292 CHAPTER 11. LIMITS OF VECTORS AND MATRICES

Assume here that p ΜΈ= q. Now it was shown above that limn→∞ pn0 j exists. Denote by Pj this

limit. Then the above becomes much simpler if written as

Pj = qPj−1 + pPj+1 for j ∈ [1,b−1] , (11.3)P0 = 1 and Pb = 0. (11.4)

It is only required to find a solution to the above difference equation with boundary con-ditions. To do this, look for a solution in the form Pj = r jand use the difference equationwith boundary conditions to find the correct values of r. Thus you need

r j = qr j−1 + pr j+1

and so to find r you need to have pr2− r+q = 0, and so the solutions for r are r =

12p

(1+√

1−4pq),

12p

(1−√

1−4pq)

Now √1−4pq =

√1−4p(1− p) =

√1−4p+4p2 = 1−2p.

Thus the two values of r simplify to

12p

(1+1−2p) =qp,

12p

(1− (1−2p)) = 1

Therefore, for any choice of Ci, i = 1,2,

C1 +C2

(qp

) j

will solve the difference equation. Now choose C1,C2 to satisfy the boundary conditions11.4. Thus you need to have

C1 +C2 = 1, C1 +C2

(qp

)b

= 0

It follows that

C2 =pb

pb−qb , C1 =qb

qb− pb

Thus Pj =

qb

qb− pb +pb

pb−qb

(qp

) j

=qb

qb− pb −pb− jq j

qb− pb =q j(qb− j− pb− j

)qb− pb

To find the solution in the case of a fair game, one could take the limp→1/2 of the abovesolution. Taking this limit, you get

Pj =b− j

b.

You could also verify directly in the case where p = q = 1/2 in 11.3 and 11.4 that Pj = 1and Pj = j are two solutions to the difference equation and proceeding as before.

292 CHAPTER 11. LIMITS OF VECTORS AND MATRICESAssume here that p 4 g. Now it was shown above that lim,_,.. pj i exists. Denote by P; thislimit. Then the above becomes much simpler if written asPi = qPj-1+pPj+1 for j € [1,b—-1], (11.3)Po = landP,=0. (11.4)It is only required to find a solution to the above difference equation with boundary con-ditions. To do this, look for a solution in the form P; = r/and use the difference equationwith boundary conditions to find the correct values of r. Thus you needr= gr! + prit}and so to find r you need to have pr? —r+q = 0, and so the solutions for r are r =| |— (14/14 ).=(1- 1-4 )5 (I+ pq) + 5, (1-v1—4pqNowV1—4pq = V1—4p (1 — p) = V1 —4p +4p? = 1 —2p.Thus the two values of r simplify to5 (1+1=2p)=4, 5 (1-(1=2p))=11’ IpTiSTherefore, for any choice of C;,i = 1,2,JjCh + (2)Ppwill solve the difference equation. Now choose C;,C2 to satisfy the boundary conditions11.4. Thus you need to havebC.+Q=1,C,4+Q (2) =0It follows that b bPp gq= , C=p’ _ @g g _ p?Thus P; =q p? (2) _ pr igi _ ai (gh i—P?)q?—p> p?-@q \p g’—p? qe—p? gq’ —p?To find the solution in the case of a fair game, one could take the lim,,_,; 2 of the abovesolution. Taking this limit, you getb-jPP} =—.J bYou could also verify directly in the case where p = g = 1/2 in 11.3 and 11.4 that P; = 1and P; = j are two solutions to the difference equation and proceeding as before.