Mean Square Optimal Hedges Using Higher Order Moments

4.1.3 Multinomial Lattice Case:Now, we show a general case where we have any given moments.Given the first m moments, there are m + 1 constraintsfor L plus 1 unknown parameters, p 1 ,...,p L , andα. If α > 0 is fixed a priori, p 1 ,...,p L can be computedby solving m + 1 linear equations. Therefore, we need atleast L = m + 1 branches to guarantee the existence of afeasible solution. In this case, p 1 ,...,p L can be parameterizedas a function of α given below. Finally, α > 0 maybe adjusted such that all the probabilities are positive. Herewe provide a parameterization for up to the 4-th moment (orcumulant), but the extension to the cases with even higherorder moments is straightforward.Suppose that we have third cumulant information correspondingto skewness, in addition to the first and the secondcumulants. This imposesC (3)n= s n τ ( √ ) 3 ( √ ) 3∑ Lσ n τ = α τ p l (L − 2l + 1) 3 ,l=1where s n τ is the skewness of X n . Let L = 4, and solve fourlinear equations for the probabilities p 1 , p 2 , p 3 , p 4 . Then,we obtain[p 1 , p 2 , p 3 , p 4 ] = 1 16 × [−1 + σ2 nα 2 (9 − σ2 nα 2 (9 + σ2 nα 2 (−1 + σ2 nα 2 (1 + s nτσ nα1 + s nτσ n3α),−1 + s )nτσ n,α1 − s )nτσ ] n.3α),If σ n is constant, i.e., σ n = σ (n = 0,...,N − 1), thechoice α = σ/2 results in the following formulas:(u 3 n = exp ν n τ + 3σ )√ (τ , u 22nd n = exp ν n τ + σ √ )τ ,2(u n d 2 n = exp ν n τ − σ √ ) (τ , d 3 n = exp ν n τ − 3σ )√ τ ,22[p 1 , p 2 , p 3 , p 4 ] =[ 316 + 1 6 s nτ,516 − s nτ8 ,516 + s nτ8 , 3 16 − 1 6 s nτIf we would like to match the 4th cumulant or “kurtosis,”we may introduce a multinomial lattice with five branches,i.e., L = 5. Let κτ denote the kurtosis of X n . Then we haveC (4)4 = κ n τ ( σ n√ τ) 4].5∑= α 4 τ 2 p l (6 − 2l) 4 − 3 ( √ ) 4σ n τ ,l=1as an additional constraint. In this case, the probabilitiesp 1 , p 2 , p 3 , p 4 , p 5 can be calculated through the solution offive linear equations, and are given by[p 1 , p 2 , p 3 , p 4 , p 5 ] = 1 96 ×[ ( σ2nα 2 −1 + s )nτσ n+ σ2 nα 4α 2 (3 + κ nτ) ,σn2 (α 2 16 − 2s )nτσ n− σ2 nα α 2 (3 + κ nτ) ,{ ()}364 + σ2 n2 α 2 −20 + σ2 nα 2 (3 + κ nτ) ,σn2 (α 2 16 + 2s )nτσ n− σ2 nα α 2 (3 + κ nτ) ,σn2 (α 2 −1 − s )]nτσ n+ σ2 nα 4α 2 (3 + κ nτ) .To understand the effect of kurtosis, assume that s n = 0 andσ n = σ (n = 0,...,N − 1), then we obtain[p 1 , p 2 , p 3 , p 4 , p 5 ]= 1 [ ()σ296 α 2 −1 + σ24α 2 (3 + κ nτ) ,σ 2 ()α 2 16 − σ2α 2 (3 + κ nτ) ,()}3{64 + σ22 α 2 −20 + σ2α 2 (3 + κ nτ) ,σ 2 ()α 2 16 − σ2α 2 (3 + κ nτ) ,σ 2 ()]α 2 −1 + σ24α 2 (3 + κ nτ) .In this case, all the probabilities are positive ifσ √ 3 + κ n τ45 − κ n τ24, 9 + κ nτ16< α < σ√ 3 + κ n τ.2Furthermore, if we choose α = σ/ √ 2, then the above probabilitiesreduce to[ 1 + κn τ[p 1 , p 2 , p 3 , p 4 , p 5 ] = ,96]. (24), 5 − κ nτ, 1 + κ nτ24 96The up-down rates corresponding to five branches can becalculated as(u 4 n = exp ν n τ + 2σ √ )2τ ,(u 3 nd n = exp ν n τ + σ √ )2τ ,u 2 nd 2 n = exp (ν n τ) ,(u n d 3 n = exp ν n τ − σ √ )2τ ,(d 4 n = exp ν n τ − 2σ √ )2τ .We first notice that the probabilities are symmetric, i.e.,p 1 = p 5 and p 2 = p 4 . In this formulation, p 1 , p 3 and p 5