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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.