Suppose a k = 1 2 p ( a k + 1 + 1 ) + 1 2 p ( a k − 1 + 1 ) + ( 1 − p ) ( a k + 1 ) for 0 &lt; k &lt; n and a 0 = a n = 0 and 0 &lt; p &lt; 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 ?

Question

Suppose       a    k    =      1    2    p  (      a          k      +      1        +  1  )  +      1    2    p  (      a          k      −      1        +  1  )  +  (  1  −  p  )  (      a    k    +  1  ) for   0  &amp;lt;  k  &amp;lt;  n and       a    0    =      a    n    =  0 and   0  &amp;lt;  p  &amp;lt;  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  ?

trantegisis · Accepted Answer

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    .

Suppose a k </msub> = 1 2 </mfrac> p ( a <

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