# 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?

• This essentially asks how to invert the Radon transform.
– 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

## 1 Answer

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.

• Interesting. My instinct was No, since this is a bit like recovering the joint 2d distribution from a set of 1d marginal distributions. But this has persuaded me otherwise Mar 13 at 14:43
• This is a nice observation and it would be directly connected to the theory of the Radon transform if you were to reframe it in terms of the characteristic function instead of the moments, because the cf always exists.
– whuber
Mar 13 at 15:33