting Model

• Assume the stock loan matures at time T <∞ andmaturity is European.

• Assume the stock price obeys the mean-reverting model.

St = eXt

dXt = a(L− Xt )dt + σdWt

where a > 0 is the rate of reversion and L is the equilibriumlevel.

Key Idea: Change of time

• Let φt ≡∫ t

0

(1/α

(φs, Γ

φs))

ds and α(t , ω) = α(t) ≡ σeat .

• Then the mean-reverting model can be written explicitly:

Xt = e−at (log x − L) + L + e−atW(

1φt

).

Solving for the Value Function (P. and Zhang, 2010)

Let u ≡ T − s. Under the mean-reverting model with Europeanmaturity,

V (u, x) =e(γ−r)u+B2

4A +C√(φu+s)−1

1√A

[1− Φ

(√A(

P − B2A

))]

− qe(γ−r)u√(φu+s)−1

1√2A

[1− Φ

(P√

2A)]

whereC ≡ e−a(u+s)(log x − L) + LB ≡ e−a(u+s)

A ≡ 12(φu+s)−1

P ≡ ea(u+s)(log q − L) + L− log x ,

Markov Chain Model for Perpetual Case

• αt denote a Markov chain with state space {1,2} and

generator Q =

(−λ1 λ1λ2 −λ2

).

• Stock obeysdSt

St= µ(αt )dt , S0 = x ≥ 0, t ≥ 0

• µ1 = µ(1) > 0 and µ2 = µ(2) < 0 are given return rates.

Markov Chain Model for Perpetual Case

• αt denote a Markov chain with state space {1,2} and

generator Q =

(−λ1 λ1λ2 −λ2

).

• Stock obeysdSt

St= µ(αt )dt , S0 = x ≥ 0, t ≥ 0

• µ1 = µ(1) > 0 and µ2 = µ(2) < 0 are given return rates.

Value Function

• Stopping Time: τ (perpetual case)• Payoff Function:

J(x , i , τ) ≡ E[e−rτ (Sτ − qeγτ )+ Iτ<∞|S0 = x , α0 = i

],

where x+ = max{0, x}.• Value Function: V (x , i) = supτ J(x , i , τ), where the sup is

taken over all stopping times τ .

Sufficient Conditions for a Closed-Form Solution

• µ2 < r < γ < µ1.• r > ρ0 where ρ0 ≡

12

(µ1 − λ1 + µ2 − λ2 +

√((µ1 − λ1)− (µ2 − λ2))2 + 4λ1λ2

)is the larger root of the equationΦ(x) = (x + λ1 − µ1)(x + λ2 − µ2)− λ1λ2.

• λi > γ − r , for i = 1,2.

Key Change of Variables

• Xt ≡ e−γtSt , so that

dXt = Xt [−γ + µ(αt )] dt .

• Letting ξ ≡ γ − r > 0, the value function becomes

V (x , i) = supτ

E[eξτ (Xτ − q)+ Iτ<∞|X0 = x , α0 = i

]

HJB Equation and Variational Inequalities

With fi ≡ µi − γ, the generator for this value function is

Ah(x , i) = xfih′(x , i) + Qh(x , ·)(i),

where Qh(x , ·)(1) = λ1(h(x ,2)− h(x ,1)), andQh(x , ·)(2) = λ2(h(x ,1)− h(x ,2)).

HJB Equation and Variational Inequalities

The associated variational inequalities are

max{ξh(x ,1) +Ah(x ,1), (x − q)+ − h(x ,1)} = 0,

max{ξh(x ,2) +Ah(x ,2), (x − q)+ − h(x ,2)} = 0.

Solution via Smooth-Fit Substitution

• Start on the region (0, x∗) with free boundary x∗, i.e. thecase in which (ξ +A)h(x , i) = 0, i = 1,2.

• Solve the case in which (ξ +A)h(x ,1) = 0 for h(x ,2) andsubstitute into (ξ +A)h(x ,2) = 0.

Solution via Smooth-Fit Substitution

• Start on the region (0, x∗) with free boundary x∗, i.e. thecase in which (ξ +A)h(x , i) = 0, i = 1,2.

• Solve the case in which (ξ +A)h(x ,1) = 0 for h(x ,2) andsubstitute into (ξ +A)h(x ,2) = 0.

Characteristic Equation

Substitution gives a 2nd order ODE with characteristic equation

φ(β) = f1f2β2+[f1(ξ − λ2) + f2(ξ − λ1)]β+[(ξ − λ1)(ξ − λ2)− λ1λ2] .

Characteristic Equation: Solutions (P. and Zhang,2014)

β1 =−D1 +

√D2

1 − 4f1f2D2

2f1f2,

β2 =−D1 −

√D2

1 − 4f1f2D2

2f1f2,

where

D1 = f1(ξ − λ2) + f2(ξ − λ1),

D2 = (ξ − λ1)(ξ − λ2)− λ1λ2.

