72 1. Streaks

the stock market exhibits streaky behavior (or perhaps other forms

of non-randomness).

One obvious way in which stock prices could be streaky concerns

their daily up-and-down motions. Most stocks have more up days

than down days, so one might model them with coins whose success

probabilities are greater than .5. Using the data, one can compute

the observed success probability and then count the number of success

(or failure) streaks of a given length, or of any length. Then one can

compare these observed values with the theoretical values.

Our data set consists of daily prices, over a period of 14 years, of

439 of the stocks that make up the S&P 500 list. These prices have

been adjusted to take into account dividends and stock splits. For this

reason, many of the stock prices are very low near the beginning of

the data set and hence are subject to rounding errors. We typically

get around this problem by using only the part of the data set in

which a stock’s price is above some cutoff value. Also, we throw out

all days for a given stock on which the stock price was unchanged.

Suppose that we want to compare the expected and the observed

number of pairs of consecutive days in which a given stock’s price

went up on both days. If we have n data points, then we can define

the random variable Xi to equal 1 if the i’th and (i + 1)’st price

changes are both positive, and 0 otherwise. Under the assumption

that the signs of the daily price changes are independent events, the

probability that Xi = 1 is just

p2,

where p is the observed long-range

probability of success for that stock. Thus, it is easy to calculate

the expected number of up-up pairs. There are n − 1 daily changes

(since there are n data points), so there are n − 2 random variables

Xi, each with the same distribution. Thus, the expected number of

up-up pairs is just

(n −

2)p2

.

Unfortunately, the Xi’s are not mutually independent. For example,

if X6 = 1 and X8 = 1, then it is not possible for X7 to equal 0.

Nonetheless, the Xi’s are m-independent, for m = 2. The sequence