Solving $p_k=\frac{1}{2}p_{k+1} + \frac{1}{2}p_{k-1}$

Question

I am trying to solve the recurrence relation

$$p_k=\frac{1}{2}p_{k+1} + \frac{1}{2}p_{k-1}.$$

The context of this recurrence relation is as follows: if I start with \$20, and I win \$1 for every head and lose \$1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate \$100.

Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!

The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent? — Michael Burr, Dec 20 '18 at 04:23

score 2 · Accepted Answer · answered Dec 20 '18 at 04:17

2

Write $p_k=\frac{1}{2}p_{k+1} + \frac{1}{2}p_{k-1}. $ in the form $p_{k+1} =2p_k-p_{k-1} $.

This has characteristic polynomial $d^2 = 2d-1$ or $d^2-2d+1 = 0$ or $(d-1)^2 = 0$.

The repeated root means that the two solutions are $r^k$ and $kr^k$.

The $r^k$ solution gives $r^{k+1} =2r^k-r^{k-1} $ or $r^2 =2r-1 $, so $r = 1$ as you got.

The $kr^k$ solution gives, since $r=1$, just $k$.

To check, if $p_k = k$ then $2p_k-p_{k-1} =2k-(k-1) =k+1 =p_{k+1} $.

Therefore the solutions are $1$ and $k$, so the general solution is $a + bk$.

answered Dec 20 '18 at 04:17

marty cohen

107,799

1

Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately. – Ankit Kumar Dec 20 '18 at 05:00
I forgot about the repeated root, thank you! – user107224 Dec 20 '18 at 09:47

score 0 · Answer 2 · answered Dec 20 '18 at 04:19

0

If $p_0$ and $p_1$ are given, then an easy proof by induction gives

$$p_n=n(p_1-p_0)+p_0.$$

answered Dec 20 '18 at 04:19

Fred

77,394

score 0 · Answer 3 · answered Dec 20 '18 at 04:45

A simple argument lets us avoid solving a recurrence relation.

The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with \$20; you end with \$100 with some unknown probability $x$, and \$0 otherwise. Therefore $20 = 100x$, which yields $x = \frac15$.

(Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)

Solving $p_k=\frac{1}{2}p_{k+1} + \frac{1}{2}p_{k-1}$

3 Answers3