mathman said:
mathman,
I think you may have provided the best possible reference for my purposes. If I understood it correctly, the general result I was looking for follows from the first theorem in that paper. The result I'm looking for is that as the number of trials N --> infinity, the total winnings divided by the total amount bet (total "action") tends to p-q almost surely under these conditions: 1) It is assumed that all bets are made in units of "chips," and that there is a maximum bet limit of K chips per trial. 2) The system for determining the bet size on the n
th trial can depend on the previous n-1 outcomes in any way. 3) The player bets at least the (positive) minimum bet on every round. 4) The trials are independent Bernoulli trials with probability of success p = 1-q, where p < 1/2 < q.
I'll need the corollary to the first theorem proved in the paper:
Let |X_n| \leq 1 almost surely and suppose that for all m, n
E[X_m X_n] = \Phi(|n-m|) where \Phi is nonnegative and
\sum_{n \geq 1 } \frac{\Phi(n)}{n} < \infty
Then the following strong law of large numbers holds:
\lim_{N \rightarrow \infty} \frac{1}{N}\sum_{n \leq N} X_n = 0 almost surely.
To get the result I want, I'll use the following trick: The gambler's chips will be kept in two separate piles. He pays his losses from one pile. These chips have nominal value p. When he wins, the casino pays him in chips of nominal value q, which go into the other pile. This makes his expected 'nominal' value for each bet zero, but at the end we will only be interested in the absolute number of chips, forgetting their nominal values.
The random variables \sigma_n indicate the outcome of trial n:
\sigma_n = \left \{ \begin{array}{cc}+q & \mbox{ if win } \\ -p & \mbox{ if loss } \end{array} \right
The number of chips wagered on trial n is W_n(\sigma_1, ..., \sigma_{n-1}), so that the 'nominal' winnings on the n
th trial is
B_n = \sigma_n W_n
Although B
m and B
n are not independent, their covariance is zero because the nominal chip values were chosen to make E[\sigma_n] = 0:
E[B_m B_n] = E[\sigma_m W_m \sigma_n W_n] = E[\sigma_n]E[\sigma_m W_m W_n] = 0 for n>m
If we set X
n = (1/K) B
n, we will have a sequence of random variables satisfying the conditions of the theorem from the paper. Therefore
\lim_{N \rightarrow \infty} \frac{1}{N}\sum_{n \leq N} X_n = 0 almost surely. Then
\lim_{N \rightarrow \infty} \frac{1}{N}\sum_{n \leq N} B_n = 0 a. s., also.
Now \sum_{n \leq N} B_n = qG_N - pL_N , where G
N is the total number of chips gained on successful trials, and L
N is the total number of chips lost on unsuccessful trials. Then the total "real" winnings in chips is S_N = G_N - L_N, while the total number of chips bet (the action) is A_N = G_N + L_N. So,
\sum_{n \leq N}B_n = \frac{q(S_N + A_N)}{2} - \frac{p(A_N - S_N)}{2} = \frac{S_N - (p-q)A_N}{2}
Putting this in the SLLN gives
\lim_{N \rightarrow \infty} \frac{1}{N}(S_N - (p-q)A_N) = 0 a. s.
Finally, since we have assumed that N <= A
N <= K*N, the fraction A
N/N remains finite and we get
\lim_{N \rightarrow \infty} \frac{S_N}{A_N} - (p-q) = 0 a. s., which is the result hoped for.
Does this make sense? If this is right, I like the fact that it does not even require that the bet sizes from different trials become less correlated as |n-m| increases, even though they will for most systems.
