Stock Loan Valuation Under Brownian-Motion Based and Markov Chain Stock Models

Stock Loan Valuation Under Brownian-Motion Based and Markov Chain Stock Models David Prager 1 1 Associate Professor of Mathematics Anderson University (SC) Based on joint work with Professor Qing Zhang, University of Georgia IMA Workshop: Financial and Economic Applications June 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

Two Examples: Borrower s Perspective Stock Apple (AAPL) Southern Co. (SO) Closing Price 6/5/17 Repayment Amount 6/5/18* $ 155.99 $172.40 $ 50.81 $ 56.15 *Assumes γ = 0.1 and q = share price.

Two Examples: Borrower s Perspective Stock Apple (AAPL) Southern Co. (SO) Closing Price 6/5/17 Repayment Amount 6/5/18* Closing Price 6/5/18 Borrower s Decision $ 155.99 $ 172.40 $ 191.83 Repay $ 50.81 $ 56.15 $ 43.93 Default *Assumes γ = 0.1 and q = share price.

Two Examples: Lender s Perspective Stock Apple (AAPL) Southern Co. (SO) Closing Price 6/5/17 Cash Paid Out* Borrower s Decision Lender s Nominal Profit $ 155.99 $ 153.99 Repay 172.40-153.99 =18.41 $ 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: (( ) ) S t = x exp r δ (σ 2 /2) t + σw t where δ is the dividend yield and x = S 0. Bank collects dividends during loan period. The loan is perpetual.

Evaluating the Stock Loan: Preliminaries Let V (x) sup E τ T 0 [e rτ ( x exp Assumption: V (x) = x q + c > 0 ) ) ) ] ((r δ σ2 τ + σw τ qe γτ. 2 +

Evaluating the Stock Loan: Preliminaries For a R +, let: τ a inf [ t 0 : e γt S t = a ] g(a) E [ e rτa (S τa qe γτa ) + ] [ ] = (a q) E e (γ r)τa I τa<.

Solving for the Value Function: Case 1 Case 1: If δ = 0 and γ r σ2 2, then (a q)x 1 g(a) = and a 2 V (x) = x.

Solving for the Value Function: Case 2 Let a 0 ( [ ]) ( ) 2 q σ 2 γ r+δ σ + 2δ + σ 2 + γ r+δ σ ( ). ( ) 2 σ 2 γ r+δ σ + 2δ σ 2 + γ r+δ σ Case 2: If δ > 0, or δ = 0 and γ r > σ2 2, and q < a 0 x, then 1 g(a) attains its maximum at a = x and 2 V (x) = x q.

Solving for the Value Function: Case 3 Let a 0 ( [ ]) ( ) 2 q σ 2 γ r+δ σ + 2δ + σ 2 + γ r+δ σ ( ). ( ) 2 σ 2 γ r+δ σ + 2δ σ 2 + γ r+δ σ Case 3: If δ > 0, or δ = 0 and γ r > σ2 2, and a 0 > x, then 1 g(a) attains its maximum on [q x, ) at a = a 0 and 2 V (x) = g(a 0 ).

Finite Maturity Stock Loans, Mean-Reverting Model Assume the stock loan matures at time T < and maturity is European. Assume the stock price obeys the mean-reverting model. S t = e X t dx t = a(l X t )dt + σdw t where a > 0 is the rate of reversion and L is the equilibrium level.

Key Idea: Change of time Let φ t t 0 ( )) 1/α (φ s, Γ φs ds and α(t, ω) = α(t) σe at. Then the mean-reverting model can be written explicitly: ( ) 1 X t = e at (log x L) + L + e at W. φ t

Solving for the Value Function (P. and Zhang, 2010) Let u T s. Under the mean-reverting model with European maturity, V (u, x) = e(γ r)u+ B 2 4A +C ( 1 A ( A [1 Φ P B ))] (φu+s ) 1 2A qe(γ r)u 1 ( [1 Φ P )] 2A (φu+s ) 1 2A where C e a(u+s) (log x L) + L B e a(u+s) 1 A 2(φ u+s ) 1 P e a(u+s) (log q L) + L log x,

Markov Chain Model for Perpetual Case α t denote a Markov ( chain with ) state space {1, 2} and λ1 λ generator Q = 1. λ 2 λ 2 Stock obeys ds t S t = µ(α t )dt, S 0 = x 0, t 0 µ 1 = µ(1) > 0 and µ 2 = µ(2) < 0 are given return rates.

Value Function Stopping Time: τ (perpetual case) Payoff Function: J(x, i, τ) E [ e rτ (S τ qe γτ ) + I τ< S 0 = x, α 0 = i ], where x + = max{0, x}. Value Function: V (x, i) = sup τ J(x, i, τ), where the sup is taken over all stopping times τ.

Sufficient Conditions for a Closed-Form Solution µ 2 < r < γ < µ 1. r > ( ρ 0 where ρ 0 ) 1 µ 1 λ 1 + µ 2 λ 2 + ((µ 1 λ 1 ) (µ 2 λ 2 )) 2 + 4λ 1 λ 2 2 is the larger root of the equation Φ(x) = (x + λ 1 µ 1 )(x + λ 2 µ 2 ) λ 1 λ 2. λ i > γ r, for i = 1, 2.