Free Boundary Solution

x∗ =

(ξ − λ1 + f1ξ − λ1

)(qβ2

β2 − 1

).

Stopping Time Solution

h(x ,1) =

{A2xβ2 if 0 ≤ x ≤ x∗,A0x + B0 if x > x∗,

h(x ,2) =

{κ2A2xβ2 if 0 ≤ x ≤ x∗,x − q if x > x∗.

A0 = − λ1

ξ − λ1 + f1

B0 =λ1qξ − λ1

κ2 =1λ1

[−(ξ − λ1)− f1β2]

A2 =A0x∗ + B0

(x∗)β2

Verification Theorem

h(x , i) = V (x , i), i = 1,2. Moreover, let

D = (0,∞)× {1} ∪ (0, x∗)× {2}

denote the continuation region. Then

τ∗ = inf{t ≥ 0; (Xt , αt ) 6∈ D}

is an optimal exercising time.

Brownian Motion

Given ε > 0, take

µ1 = r − σ2

2+

σ√ε,

µ2 = r − σ2

2− σ√

ε,

λ1 = λ2 =1ε.

Brownian Motion

As ε→ 0,• St = Sε

t converges weakly to

S0t = S0 exp

((r − σ2

2

)t + σWt

).

• x∗ = x ε,∗ → x0 ≡ β0q/(β0 − 1)

• V (x ,1) and V (x ,2) both converge to

V 0(x) =

{A0xβ0 if x < x0,

x − q if x ≥ x0,

where A0 ≡ (β0 − 1)β0−1q1−β0

(β0)β0.

Numerical Examples: Default Parameters and InitialConditions

Parameter Valuer 0.05q 30S0 33γ 0.1λ1 135.25λ2 130.95µ1 4.89µ2 −5.13

Numerical Examples: γ versus S0

0.05

0.1

0.15

0.2

0.25 2040

6080

100

0

20

40

60

80

Initial Stock Price

Loan Interest Rate

Sta

te 1

Loa

n V

alue

Numerical Examples: γ versus S0

0.05

0.1

0.15

0.2

0.25 2040

6080

100

0

20

40

60

80

Initial Stock Price

Loan Interest Rate

Sta

te 2

Loa

n V

alue

Numerical Examples: q versus λ2

110120

130140

1020

3040

50

0

5

10

15

20

25

Loan Principal

State 2 Switching Rate

Sta

te 1

Loa

n V

alue

Numerical Examples: q versus λ2

110115

120125

130135

1020

3040

50

0

5

10

15

20

25

Loan Principal

State 2 Switching Rate

Sta

te 2

Loa

n V

alue

Numerical Examples: γ versus µ2

0.050.1

0.150.2

0.25

−7

−6.5

−6

−5.5

−54

4.5

5

5.5

6

6.5

Loan Interest Rate

State 2 Return Rate

Sta

te 1

Loa

n V

alue

Numerical Examples: γ versus µ2

0.050.1

0.150.2

0.25

−7−6.5

−6−5.5

−53

3.5

4

4.5

5

5.5

6

Loan Interest Rate

State 2 Return Rate

Sta

te 2

Loa

n V

alue

Numerical Examples: λ1 versus µ2

130

140

150

160

−7−6.5

−6−5.5

−54

4.5

5

5.5

6

6.5

7

7.5

State 1 Switching Rate

State 2 Return Rate

Sta

te 1

Loa

n V

alue

Numerical Examples: λ1 versus µ2

130140

150160

−7−6.5

−6−5.5

−53

3.5

4

4.5

5

5.5

6

6.5

7

State 1 Switching Rate

State 2 Return Rate

Sta

te 2

Loa

n V

alue

Conclusions

• Combined continuous and discrete properties• Closed-form formulas for optimal stopping time and value

function• The stock loan valuation can be determined by the

corresponding exercise time, which is given in terms of asingle threshold level.

Directions for Further Study

• Model calibration• Time variables

References

• Xia, Jianming and Xun Yu Zhou, Stock Loans,Mathematical Finance, April 2007, 307-317.

• Norberg R, The Markov chain market, ASTIN Bulletin,2003, 33: 265–287.

• Prager D and Zhang Q, Stock loan valuation under aregime-switching model with mean-reverting and finitematurity, Journal of Systems Science and Complexity,2010: 572–583.

• Prager D and Zhang Q, Valuation of Stock Loans under aMarkov Chain Model, Journal of Systems Science andComplexity, 2014: 1–17.

• All stock market data is from Yahoo! Finance.• All figures were produced using MATLAB.