Suppose a k = 1 2 p ( a k + 1 + 1 )...

DIAMMIBENVERMk1

DIAMMIBENVERMk1

Answered

2022-07-02

Suppose a k = 1 2 p ( a k + 1 + 1 ) + 1 2 p ( a k 1 + 1 ) + ( 1 p ) ( a k + 1 ) for 0 < k < n and a 0 = a n = 0 and 0 < p < 1. How do we find a closed form of a k ? For p = 1, I know the solution is a k = k ( n k ). But I have no idea about this case. It is a system of linear equations, so I think it will a unique solution. In fact, the problem comes from random walk on { 0 , 1 , 2 , . . . , n } with absorbing states 0 , n. And a k is the expectation of number of steps reaching absorbing states when starting at k.
a k = 1 2 p ( a k + 1 + 1 ) + 1 2 p ( a k 1 + 1 ) + ( 1 p ) ( a k + 1 ) p a k = 1 + 1 2 p a k 1 + 1 2 p a k + 1 . Thus if we let b k = p a k ,, then it is equivalent to b k = 1 + 1 2 b k 1 + 1 2 b k + 1 , which has solution as above. But how to solve this one about b k ?

Answer & Explanation

trantegisis

trantegisis

Expert

2022-07-03Added 20 answers

Hint: sum a j + 1 2 a j + a j 1 = 2 p for j = 1 , , k and watch most terms telescope, leaving a k + 1 a k a 1 + a 0 = 2 k p a k + 1 a k = a 1 2 k p since a 0 = 0 .
Sum a j + 1 a j = a 1 2 j p for j = 1 , , k and telescope again: a k + 1 a 1 = k a 1 2 p k ( k + 1 ) 2 , or a k = k a 1 k ( k 1 ) p .
Lastly, let k = n 1 and determine a 1 from the condition 0 = a n = n a 1 n ( n 1 ) p , which gives a 1 = n 1 p , so in the end a k = k a 1 k ( k 1 ) p = k ( n 1 ) p k ( k 1 ) p = k ( n k ) p .

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