Key Change of Variables X t e γt S t, so that dx t = X t [ γ + µ(α t )] dt. Letting ξ γ r > 0, the value function becomes [ ] V (x, i) = sup E e ξτ (X τ q) + I τ< X 0 = x, α 0 = i τ

HJB Equation and Variational Inequalities With f i µ i γ, the generator for this value function is Ah(x, i) = xf i h (x, i) + Qh(x, )(i), where Qh(x, )(1) = λ 1 (h(x, 2) h(x, 1)), and Qh(x, )(2) = λ 2 (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. the case 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) and substitute into (ξ + A)h(x, 2) = 0.

Characteristic Equation Substitution gives a 2nd order ODE with characteristic equation φ(β) = f 1 f 2 β 2 +[f 1 (ξ λ 2 ) + f 2 (ξ λ 1 )] β+[(ξ λ 1 )(ξ λ 2 ) λ 1 λ 2 ].

Characteristic Equation: Solutions (P. and Zhang, 2014) where β 1 = D 1 + D1 2 4f 1f 2 D 2, 2f 1 f 2 β 2 = D 1 D 2 1 4f 1f 2 D 2 2f 1 f 2, D 1 = f 1 (ξ λ 2 ) + f 2 (ξ λ 1 ), D 2 = (ξ λ 1 )(ξ λ 2 ) λ 1 λ 2.

Free Boundary Solution ( ) ( ) ξ x λ1 + f 1 qβ2 =. ξ λ 1 β 2 1

Stopping Time Solution { A2 x h(x, 1) = β 2 if 0 x x, A 0 x + B 0 if x > x, { κ2 A h(x, 2) = 2 x β 2 if 0 x x, x q if x > x. λ 1 A 0 = ξ λ 1 + f 1 B 0 = λ 1q ξ λ 1 κ 2 = 1 λ 1 [ (ξ λ 1 ) f 1 β 2 ] A 2 = A 0x + B 0 (x ) β 2

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; (X t, α t ) D} is an optimal exercising time.

Brownian Motion Given ɛ > 0, take µ 1 = r σ2 2 + σ ɛ, µ 2 = r σ2 2 σ ɛ, λ 1 = λ 2 = 1 ɛ.

Brownian Motion As ɛ 0, S t = S ɛ t converges weakly to S 0 t = S 0 exp x = x ɛ, x 0 β 0 q/(β 0 1) ) ) ((r σ2 t + σw t. 2 V (x, 1) and V (x, 2) both converge to V 0 (x) = { A 0 x β 0 if x < x 0, x q if x x 0, where A 0 (β 0 1) β0 1 q 1 β 0 (β 0 ) β. 0

Numerical Examples: Default Parameters and Initial Conditions Parameter Value r 0.05 q 30 S 0 33 γ 0.1 λ 1 135.25 λ 2 130.95 µ 1 4.89 µ 2 5.13

Numerical Examples: γ versus S 0 80 State 1 Loan Value 60 40 20 0 0.05 0.1 0.15 0.2 Loan Interest Rate 0.25 20 40 60 Initial Stock Price 80 100

Numerical Examples: γ versus S 0 80 State 2 Loan Value 60 40 20 0 0.05 0.1 0.15 0.2 Loan Interest Rate 0.25 20 40 60 Initial Stock Price 80 100

Numerical Examples: q versus λ 2 25 20 State 1 Loan Value 15 10 5 0 140 130 120 State 2 Switching Rate 110 50 40 30 Loan Principal 20 10

Numerical Examples: q versus λ 2 25 20 State 2 Loan Value 15 10 5 0 135 130 125 120 115 State 2 Switching Rate 110 50 40 30 20 Loan Principal 10

Numerical Examples: γ versus µ 2 6.5 6 State 1 Loan Value 5.5 5 4.5 4 5 5.5 6 State 2 Return Rate 6.5 7 0.05 0.1 0.15 0.2 Loan Interest Rate 0.25

Numerical Examples: γ versus µ 2 6 5.5 State 2 Loan Value 5 4.5 4 3.5 3 5 5.5 6 State 2 Return Rate 6.5 7 0.05 0.1 0.25 0.2 0.15 Loan Interest Rate

Numerical Examples: λ 1 versus µ 2 7.5 7 State 1 Loan Value 6.5 6 5.5 5 4.5 5 4 5.5 6 State 2 Return Rate 6.5 7 130 160 150 140 State 1 Switching Rate

Numerical Examples: λ 1 versus µ 2 7 6.5 6 State 2 Loan Value 5.5 5 4.5 4 3.5 5 3 5.5 6 State 2 Return Rate 6.5 7 130 160 150 140 State 1 Switching Rate

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 a single 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 a regime-switching model with mean-reverting and finite maturity, Journal of Systems Science and Complexity, 2010: 572 583. Prager D and Zhang Q, Valuation of Stock Loans under a Markov Chain Model, Journal of Systems Science and Complexity, 2014: 1 17. All stock market data is from Yahoo! Finance. All figures were produced using MATLAB.