Z is a finite additive group with a fixed symmetric non-degenerate bilinear form.
Let $x$ be an element of $Z$ chosen uniformly at random. Show that the random variables $\{e_{\xi} (x): \xi \in Z\}$ are pairwise independent, and have variance $1$ and mean $0$ for $\xi \ne 0$ and mean $1$ for $\xi = 0$.
I wish to know how to define the notion of Random variables and the notion of $P(X=x)$. Because the notion defined here doesn't go well with the equation. Since if $\xi \neq 0$ and if we consider the same example by choosing $Z= \mathbb{Z}/5\mathbb{Z}$. Suppose $\xi =2$:
$\mathbb{E}=\sum_{i \in \{0,1,2,3,4\}}e(i)\mathbb{P}(e_2(x) =e(i))=1$ (since $Z$ here is a field $\mathbb{P(e_2(x) =e(i))}=1/5\;\; \forall\; i $
Can someone please help to solve this question. Thanks
Define $e: \mathbb{R}/\mathbb{Z} \to \mathbb{C}$ by $e(\theta) := e^{2 \pi i \theta}$
Define $e_\xi: Z \to \mathbb{C}$ by $e_\xi (x) := e(\xi \cdot x)= e^{2 \pi i \xi \cdot x}$
$e_\xi$ is called a character.
I responded to the other thread here
Basically, I think the authors made a mistake. The other post's interpretation of "independent random variables" is correct, but we don't actually need the random variables to be independent, just to have zero covariance pairwise. Pairwise independent events have zero covariance, but the other way around is not always true (as in this case).
Слава Україні!