What is the domain and range of the sum of two random variables? Let $(\Omega, \mathcal{F}, P)$ be  a probability space s.t. $\Omega = \{0,1\}$. Let $X_1: \Omega \rightarrow \{0,1\}$, $X_2: \Omega \rightarrow \{0,1\}$ be two random variables over $\Omega$ (i.e., we're in the context of a Bernouli trial with, say, a coin flip). In fact, we can let $X_1$ and $X_2$ be $id$ in this context. We can suppose further that they are $i.i.d.$
Question: When people write $X_1 + X_2$, what do they mean? In particular, what is the domain and range of $X_1 + X_2$? I'm assuming that its probability distribution forms a convolution of $X_1$ and $X_2$. As I see it, the options for the domain and range are as follows:


*

*$X_1 +X_2$ is a function from $\Omega = \{0, 1\}$ to $\{0, 2\}$ s.t. $(X_1 + X_2)(0) = 0 + 0 = 0$ and $X_1 \rightarrow X_2(1) = 1 + 1 = 0$.

*$X_1 + X_2$ is a function from $\{(0,0), (0,1), (1,0), (1,1)\}$ to $\{0, 1, 2\}$ s.t. $(X_1 + X_2)(i,k) = i + k$.
Notice how in (2) the domain of $X_1 + X_2$ is no longer equal to the domain of $X_1$ and $X_2$.
So, with those options laid out, is $X_1 + X_2$ (1), (2), or something else entirely?
 A: The sum of two functions $X_1,X_2:\Omega\to\mathbb{R}$ is defined pointwise by $(X_1+X_2)(\omega) = X_1(\omega) + X_2(\omega)$.
In your example, there are only four possible functions $X_1,X_2:\{0,1\}\to\{0,1\}$, and contrary to what you said, this prevents them from being independent unless one of them is constant (the non-constant cases are $X_2 = X_1$ or $X_2 = 1 - X_1$). To see this, suppose $X_1,X_2:\{0,1\}\to\{0,1\}$ are the functions $X_1(\omega) = X_2(\omega) = \omega$ for $\omega \in \{0, 1\}$ (the other non-constant case is similar). For $X_1$ and $X_2$ to be independent, we need
$$
P(X_1 \in A, X_2 \in B)
= P(X_1 \in A) P(X_2 \in B)
$$
for all Borel subsets $A, B \subseteq \mathbb{R}$. This does not hold, since $X_1$ and $X_2$ are literally equal functions, so
$$
P(X_1 = 0, X_2 = 1)
= P(\emptyset)
= 0
\neq \frac{1}{4}
= P(X_1 = 0) P(X_2 = 1).
$$
The problem is that the sample space $\{0,1\}$ is not rich enough to handle non-constant independent random variables. Intuitively, $\{0,1\}$ only knows about a single coin flip, not multiple coin flips.
One way to create independent Bernoulli random variables is to consider the product space $\{0,1\}\times\{0,1\}$ (with the corresponding product probability measure), where each component represents a separate coin flip. Now if $X_1$ represents the first coin flip and $X_2$ the second, we can define $X_1((\omega_1,\omega_2)) = \omega_1$ and $X_2((\omega_1,\omega_2)) = \omega_2$. Now $X_1$ and $X_2$ truly represent independent coin flips, and their sum $X_1+X_2$ coincides with the definition above:
$$
(X_1 + X_2)((\omega_1, \omega_2))
= X_1((\omega_1,\omega_2)) + X_2((\omega_1,\omega_2))
= \omega_1 + \omega_2.
$$
A: I think your second answer is closer. Let $S = X_1+X_2$, then $S$ is certainly has a 2-dimensional domain, i.e. $S : \Omega^2 \to \{0,1,2\}$ with
$$
S(\omega) = X_1(\omega_1) + X_2(\omega_2), \quad \forall \ \omega  = (\omega_1, \omega_2) \in \Omega^2.
$$
