{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol7 - STA3007 Applied probability(08-09 Chapter 5 Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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) = 1 0 [1 - Pr( Y 1 - x | X = x 1)] f k ( x ) dx = 1 0 [1 - F (1 - x )] f k ( x ) dx = 1 0 e - λ (1 - x ) λ k ( k - 1)! e - λx x k - 1 dx = λ k e - λ ( k - 1)! 1 0 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
So, Pr( X = k ) = Pr( X 1 < X + Y ) = Pr( X 1 and Y > 1 - X ) = 1 0 1 - x f ( x, y ) dxdy = 1 0 1 - x f ( y ) dy f k ( x ) dx = 1 0 e - λ (1 - x ) λ k ( k - 1)! e - λx x k - 1 dx = λ k e - λ k ! . Problem 1.2 Let X 1 ( t ) , X 2 ( t ) be the Poisson process corresponding to minor and major defects, respectively, then X ( t ) = X 1 ( t ) + X 2 ( t ). Since X 1 ( t ) and X 2 ( t ) are independent, we have (1) for any time points t 0 = 0 < t 1 < t 2 · · · < t n , the process increments X ( t k ) - X ( t k - 1 ) = ( X 1 ( t k ) - X 1 ( t k - 1 )) + ( X
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}