Krdytkyu

Let $n in mathbb N$, and $mathcal{K}$ be the set of permutations possible for a set ${1,...,n}$. Let $sigma in mathcal{K}, $ such that $sigma: [n] to [n]$ is a randomly selected permutation.

Define/Find: the appropriate probability space and define the random variable $X$ as the number of fixed points (note: $i$ is a fixed point under $sigma$ if $sigma(i) = i$ for $i in [n]$). Moreover, find $mathbb E[X]$.

My ideas:
$Omega:={(1,sigma(1))times...times(n,sigma(n))in ({1,...,n},{1,...,n})^{n}:sigma in mathcal{K}}$ (not sure in this case; any alternatives?).

I am having difficulties to define random variable $X$ as one "function": As multiple functions I would take $sum_{i=1}^{n}X_{i}=X$, where
$X_{i}(omega)=begin{cases}
text{1,} &quadtext{if } text{$i=sigma(i)$}\
text{0,} &quadtext{otherwise.} \
end{cases}$

Is there anyway of defining $X$ as simply one "Function"?

Now onto my greatest worry, the expected value $mathbb{E}[X]$:

Initially I thought to myself, the probability measure $mathbb{P}$ needs to be a binomial distribution, given that we're looking at a given number of "successes" within $n$ particular instances. Then I thought about the parameter $p$ and knew it would not be constant, which changed my thinking (Does this mean $Binom(n) to X$ is only valid if $(X_{i})_{i =1,...,n}$ are I.I.D?).

With this in mind, we know $mathbb{E}[X]=sum_{i=1}^{n}mathbb{E}{[X_{i}]}$. Individually, this means $mathbb{E}[X_{1}]=0times frac{n-1}{n}+1timesfrac{1}{n}$. But then for $X_{2},...,X_{n}$ surely it would differ greatly on the basis of what "happened" in the previous random variable? How can I calculate this expected value?

You can take $Omega=mathcal K$ equipped with $sigma$-algebra $wp(mathcal K)$ and where all outcomes are equiprobable, so that $P({sigma})=frac1{n!}$ for every $sigmainmathcal K$.

Then $X:Omega=mathcal Ktomathbb R$ is prescribed by $sigmamapsto|{iin[n]mid i=sigma(i)}|$.

$X$ has no binomial distribution (e.g. note that $P(X=n-1)=0$).

To find $mathbb EX$ you can use linearity of expectation, and your setup for this (introducing $X_i$) is okay. The $X_i$ are not independent, but that does not matter. Independence is not needed for linearity of expectation. The $X_i$ all have the same distribution, allowing us to make use of symmetry as well.

Then we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(sigma(1)=1)=nfrac1n=1$$