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# hw1-3 - 2 Basic Options 9 b Due to the fact that the sum of...

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2 Basic Options 9 b) Due to the fact that the sum of two normal random variables is a normal variable, it is easy to show that the sum of n normal variables, φ 1 Δt + · · · + φ n Δt , and the limit of the sum as n → ∞ are normal variables. Because E [ X ( t )] = lim n →∞ E " n X i =1 φ i Δt # = lim n →∞ n X i =1 E [ φ i ] Δt = 0 , and Var [ X ( t )] = lim n →∞ n X i =1 Var [ φ i ] Δt = lim n →∞ t = t, X ( t ) is a normal random variable with mean zero and variance t . c) From a) we know that dX = X ( t + dt ) - X ( t ) is a normal variable. For dX we have E [ dX ] = E [ X ( t + dt )] - E [ X ( t )] = 0 and Var [ dX ] = E h ( dX ) 2 i = E h ( X ( t + dt ) - X ( t )) 2 i = E £ X 2 ( t + dt ) - 2 X ( t + dt ) X ( t ) + X 2 ( t ) / = t + dt - 2E £ X 2 ( t ) / - 2E [ dX · X ( t )] + t = dt, so dX is a normal random variable with mean zero and variance dt . Here we have used the fact that X ( t ) and dX are independent. d) From S = e μt + σX ( t ) , we have ln S = μt + σX ( t ). Because ln S ∂t = μ, ln S ∂X = σ , 2 ln S ∂X 2 = 0 and dX = dX , using Itˆ o’s lemma, we have d ln S = μdt + σdX. 4. Suppose dS = a ( S, t ) dt + b ( S, t ) dX , where dX is a Wiener process. Let f be a function of S and t . Show that df = ∂f ∂S dS + ∂f ∂t + 1 2 b 2 2 f ∂S 2 dt = b ∂f ∂S dX + ∂f ∂t + 1 2 b 2 2 f ∂S 2 + a ∂f ∂S dt. Solution : If we vary S and t by a small amount dS and dt , then f also varies by a small amount. From the Taylor series expansion we can write

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10 2 Basic Options df = ∂f
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hw1-3 - 2 Basic Options 9 b Due to the fact that the sum of...

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