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 $X_{i}(\omega)=\begin{cases} \text{1,} &\quad\text{if } \text{$i=\sigma(i)$}\\ \text{0,} &\quad\text{otherwise.} \\ \end{cases}$
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 $\mathbb{E}[X_{1}]=0\times \frac{n-1}{n}+1\times\frac{1}{n}$. 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 $\sigma\in\mathcal K$.
Then $X:\Omega=\mathcal K\to\mathbb R$ is prescribed by $\sigma\mapsto|\{i\in[n]\mid i=\sigma(i)\}|$.
$X$ 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:$$\mathbb EX=\sum_{i=1}^n\mathbb EX_i=n\mathbb EX_1=nP(\sigma(1)=1)=n\frac1n=1$$