hw2-3 - 2 Basic Options 29 19 Suppose V(S t is the solution...

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2 Basic Options 29 19. Suppose V ( S, t ) is the solution of the problem ∂V ∂t + 1 2 σ 2 ( S ) S 2 2 V ∂S 2 + ( r - D 0 ) S ∂V ∂S - rV = 0 , 0 S, t T, V ( S, T ) = V T ( S ) , 0 S. Let ξ = S S + P m , τ = T - t and V ( S, t ) = ( S + P m ) V ( ξ, τ ), where P m is a positive constant. a) Show that V ( ξ, τ ) is the solution of the problem V ∂τ = 1 2 ¯ σ 2 ( ξ ) ξ 2 (1 - ξ ) 2 2 V ∂ξ 2 + ( r - D 0 ) ξ (1 - ξ ) V ∂ξ - [ r (1 - ξ ) + D 0 ξ ] V , 0 ξ 1 , 0 τ, V ( ξ, 0) = (1 - ξ ) P m V T P m ξ 1 - ξ , where ¯ σ ( ξ ) = σ P m ξ 1 - ξ . b) What are the advantages of reformulating the problem on a finite domain? Solution : a) Noticing S = P m ξ 1 - ξ , S + P m = P m 1 - ξ , P m S + P m = 1 - ξ, dS = P m ( S + P m ) 2 , we have ∂V ∂t = ( S + P m ) V ∂τ · dt = - ( S + P m ) V ∂τ = - P m 1 - ξ V ∂τ , ∂V ∂S = V ( ξ, τ ) + ( S + P m ) V ∂ξ · dS = V ( ξ, τ ) + P m S + P m V ∂ξ = V ( ξ, τ ) + (1 - ξ ) V ∂ξ , 2 V ∂S 2 = V ∂ξ dS - dS V ∂ξ + (1 - ξ ) 2 V ∂ξ 2 dS = (1 - ξ ) 3 P m 2 V ∂ξ 2 . Substituting these expressions into the equation yields - P m 1 - ξ V ∂τ + 1 2 σ 2 ( ξ ) P m ξ 1 - ξ 2 (1 - ξ ) 3 P m 2 V ∂ξ 2 +( r - D 0 ) P m ξ 1 - ξ V + (1 - ξ ) V ∂ξ - r P m 1 - ξ V ( ξ, τ ) = 0
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30 2 Basic Options or V ∂τ = 1 2 σ 2 ( ξ ) ξ 2 (1 - ξ ) 2 2 V ∂ξ 2 + ( r - D 0 ) ξ (1 - ξ ) V ∂ξ - [ r (1 - ξ ) + D 0 ξ ] V , where we have defined ¯ σ ( ξ ) = σ ( S ). Noticing S = ξP m 1 - ξ , ¯ σ ( ξ ) can be written as ¯
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