Suppose that $(X_n,Y_n)$ is a sequence of random variables with $X_n \geq 0$ and $Y_n \geq 0$ for all $n$. Also, suppose that $a X_n + b Y_n$ converges in distribution for every $a \geq 0$ and $b \geq 0$. Does this imply that $(X_n,Y_n)$ converges in distribution?
The result is obviously true if the weak convergence of $a X_n + b Y_n$ holds for all $a$ and $b$. However, we are given this information only for all non-negative $a$ and $b$, but note that our random variables in question are also non-negative.
I am curious to know if the result is true, and if so, why.
The result is true. Since $\{X_n\}$ and $\{Y_n\}$ are tight so is the $\mathbb R^{2}$-valued sequence $\{(X_n,Y_n)\}$. Hence, we only have to show that any two sub-sequential limits are the same. So let $\mu$ and $\nu$ be sub-sequential limits. These are Borel probability measures on the locally compact space $S=[0,\infty) \times [0,\infty)$. The hypothesis implies that $\int e^{-(ax+by)} d\mu (x,y)=\int e^{-(ax+by)} d\nu (x,y)$ for $a,b \geq 0$. Consider the collection of all finite linear combinations of functions of the type $(x,y) \to e^{-(ax+by)}$ with $a,b > 0$. This is an algebra contained in the space $C_0 (S)$ of all continuous functions on $S$ vanishing at $\infty$ with the supremum norm. It separates points. By the (locally compact version of ) Stone - Weierstrass Theorem it is dense in $C_0 (S)$. It follows that $\int f d\mu =\int f d\nu$ for any $f \in C_0 (S)$. Hence $\mu =\nu$.