| If the bin of the current data point does not affect the bin of the next data point,
then |
| Prob(address ij is occupied) |
= Prob(current data point is in bin i, given the previous was in bin j) |
| = P(i|j) (conditional probability) |
| = P(i)⋅P(j) (independence) |
|
| Denote by N(i) the number of driven IFS points with address i, and by N(ij) the
number with address ij. |
| Then in a data set of M points, the probabilities are given by |
| P(i) = N(i)/M and P(ij) = N(ij)/M |
| This gives a simple test of independence: use Address Stats to find N(1) through
N(4), and N(11) through N(44). Then compare each P(ij) with P(i)⋅P(j). |
| By how how much must they differ to deduce the difference is significant? |
| On the right are the numbers N(11) through N(44) for the previous random example. |
| For ease of comparison, on the left are the products of the probabilities, expressed in the same scale. That
these numbers sum to 9999 instead of 10000 is a consequence of rounding. |
|
| Although the corresponding numbers are relatively close, they are not identical. |
| Nevertheless, these data points are independent, so corresponding values do not differ
significantly. By how much they must differ to claim statistical significance is a more
delicate point. |