Let $X$ be some random vector in $\mathbb{R}^d$. What are minimal assumptions on the distribution of $X$, so that we can claim the following $$\langle X,a\rangle = \langle X,b\rangle \implies a=b$$
My first thought was that $X$ should be supported on the entire $\mathbb{R}^d$. Because we know that $$\langle x,a\rangle=\langle x,b\rangle,\forall x\in\mathbb{R}^d \implies a=b.$$
Is this correct?
If $X$ satisfies $\langle X, a \rangle = \langle X, b \rangle \implies a = b$ then in particular $\langle X, a \rangle = 0 \implies a = 0$.
Therefore $a \perp \operatorname{Im} X \implies a = 0$ so $(\operatorname{Im} X)^\perp = \{0\}$. Taking orthogonal complements implies $\operatorname{span}(\operatorname{Im X}) = \mathbb{R}^d$.
Conversely, if $\operatorname{span}(\operatorname{Im X}) = \mathbb{R}^d$ then clearly $(\operatorname{Im} X)^\perp = \{0\}$.
Assume $\langle X, a \rangle = \langle X, b \rangle$. We have $$\langle X, a \rangle = \langle X, b \rangle \implies \langle X, a - b\rangle = 0\implies a - b \in (\operatorname{Im} X)^\perp = \{0\} \implies a = b$$
We conclude that for the condition to hold, it is necessary and sufficient that the image of $X$ spans $\mathbb{R}^d$, or equivalently, that it contains $d$ linearly independent vectors.