when does inner product with fixed vectors determine joint distribution? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:06:47Z http://mathoverflow.net/feeds/question/54130 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54130/when-does-inner-product-with-fixed-vectors-determine-joint-distribution when does inner product with fixed vectors determine joint distribution? John Jiang 2011-02-02T20:13:51Z 2011-02-02T21:35:19Z <p>Given a random vector $(X_1,X_2)$. If $aX_1 + bX_2$ is Gaussian for all pairs $a,b$, then $(X_1,X_2)$ is jointly normal. More generally, is the following statement true? If $aX_1 + bX_2$ has the same distribution as $aY_1 + bY_2$, for all $a,b$, then $(X_1,X_2)$ has the same distribution as $(Y_1,Y_2)$. I know this is true if $(Y_1,Y_2)$ is the pullback under some diffeomorphism of a jointly normal vector. But is it true in general?</p> <p>Also if I have total variation bound on the difference between $aX_1 + bX_2$ and $aY_1 + bY_2$ for all $a,b$, say above by $\epsilon$, does that give any total variation bound between the joint distributions?</p> http://mathoverflow.net/questions/54130/when-does-inner-product-with-fixed-vectors-determine-joint-distribution/54133#54133 Answer by Gideon Schechtman for when does inner product with fixed vectors determine joint distribution? Gideon Schechtman 2011-02-02T21:04:00Z 2011-02-02T21:04:00Z <p>The first question has an affirmative answer. If two 2-dimensional distributions have the same (2-dim) characteristic functions they coincide. The characteristic function of the 2-dimensional distribution of (X,Y) is determined only by the distributions of aX+bY for all a and b. The proof of that fact (Fourier inversion formula) may help to solve the second question as well.</p> http://mathoverflow.net/questions/54130/when-does-inner-product-with-fixed-vectors-determine-joint-distribution/54135#54135 Answer by Mark Meckes for when does inner product with fixed vectors determine joint distribution? Mark Meckes 2011-02-02T21:33:43Z 2011-02-02T21:33:43Z <p>The answer to the second question is no. I heard the following counterexample indirectly from Yitzhak Katznelson. Let $X=(X_1,X_2)$ be a standard Gaussian random vector. One can construct $Y=(Y_1,Y_2)$ with a rotationally symmetric distribution supported on (many) circles centered at the origin such that the total variation distance between $X_1$ and $Y_1$ is arbitrarily small (which by symmetry implies the same for all 1-dimensional marginals). But the distributions of $X$ and $Y$ are mutually singular, so their total variation distance is 2 (or 1, depending on your normalization).</p>