I have three random variables $\theta_1, \theta_2, \theta_3$ all are i.i.d uniformly over $[-\pi,\pi]$. These in reality represent angles in my problem that I am trying to solve. I have a linear combination as follows $$\theta_1-\theta_2 \sim \mathcal{U}(0,2\pi)$$
NOTE THAT THIS DIFFERENCE IS WRAPPED TO AROUND TO BE ALWAYS BETWEEN $[0,2\pi]$.
Then the linear combination uniformly distributed over $[-\pi,\pi]$ due to my special condition. My second linear combination is $$\theta_1-\theta_3 \sim \mathcal{U}(0,2\pi)$$ also uniformly distributed over $[0,2\pi]$, can I say that
$$ \theta_1-\theta_2 \text{ and } \theta_1-\theta_3 $$ are independent, if yes why ???
Let $\ell$ denote the Lebesgue measure on $\mathbb R/2\pi\mathbb Z$ divided by $2\pi$ and consider $(\theta_1,\theta_2,\theta_3)$ i.i.d. with distribution $\ell$. Fix some Borel subsets $A$ and $B$ of $\mathbb R/2\pi\mathbb Z$ and define $$u(x)=P(\theta_1-\theta_2\in A,\theta_1-\theta_3\in B\mid\theta_1=x),$$ for every $x$ in $\mathbb R/2\pi\mathbb Z$, then $$u(x)=P(\theta_2\in x-A,\theta_3\in x-B\mid\theta_1=x)=P(\theta_2\in x-A,\theta_3\in x-B),$$ that is, $$u(x)=\ell(x-A)\ell(x-B)=\ell(A)\ell(B).$$ This implies that $$P(\theta_1-\theta_2\in A,\theta_1-\theta_3\in B)=E(u(\theta_1))=\ell(A)\ell(B),$$ that is, that $(\theta_1-\theta_2,\theta_1-\theta_3)$ is i.i.d. with distribution $\ell$. Note that the method actually proves that $(\theta_1,\theta_1-\theta_2,\theta_1-\theta_3)$ is i.i.d. with distribution $\ell$.