HW10sol

# HW10sol - ORIE 3500/5500 Fall Term 2008 Assignment 10...

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Unformatted text preview: ORIE 3500/5500 Fall Term 2008 Assignment 10 Solution 1. (a) The final score is S = 0 . 4 X + 0 . 6 Y . So E ( S ) = 0 . 4 μ X + 0 . 6 μ Y = 69 and var ( S ) = 0 . 4 2 σ 2 X + 0 . 6 2 σ 2 Y + 2 · . 4 · . 6 ρσ X σ Y = 181 . 44 . So S ∼ Normal (69 , 181 . 44). (b) Since ( X- 60) / 18 ∼ N (0 , 1), P ( X > 70) = P [( X- 60) / 18 > 5 / 9] = 1- Φ(5 / 9) = 0 . 2893 . So the expected number is 300 · . 2893 = 86 . 8. (c) We have E ( Y | X = 70) = μ Y + ρ σ Y σ X (70- μ X ) = 80 ,var ( Y | X = 70) = σ 2 Y (1- ρ 2 ) = 63 . So P ( Y > 80 | X = 70) = P [( Y- 80) / √ 63 > | Y = 70] = 1- Φ(0) = 1 / 2 . Hence the expected number is 50 · (1 / 2) = 25. 2. (a) We see that P ( X < αλ ) = P ( e- sX > e- sαλ ) ≤ E [ e- sX ] e- sαλ = exp λ ( e- s- 1) + sαλ , as M X ( s ) = exp( λ ( e s- 1)). (b) Minimizing the bound is equivalent to minimize f ( s ) = λ ( e- s- 1) + sαλ . Now f ( s ) = 0 ⇒ - λe- s + αλ = 0 ⇒ e- s = α ⇒ s = log(1 /α ) ....
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HW10sol - ORIE 3500/5500 Fall Term 2008 Assignment 10...

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