sol7 - STA3007 Applied probability (08-09) Solution to...

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Unformatted text preview: STA3007 Applied probability (08-09) Solution to Assignment 7 Chapter 5 Problem 1.1 Let X = ξ 1 + ··· + ξ k and Y = ξ k +1 , then X and Y are independently distributed as Gamma distribution with pdf f k ( x ) = λ k ( k- 1)! e- λx x k- 1 and exponential distribution with parameter λ , respectively. Method I: Pr( X = k ) = Pr( X ≤ 1 < X + Y ) = Pr( Y > 1- x | X = x ≤ 1)Pr( X ≤ 1) = Z 1 [1- Pr( Y ≤ 1- x | X = x ≤ 1)] f k ( x ) dx = Z 1 [1- F (1- x )] f k ( x ) dx = Z 1 e- λ (1- x ) λ k ( k- 1)! e- λx x k- 1 dx = λ k e- λ ( k- 1)! Z 1 x k- 1 dx = λ k e- λ k ! . Method II: Since X and Y are independent, the joint distribution of X and Y is f ( x, y ) = f k ( x ) f ( y ) = λ k ( k- 1)! e- λx x k- 1 λe- λy . 1 So, Pr( X = k ) = Pr( X ≤ 1 < X + Y ) = Pr( X ≤ 1 and Y > 1- X ) = Z 1 Z ∞ 1- x f ( x, y ) dxdy = Z 1 Z ∞ 1- x f ( y ) dy f k ( x ) dx = Z 1 e- λ (1- x ) λ k ( k- 1)!...
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sol7 - STA3007 Applied probability (08-09) Solution to...

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