m48-ch21

# m48-ch21 - Chapter 21 The Black-Scholes Equation Question...

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Unformatted text preview: Chapter 21 The Black-Scholes Equation Question 21.1. If V (S, t) = e − r(T − t) then the partial derivatives are V S = V SS = 0 and V t = rV . Hence V t + (r − δ) SV S + S 2 σ 2 V SS / 2 = rV . Question 21.2. If V (S, t) = AS a e γt then V t = γV , V S = aS a − 1 e γt = aV/S , and V SS = a (a − 1 ) S a − 2 e γt = a (a − 1 ) V/S 2 . Therefore the left hand side of the Black-Scholes equation (21.11) is V t + (r − δ) V S S + V SS S 2 σ 2 / 2 − rV = µ γ − r + (r − δ) a + σ 2 2 a (a − 1 ) ¶ V. (1) We can rewrite the coefFcient of V as γ + (r − δ) a + σ 2 2 a (a − 1 ) = σ 2 2 a 2 + µ r − δ − σ 2 2 ¶ a + γ − r. (2) ±rom the quadratic formula, this has roots a = − ³ r − δ − σ 2 2 ´ σ 2 ± r ³ r − δ − σ 2 2 ´ 2 − 4 σ 2 2 (γ − r) σ 2 . (3) Simplifying, a = µ 1 2 − r − δ σ 2 ¶ ± s µ r − δ σ 2 − 1 2 ¶ 2 + 2 (r − γ) σ 2 . (4) Note, for a given γ , these are the only values for a that will satisfy the PDE. Question 21.3. If V (S, t) = e − r(T − t) S a exp (( a (r − δ) + 1 2 a (a − 1 ) σ 2 ) (T − t) ) ,wehave V (S, T ) = S a T ,hence the boundary condition is satisFed. Note that V is of the form AS a e γt , where γ = r − a (r − δ) − 1 2 a (a − 1 ) σ 2 . The previous problem’s result shows γ must solve a = µ 1 2 − r − δ σ 2 ¶ ± s µ r − δ σ 2 − 1 2 ¶ 2 + 2 (r − γ) σ 2 . (5) 264 Chapter 21 The Black-Scholes Equation Letting k = ³ 1 2 − r − δ σ 2 ´ , we have to check a ? = k ± r k 2 + 2 (r − γ) σ 2 . (6) This is equivalent to checking k 2 + 2 (r − γ) σ 2 ? = (a − k) 2 . (7) Expanding, this becomes 2 (r − γ) σ 2 ? = a 2 − 2 a µ 1 2 − r − δ σ 2 ¶ . (8) Solving for γ , γ ? = r − σ 2 a 2 2 + a µ σ 2 2 − (r − δ) ¶ = r − a (r − δ) − σ 2 2 a (a − 1 ) (9) which is conFrmed. One could also do this as a partial derivative exercise. Question 21.4. DeFning V (S, t) = Ke − r(T − t) + Se − δ(T − t) wehave V t = rKe − r(T − t) + δSe − δ(T − t) , V S = e − δ(T − t) and V SS = 0. The Black-Scholes equation is satisFed for V t + (r − δ)V S S + V SS S 2 σ 2 / 2 is rKe − r(T − t) + δSe − δ(T − t) + (r − δ) e − δ(T − t) S (10) = r ³ Ke − r(T − t) + Se − δ(T − t) ´ = rV. (11) This also follows from the result that linear combinations of solutions of the PDE are also solutions. The boundary condition is V (S, T ) = K + S T , i.e. we receive one share and K dollars. Similarly, a long forward contract with value Se − δ(T − t) − Ke − r(T − t) will solve the PDE....
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m48-ch21 - Chapter 21 The Black-Scholes Equation Question...

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