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Let $X=(X_1,\dots,X_d)$ be a random vector, and a.s. $X \in [0,1]^d$. Suppose that for every $a \in \mathbb{R}^d$, we know the probability distribution of the random variable $Y_a = <a,X>$. My question is that can one uniquely determine the distribution of $X$ (i.e., the joint distribution of $X_1,\dots,X_d$)?

If the answer of the above question is yes, here is a further question. Are there a positive integer $n$, and $a_1,\dots,a_n$ ($a_i \in \mathbb{R}^d$) so that if we know the distributions of $Y_{a_1},\dots,Y_{a_n}$, where $Y_{a_i}=<a_i, X>$, then the distribution of $X$ can be uniquely determined?

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  • $\begingroup$ Certainly not - check out couplings. $\endgroup$ – Anthony Quas Mar 6 '14 at 16:02
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The answer to the first question is yes. Think of the characteristic function of X, namely $$ \varphi_X(a) = \mathbb{E}[e^{i\langle a,X\rangle}] = \mathbb{E}[e^{i Y_a}] $$ For the second question, I would say the answer is no. If you know $(Y_{a_i})_{i=1,\dots,n}$, it amounts to know $\varphi_X$ along the directions $a_1,\dots,a_n$. Since the characteristic function is no more than a Fourier transform, you may look at the problem from an analytic point of view: is the Fourier transform determined by the values on a finite number of directions? I think it is now easy to find counterexamples.

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This is a consequence of what is called Cramer-Wold device.

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