# hw2-2 - 24 2 Basic Options b Because V = Erer(T t t we have...

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24 2 Basic Options b) Because ∂V ∂t = Ere - r ( T - t ) , ∂V ∂S = 0 , and 2 V ∂S 2 = 0 , we have ∂V ∂t + 1 2 σ 2 S 2 2 V ∂S 2 + ( r - D 0 ) S ∂V ∂t - rV = Ere - r ( T - t ) - rEe - r ( T - t ) = 0 . If E is an amount of money you want to have at time T . Ee - r ( T - t ) is the amount of money you need to deposit in a bank at time t . 16. Suppose V ( S, t ) is the solution of the problem ∂V ∂t + 1 2 σ 2 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 x = 2 σ £ ln S + ( r - D 0 - σ 2 / 2)( T - t ) / , τ = T - t and V ( S, t ) = e - r ( T - t ) u ( x, τ ). Show that u ( x, τ ) is the solution of the problem ∂u ∂τ = 2 u ∂x 2 , -∞ < x < , 0 τ, u ( x, 0) = V T e σx/ 2 · , -∞ < x < . Solution : Noticing ∂x ∂S = 2 σ 1 S , ∂x ∂t = - 2 σ ( r - D 0 - σ 2 / 2) , dt = - 1 , we have ∂V ∂t = e - r ( T

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hw2-2 - 24 2 Basic Options b Because V = Erer(T t t we have...

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