Random Permutation Statistics - Expected Number of Cycles of A Given Size m

Expected Number of Cycles of A Given Size m

In this problem we use a bivariate generating function g(z, u) as described in the introduction. The value of b for a cycle not of size m is zero, and one for a cycle of size m. We have

$\frac{\partial}{\partial u} g(z, u) \Bigg|_{u=1} = \frac{1}{1-z} \sum_{k\ge 1} b(k) \frac{z^k}{k} = \frac{1}{1-z} \frac{z^m}{m}$

$\frac{1}{m} z^m \; + \; \frac{1}{m} z^{m+1} \; + \; \frac{1}{m} z^{m+2} \; + \; \cdots$

This means that the expected number of cycles of size m in a permutation of length n less than m is zero (obviously). A random permutation of length at least m contains on average 1/m cycles of length m. In particular, a random permutation contains about one fixed point.

The OGF of the expected number of cycles of length less than or equal to m is therefore

$\frac{1}{1-z} \sum_{k=1}^m \frac{z^k}{k} \mbox{ and } \frac{1}{1-z} \sum_{k=1}^m \frac{z^k}{k} = H_m \mbox{ for } n \ge m$

where H_m is the mth harmonic number. Hence the expected number of cycles of length at most m in a random permutation is about ln m.

Read more about this topic: Random Permutation Statistics

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