Sigma algebra generated by a random variable

Question

I am not sure if I understand the logic behind sigma algebras generated by random variables. I will try to present my thinking on uniform distribution. Let X be uniformly distributed on $[0,2]$, so for instance $X^{-1}(A)$ where A is an interval $[0,\frac{1}{2}]$ would be a set of intervals on $[0,2]$ which have length less or equal to $\frac{1}{2}$ and if B was an interval $[-5,-1]$ then $X^{-1}(B)$ would be an empty set. Now I think that this far I am correct, but let's add some other variable lets say $Y$ where $P(Y=1)=P(Y=-1)=\frac{1}{2}$. Now, what is a delta algebra generated by those two variables $\sigma(X,Y)$? Even more far into the depths: Let's create another random variable $S_n=\sum_1^nY_i$, where Y is defined as earlier. What are the sets generated by $(S_1,S_2,...,S_n)$? Any intuition appreciated.

Answer 1

If $X$ and $Y$ are independent, then $(,) = ()\times()$.

Answer 2

In general if $(\Omega,\mathcal A,P)$ denotes a probability space and $X:\Omega\to\mathbb R$ is a random variable then the $\sigma$-algebra generated by $X$ is the collection:$$\sigma(X)=X^{-1}(\mathcal B)=\{X^{-1}(B)\mid B\in\mathcal B\}\subseteq\mathcal A$$where $\mathcal B$ denotes the Borel-$\sigma$-algebra on $\mathbb R$.

Similarly this is true for a random vector $X:\Omega\to\mathbb R^n$:$$\sigma(X)=X^{-1}(\mathcal B^n)=\{X^{-1}(B)\mid B\in\mathcal B^n \}\subseteq\mathcal A$$where $\mathcal B^n$ denotes the Borel-$\sigma$-algebra on $\mathbb R^n$.

It is straightforward to prove that these collections are $\sigma$-algebra, and actually it is the smallest $\sigma$-algebra on $\Omega$ such that $X$ is Borel-measurable.