What is the expected number of fixed points in a random permutation?

What is the expected number of fixed points in a random permutation?

So the probability of particular tuple of j values which defines a unique permutation is: 1/n × 1/(n − 1) × 1/(n − 2)··· = 1/n! So we get the rather surprising fact that the expected number of fixed points in a random permutation is 1, independent of how many elements are being permuted.

What is the average number of fixed points for a permutation of an n element set?

Here it is: since the probability that a random permutation of an n-element set fixes any given point is 1/n, the expected number of fixed points is n⋅1n=1.

What is a fixed point in a permutation?

Fixed points of permutations. Let f : S → S be a permutation of a set S. An element s ∈ S is a fixed point of f if f(s) = s. That is, the fixed points of a permutation are the points not moved by the permutation.

What is the expected number of cycles in a random permutation of n numbers?

n! I.e., the expected number of cycles in a random permutation on n elements as n ! 1 is ln n + constant.

What is fixed for every element?

The number of protons in the nucleus of an atom determines what element it is, so the number of protons is fixed. But, the number of neutrons can vary. Nuclei with the same number of protons but different numbers of neutrons are called isotopes.

How to calculate the number of fixed points in random permutations?

If your problem asks about the number of commonfixed points of two random permutations, define $Y_i=1$ if the two random permutations have $i$ as a common fixed point. Then $\\pr(Y_i=1)=\\frac{1}{n^2}$, so the expected number is $n\\cdot\\frac{1}{n^2}$.$\\endgroup$– André NicolasMay 6 ’14 at 15:50 $\\begingroup$Got it.

How to find the expected number of fixed points?

Thus the expected number of fixed points is $n\\cdot\\frac{1}{n}$, that is, $1$. Remarks:$1$. If we want to find the expected number of commonfixed points of two random permutations, the same technique works. Let $Y_i=1$ if $i$ is a common fixed point of the two permutations. Then the number of common fixed points is $Y_1+\\cdots +Y_n$.

Is the distribution of a random variable practically accessible?

However, there are quite a few instances where the distribution of a random variable is not practically accessible, but the method of indicator random variables works smoothly.$\\endgroup$– André NicolasMay 6 ’14 at 15:59 3