Showing independence of random variables The configuration space is given by the
binary sequences of length N, i.e.,
$Ω_N =\{ω = (ω_1,...,ω_N ) ∈ \{−1,+1\}^N\}$
Write $X_k (ω) = ω_k , 1 ≤ k ≤ N, ω ∈ Ω_N$
to denote the projection on the k-th component of ω, which is to be thought of as the step of the random walk
at time k. As probability
distribution on $Ω_N$ we take the uniform distribution, i.e., $P_N (A) = |A|2^{−N}, A ⊆ Ω_N$. It follows that for $1 ≤ k_1 < ··· < k_n ≤ N$ and $x_{k_i}
∈ \{−1, 1\}, i = 1,...,n, P(X_{k_1} = x_{k_1},...,X_{k_n} = x_{k_n}) = 2^{N−n}2^{−N} = 2^{−n}$

Q: Use this to conclude that $X_1, X_2, . . . , X_n$ are independent and identically distributed with $P(X_k =1) = P(X_k = −1) =\frac{1}{2}$

The definition I've been asked to use is: A finite collection of random variables $X_1, X_2, . . . , X_n$ is mutually independent if the sets $(X_j ∈ A_j )$ are mutually independent for all events $A_j$ in the ranges of the corresponding $X_j$.
I'm struggling to understand and use this definition, what exactly are the sets "$(X_j ∈ A_j )$", and are the events $A_j=\{-1,1\}$?
So far, I've understood that with reference to my question, $(X_j ∈ A_j )=\{ω \in Ω_N  | X_j(ω) \in A_j\}= \{ω \in Ω_N  | ω_j \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\}$. To prove that these sets are mutually independent I have done this: $P(\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\} \cap \{ω \in Ω_N  | ω_{j} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})=P(\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})P(\{ω \in Ω_N  | ω_{j} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})$ for all pairs $(i, j), i,j, \in \{1,2,...,N\}, i \neq j. -(1)$
$ P(\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\} \cap \{ω \in Ω_N  | ω_{k} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\} \cap  \{ω \in Ω_N  | ω_{j} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})=P(\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})P (\{ω \in Ω_N  | ω_{k} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})P(\{ω \in Ω_N  | ω_{j} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\}) \forall (i,j,k), i,j,k \in {1,2,...,N}, i \neq j \neq k$ and so on till
$P(\cap_{i \in \{1,2,...,N\}}\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})=\prod_{i=1}^NP(\{ω \in Ω_N  | ω_{i} \in \{\phi, \{1\},\{-1\},\{-1,1\}\}\})$
Is this okay? What justification should I give for the equalities in these equations? Intuitively I understand that whether the ith R.V. takes the value $1$ or $-1$ has no influence on what the jth R.V. takes, but I'm struggling to write it mathematically. Can someone please help?

The definition of independence involves two concepts: events and probabilities.  This is the case even for the definition of independent random variables.
Let's review.  A random variable $X:\Omega\to R$ with values in some measure space $R$ determines a collection of relevant events; namely, the inverse images of measurable sets $\mathcal B \subset R$ of possible values $X$ can attain, $X^{-1}(\mathcal B) = \{\omega\in\Omega \mid X(\omega)\in \mathcal B\}.$  This collection is often denoted $\sigma(X).$
A pair of random variables $(X,Y)$ is independent when for any $\mathcal E\in \sigma(X)$ and $\mathcal F\in\sigma(Y)$ the probabilities multiply:
$$\Pr(\mathcal E \cap \mathcal F) = \Pr(\mathcal E)\Pr(\mathcal F).$$
The problem you face in applying this definition is two-fold: first, you have $n$ variables, not just two.  Second, there can be a lot of events to consider.  To address the first problem let's consider the two-variable case and hope the ideas generalize.  To address the second problem we need a shortcut.
This shortcut is available because all your variables are binary: with 100% probability their values are in $\{-1,1\}\subset \mathbb R.$  Thus, although there are uncountable many $\mathcal B \subset \mathbb R$ you might need to check, there are really only four kinds:

*

*When $-1\notin \mathcal B$ and $1\notin \mathcal B,$ $X_j^{-1}(\mathcal B) = \emptyset.$


*When $-1\in \mathcal B$ and $1\notin \mathcal B,$ $X_j^{-1}(\mathcal B) = X^{-1}(\{-1\}) = \{\omega\in\Omega\mid X_j(\omega) = -1\}.$ This is the set of all sequences with $-1$ in the $j^\text{th}$ position.


*Similarly when $-1\notin \mathcal B$ and $1\in \mathcal B,$ $X_j^{-1}(\mathcal B)$ is the set of all sequences with $1$ in the $j^\text{th}$ position.


*When $\{-1,1\}\subset \mathcal B,$ $X^{-1}(\mathcal B) = \Omega.$
Because axiomatically $\Pr(\emptyset)=0$ and $\Pr(\Omega)=1,$ there's nothing to check in cases (1) and (4): the equation in the definition of independence reduces to $0=0$ (in the first instance) or $\Pr(\mathcal E) = \Pr(\mathcal E)$ or $\Pr(\mathcal F) = \Pr(\mathcal F)$ (in the second instance), all of which are guaranteed by the reflexive property of equality.
To deal with cases (2) and (3) (and anticipating the generalization to more than two variables), prove the following lemma:
Given a collection of positions $1\le i_1 \lt i_2\lt \cdots \lt i_k\le n$ and specified values $b_{i_j}\in\{-1,1\},$ $j=1,2,\ldots, k,$ the chance that a sequence has those values in those positions is $2^{-k}.$
The demonstration that any two of the $X_j$ are independent is this:

Let $i\ne j.$ The chance that a sequence has a specified value $b_i$ in position $i$ is $2^{-1}.$  The chance it has a specified value $b_j$ in position $j$ is also $2^{-1}.$  The chance that it has both the value $b_i$ in position $i$ and $b_j$ in position $j$ is $2^{-2} = 2^{-1}\times 2^{-1}.$

I leave the generalization to $k\gt 2$ variables to you.
