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m48-ch18 - Chapter 18 The Lognormal Distribution Question...

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Chapter 18 The Lognormal Distribution Question 18.1. The five standard normals are 7 + 8 15 = . 2582, 11 + 8 15 = − . 7746, 3 + 8 15 = 1 . 291, 2 + 8 15 = 2 . 582, and 15 + 8 15 = − 1 . 8074. Question 18.2. If z is standard normal, µ + σ × z is N ( µ, σ 2 ) hence our five standard normals can be use to create the desired properties: . 8 + 5 ( 1 . 7 ) = − 7 . 7, . 8 + 5 (. 55 ) = 3 . 55, . 8 + 5 ( . 3 ) = − 0 . 7, . 8 + 5 ( . 02 ) = . 7, and . 8 + 5 (. 85 ) = 5 . 05. Question 18.3. x 1 + x 2 is normally distributed with mean 1 and variance 5 + 2 + 2 ( 1 . 3 ) = 9 . 6. x 1 x 2 is nor- mally distributed with mean 3 and variance 5 + 2 2 ( 1 . 3 ) = 4 . 4. Question 18.4. Sums and differences of two random variables are normally distributed hence x 1 + x 2 is normally distributed with mean µ 1 + µ 2 = 10 and variance σ 2 1 + σ 2 2 + 2 ρσ 1 σ 2 = 0 . 5 + 14 + 2 × ( 0 . 3 ) × 0 . 5 × 14 = 10 . 3 The difference is normally distributed with mean µ 1 µ 2 = − 6 and (higher) variance σ 2 1 + σ 2 2 2 ρσ 1 σ 2 = 0 . 5 + 14 2 × ( 0 . 3 ) × 0 . 5 × 14 = 18 . 7 . Question 18.5. x 1 + x 2 + x 3 ˜ N ( 5 . 5 , 47 . 8 ) , x 1 + 3 x 2 + x 3 ˜ N ( 9 . 5 , 123 . 4 ) , and x 1 + x 2 + . 5 x 3 ˜ N ( 4 . 25 , 30 . 65 ) . Question 18.6. If x ˜ N ( µ, σ 2 ) then E (e x ) = ex µ + . 5 σ 2 ; using the given numbers, E (e x ) = e 2 + 2 . 5 = 90 . 017.There is a 50% probability x is below its mean of 2 hence the median of e x is e 2 = 7 . 3891. 244
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Chapter 18 The Lognormal Distribution Question 18.7. Denote the stock price of month i by S i and let the continuous return of month i be denoted as R i = ln (S i + 1 /S i ) .
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