Originally from Tao's note: https://terrytao.wordpress.com/2015/10/12/275a-notes-2-product-measures-and-independence/comment-page-2/#comment-682638.
Let ${V}$ be a finite-dimensional vector space over a finite field ${F}$, and let ${X}$ be a random variable drawn uniformly at random from ${V}$. Let ${\langle, \rangle: V \times V \rightarrow F}$ be a non-degenerate bilinear form on ${V}$, and let ${v_1,\dots,v_n}$ be non-zero vectors in ${V}$. Show that the random variables ${\langle X, v_1 \rangle, \dots, \langle X, v_n \rangle}$ are jointly independent if and only if the vectors ${v_1,\dots,v_n}$ are linearly independent.
Any suggestion for either direction would be appreciated.
Best Answer
The cardinality of any $\ d$-dimensional affine subspace $\ S\ $ of $\ V\ $ is $\ q^d\ ,$ where $\ |F|=q\ .$ Therefore \begin{align} \mathbb{P}(X\in S)&=\frac{q^d}{q^r}\ ,\\ \mathbb{P}\big(\big\langle X,v_i\big\rangle=f_i\big)&=\frac{q^{r-1}}{q^r}=\frac{1}{q}\ ,\ \text{and}\\ \mathbb{P}\left(\bigcap_{i=1}^n\big\{\big\langle X,v_i\big\rangle=f_i\big\}\right)&=\frac{q^{r-s}}{q^r}=\frac{1}{q^s}\ , \end{align} where $\ r\ $ is the dimension of $\ V\ ,$ and $\ s\ $ the dimension of the subspace spanned by $\ v_1,v_2,\dots,v_n\ .$ Thus $$ \mathbb{P}\left(\bigcap_{i=1}^n\big\{\big\langle X,v_i\big\rangle=f_i\big\}\right)=\prod_{i=1}^n\mathbb{P}\big(\big\langle X,v_i\big\rangle=f_i\big) $$ if and only if $\ s=n\ $—that is, if and only if $\ v_1,v_2,\dots,v_n\ $ are linearly independent.