Hw9-2 - 184 3 Exotic Options ¯ C ¯ C = 1 ¯ C α ¯ C α = 0 ¯ C ¯ ξ α f 2 ¯ C ¯ ξ α f 2 = ξ f 2 ¯ C α ¯ ξ α f 2 ¯ C α ¯ ξ α f

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Unformatted text preview: 184 3 Exotic Options ¯ C + + ¯ C- = 1 , ¯ C + α + + ¯ C- α- = 0 , ¯ C + ¯ ξ α + f 2 + ¯ C- ¯ ξ α- f 2 = ξ f 2 , ¯ C + α + ¯ ξ α + f 2 + ¯ C- α- ¯ ξ α- f 2 = ξ f 2 . From the first two conditions, we have ¯ C + = α- α-- α + , and ¯ C- = α + α +- α- and W ∞ = ¯ C + µ ξ ξ f 1 ¶ α + + ¯ C- µ ξ ξ f 1 ¶ α- = α- α-- α + µ ξ ξ f 1 ¶ α + + α + α +- α- µ ξ ξ f 1 ¶ α- . Subtracting the third condition from the fourth condition yields ¯ C + ( α +- 1) ¯ ξ α + f 2 + ¯ C- ( α-- 1) ¯ ξ α- f 2 = 0 . Hence we have ¯ ξ f 2 = µ- ¯ C- ( α-- 1) ¯ C + ( α +- 1) ¶ 1 α +- α- = µ α + ( α-- 1) α- ( α +- 1) ¶ 1 α +- α- . From the third condition, we further have ξ f 2 = ¯ ξ α- f 2 ‡ ¯ C + ¯ ξ α +- α- f 2 + ¯ C- · = ¯ ξ α- f 2 µ- α- α +- α- · α + ( α-- 1) α- ( α +- 1) + α + α +- α- ¶ = µ α + ( α-- 1) α- ( α +- 1) ¶ α- α +- α- α + α +- α- µ 1- α- α +- 1 + 1 ¶ = α + α +- 1 µ α + ( α-- 1) α- ( α +- 1) ¶ α- α +- α- . Finally we reach at ξ f 1 = ξ f 2 / ¯ ξ f 2 = α + α +- 1 µ α + ( α-- 1) α- ( α +- 1) ¶ α-- 1 α +- α- . 21. Consider the following problem 3 Exotic Options 185 ∂V ∂t + 1 2 n ∑ i =1 n ∑ j =1 ρ ij σ i σ j S i S j ∂ 2 V ∂S i ∂S j + n ∑ i =1 ( r- D i ) S i ∂V ∂S i- rV = 0 , ≤ S , ≤ t ≤ T, V ( S , T ) = V T ( S 1 , S 2 , ··· , S n ) , ≤ S . a) Let V ( S , t ) = e- r ( T- t ) V ( S , t ). Show that V ( S , t ) is the solution of the problem ∂ V ∂t + 1 2 n ∑ i =1 n ∑ j =1 ρ ij σ i σ j S i S j ∂ 2 V ∂S i ∂S j + n ∑ i =1 ( r- D i ) S i ∂ V ∂S i = 0 , ≤ S , ≤ t ≤ T, V ( S , T ) = V T ( S 1 , S 2 , ··· , S n ) , ≤ S . b) Let ( y i = a i [ln S i + b i ( T- t )] , i = 1 , 2 , ··· , n, τ = T- t, and V 1 ( y , τ ) = V ( S , t ), y standing for ( y 1 , y 2 , ··· , y n ) T . Find a i and b i such that V 1 ( y , τ ) satisfies ∂ V 1 ( y , τ ) ∂τ = n ∑ i =1 n ∑ j =1 ρ ij ∂ 2 V 1 ( y , τ ) ∂y i ∂y j ,-∞ < y < ∞ , ≤ τ ≤ T, V ( y , 0) = V 1 T ( y ) ,-∞ < y < ∞ , where V 1 T ( y ) ≡ V T ‡ e σ 1 y 1 / √ 2 , e σ 2 y 2 / √ 2 , ··· , e σ n y n / √ 2 · . c) Let r ij denote the element on the i-th row and the j-th column of a matrix R , where R represents any letter. A , B , and C are M × M matrices. Define D = AB and E = ABC . According to the definition of multification of two matrices, we have d ij = M X k =1 a ik b kj . Show e ij = M X l =1 M X k =1 a ik b kl c lj ....
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This note was uploaded on 02/11/2010 for the course MATH 6203 taught by Professor Zhu during the Spring '10 term at University of North Carolina Wilmington.

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Hw9-2 - 184 3 Exotic Options ¯ C ¯ C = 1 ¯ C α ¯ C α = 0 ¯ C ¯ ξ α f 2 ¯ C ¯ ξ α f 2 = ξ f 2 ¯ C α ¯ ξ α f 2 ¯ C α ¯ ξ α f

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