CommentSolution4_18AnotherWay

CommentSolution4_18AnotherWay - 1 The probability of ever...

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1 The probability of ever suspending at 1 is P (the jumps ever suspending at 1) = λ The probability of ever suspending at 2 is P (the jumps ever suspending at 2) = λ (1 - λ ) + λ λ - λ 1 λ λ = λ - λλ 1 = λ (1 - λ 1 ) We assume for k = 1 , 2 , . . . , n - 1, we alway have P (the jumps ever suspending at k ) = λ (1 - λ 1 )(1 - λ 2 ) · · · (1 - λ k - 1 ) then for k = n , we have P (the jumps ever suspending at n ) = λ (1 - λ ) n - 1 + n - 1 i =1 { [ λ i - 1 j =1 (1 - λ j )][ λ - λ i λ λ (1 - λ ) n - i - 1 ] } = λ (1 - λ ) n - 1 + n - 1 i =1 [ λ ( λ - λ i )(1 - λ ) n - i - 1 i - 1 j =1 (1 - λ j )] Indeed, we are expecting λ (1 - λ ) n - 1 + n - 1 i =1 [ λ ( λ - λ i )(1 - λ ) n - i - 1 i - 1 j =1 (1 - λ j )] = λ n - 1 j =1 (1 - λ j ) (1) and when k = n - 1, we already have λ (1 - λ ) n - 2 + n -
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