Show that distributions of linear combinations of two variables determine couple law Let $X=(X_{1},X_{2})$ and $Y=(Y_{1},Y_{2})$ be two $\mathbb{R}^{2}$-valued variables. Show that if $aX_{1}+bX_{2}$ has same distribution as $aY_{1}+bY_{2}$ for all real numbers $a,b$ then $X$ has same distribution as $Y$.
I know this can be made to follow by the fact that knowledge of  the characteristic function $v\mapsto\mathbb{E}(\exp(i\langle v, X \rangle))$ determines the distribution of $X$. However I don't know how to prove that (and I know it is tedious), so I was wondering if it could be made to follow from the fact that $t\mapsto \mathbb{E}(\exp(itX_{1}))$ determines $X_{1}$, or even better if it is possible to prove this fact without characteristic functions.
 A: This result is a special case of the so-called Cramer-Wold device. 
Let $X$ and $Y$ be $n$-dimensional random vectors such that, for each $s\in\mathbb{R}^n$, $s^TX \stackrel{dist}= s^TY $. We wish to show that this implies $X\stackrel{dist}=Y$. Now, Levy's continuity theorem guarantees that
$$
\phi_{X}(s) = \phi_{Y}(s),\ \forall\ s\in\mathbb{R}^n \implies X\stackrel{dist}=Y\,.
$$ 
So, we need only show that $\phi_{X}(s) = \phi_{Y}(s)$. But, this follows, because each pair $s^TX,\ s^TY$ are univariate identically distributed random variabes (so their characteristic functions agree), therefore  

$$
\phi_{X}(s) = \mathbb E\left[e^{is^TX}\right] = \phi_{s^TX}(1) = \phi_{s^TY}(1) = \mathbb E\left[e^{is^TX}\right] = \phi_{Y}(s)\,.
$$

According to Billingsley, at the time of the print of the second edition, "This result...seems to require Fourier methods - no elementary proof is known". I am aware of Walther's 1997 proof which does not use Fourier methods. Whether it is "elementary" is another matter entirely :).
