# Can we recover joint distribution from a continuous range of convolutions?

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?

– whuber
Jan 23, 2022 at 17:01
• @whuber, interesting! And this transformation seems to have some relevant applications: The Radon transform is widely applicable to tomography, the creation of an image from the projection data associated with cross-sectional scans of an object. Jan 23, 2022 at 17:39

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.