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

Transcription

1 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

2 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

3 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).

4 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

5 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

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

7 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 $ $ $ Repay $ $ $ Default *Assumes γ = 0.1 and q = share price.

8 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 $ $ Repay =18.41 $ $ Default =(4.88) *Assumes c = $2 and q = share price.

9 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.

10 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 +

11 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<.

12 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.

13 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.

14 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 ).

15 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.

16 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.

17 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

18 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,

19 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.

20 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.

21 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.

22 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.

23 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

24 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)).

25 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.

26 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.

27 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.

28 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 ].

29 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.

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

31 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

32 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.

33 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.

34 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

35 Numerical Examples: Default Parameters and Initial Conditions Parameter Value r 0.05 q 30 S 0 33 γ 0.1 λ λ µ µ

36 Numerical Examples: γ versus S 0 80 State 1 Loan Value Loan Interest Rate Initial Stock Price

37 Numerical Examples: γ versus S 0 80 State 2 Loan Value Loan Interest Rate Initial Stock Price

38 Numerical Examples: q versus λ State 1 Loan Value State 2 Switching Rate Loan Principal 20 10

39 Numerical Examples: q versus λ State 2 Loan Value State 2 Switching Rate Loan Principal 10

40 Numerical Examples: γ versus µ State 1 Loan Value State 2 Return Rate Loan Interest Rate 0.25

41 Numerical Examples: γ versus µ State 2 Loan Value State 2 Return Rate Loan Interest Rate

42 Numerical Examples: λ 1 versus µ State 1 Loan Value State 2 Return Rate State 1 Switching Rate

43 Numerical Examples: λ 1 versus µ State 2 Loan Value State 2 Return Rate State 1 Switching Rate

44 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.

45 Directions for Further Study Model calibration Time variables

46 References Xia, Jianming and Xun Yu Zhou, Stock Loans, Mathematical Finance, April 2007, Norberg R, The Markov chain market, ASTIN Bulletin, 2003, 33: 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: Prager D and Zhang Q, Valuation of Stock Loans under a Markov Chain Model, Journal of Systems Science and Complexity, 2014: All stock market data is from Yahoo! Finance. All figures were produced using MATLAB.