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Given a correct model of the joint distribution of a random vector $(X,Y)$, we can derive (though not necessarily in closed form) the correct distribution model for $pX+(1-p)Y$ for any $p\in[0,1]$.
What about the reverse? Is having correct models for $pX+(1-p)Y$ for any $p\in[0,1]$ sufficient for recovering the joint distribution of a random vector $(X,Y)$? If not, (1) what would be a counterexample and (2) what are some examples of additional conditions needed for that?
I am not sure I understand the question, because I don't understand the relation with convolutions. If we assume that $X$ and $Y$ are independent, then $pX+(1-p)Y$ is indeed a convolution, but what if there are not assumed independent? If $X$ and $Y$ are assumed independent, then taking $p=1$ and $p=0$ allows us to recover the law of $X$ and $Y$, and therefore of $(X,Y)$.
If the law of $(X,Y)$ is uniquely determined by its moments (for example if the law of $(X,Y)$ has compact support), then from the knowledge of the law of $pX+(1-p)Y$ for each $p \in [0,1]$, we can recover all the moments $\mathbb{E}[X^nY^m]$ (since the map $p \mapsto \mathbb{E}[(pX+(1-p)Y)^{n+m}]$ is polynomial, and $\mathbb{E}[X^nY^m]$ can be recovered from the coefficients of said polynomial) and we can therefore recover the law of $(X,Y)$.
Following the remark of @whuber, we can also try to recover the characteristic function of $(X,Y)$. Indeed, we have that $\phi_{pX+(1-p)Y}(t) = \phi_{(X,Y)}(tp, t(1-p))$. If we only know the law of $pX +(1-p)Y$ for $p \in [0,1]$, we can recover half of the values (that is, on all pairs $(\lambda_1,\lambda_2)$ that have the same sign) of the characteristic function. If we know the law of $pX + (1-p)Y$ for all $p \in \mathbb{R}$, then we can recover the characteristic function of $(X,Y)$ on all the plane but one line (the line $\{(\lambda, - \lambda) \ \vert \ \lambda \in \mathbb{R}\}$), but by continuity, we can deduce the values of the characteristic function on the missing line.
