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# hw3-1 - 2 Basic Options 33 21 Suppose that S is a random...

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2 Basic Options 33 is the solution of the initial-value problem ∂u ¯ τ = 2 u ∂x 2 , -∞ < x < , ¯ τ > 0 , u ( x, 0) = u 0 ( x ) , -∞ < x < . 21. Suppose that S is a random variable which is deFned on [0 , ) and whose probability density function is G ( S ) = 1 2 πbS e - [ ln( S/a )+ b 2 / 2 ] 2 / 2 b 2 , a and b being positive numbers. Show that a) Z c 0 G ( S ) dS = N µ ln( c/a ) + b 2 / 2 b ; b) Z c 0 SG ( S ) dS = aN µ ln( c/a ) - b 2 / 2 b ; c) for any real number n Z c 0 S n G ( S ) dS = a n e ( n 2 - n ) b 2 / 2 N µ ln( c/a ) + b 2 / 2 b - nb ; d) for any real number n E [ S n ] = a n e ( n 2 - n ) b 2 / 2 ; e) for any real number n Z c S n G ( S ) dS = a n e ( n 2 - n ) b 2 / 2 N µ - ln( c/a ) + b 2 / 2 b + nb ; f) Z c 0 ln S G ( S ) dS = - b 2 π e - [ ln( c/a )+ b 2 / 2 ] 2 / 2 b 2 + ( ln a - b 2 / 2 ) N µ ln ( c/a ) + b 2 / 2 b ; g) Z c ln S G ( S ) dS = b 2 π e - [ ln( c/a )+ b 2 / 2 ] 2 / 2 b 2 + ( ln a - b 2 / 2 ) N µ - ln ( c/a ) + b 2 / 2 b ,

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34 2 Basic Options where N ( z ) = 1 2 π Z z -∞ e - ξ 2 / 2 dξ. Solution : Let η ( S ) = ln( S/a ) + b 2 / 2 b i.e., S = ae - b 2 / 2 . Thus
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hw3-1 - 2 Basic Options 33 21 Suppose that S is a random...

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