可以通过2D边际重建3D联合分布吗？

假设我们知道p（x，y），p（x，z）和p（y，z），那么联合分布p（x，y，z）是否可识别是真的吗？即，只有一个可能的p（x，y，z）的边际以上？

distributions mathematical-statistics

— 用户名
source

相关：是否可能有一对联合分布不是高斯的高斯随机变量？（这涉及2D联合与1D边缘，但答案和直觉最终是相同的，再加上@Cardinal的答案中的图片很漂亮。）

— gung-恢复莫妮卡

@gung关系有点遥远。这个问题背后的微妙之处在于，系动词向我们展示了如何在给定边际的情况下发展双变量分布。但是，如果我们为三变量分布指定三个双变量边际，则在该三变量分布上必须有相当严格的附加约束：单变量边际必须一致。那么问题是这些约束条件是否足以确定三变量分布。这使得它本质上是一个二维以上的问题。

— ub

@whuber，我了解您在说2D边际比1D边际更具约束力，这是合理的。我的观点是，在两个答案中，边际都不能充分约束联合分布，并且那里的枢机主教的答案使问题很容易看出。如果您认为这太让人分心，可以删除这些评论。

— gung-恢复莫妮卡

@gung我要说的是完全不同的东西，而且很难看到（除非您非常擅长3D可视化）。您还记得霍夫施塔特的戈德尔，埃舍尔，巴赫的封面吗？（可以通过Googling轻松找到它；也许我会扩大答案以包括它。）在坐标轴上具有相同投影集的这两种不同实体的存在是相当令人惊讶的。这捕获了3D对象的完整正交2D“视图”集合不一定确定对象的想法。这就是问题的症结所在。

— ub

@Gung让我再尝试一次。是的，两种情况都普遍认为边际不能完全确定分布。这个问题的复杂性（我认为一个与另一个是如此不同）是，当前情况下的边际绝不是独立的：每个2D边际确定两个1D边际，以及这些边际之间的牢固关系边缘的。从概念上讲，这个问题可能会被重述为“为什么在确定完整3D分布的意义上，二维边际中的依存关系不是“传递性”还是“累积性”？

— ub

Answers:

号也许最简单的反关注三个独立的分布的变量，为此，从所有八个可能的结果通过是等可能的。这使得在所有四个边缘分布均匀 $\text{Bernoulli}(1/2)$ $X_i$ $(0,0,0)$ $(1,1,1)$ 。 $\{(0,0),(0,1),(1,0),(1,1)\}$

考虑随机变量，由在组均匀分布。这些边距与 $(Y_1,Y_2,Y_3)$ $\{(1,0,0),(0,1,0), (0,0,1),(1,1,1)\}$ 。 $(X_1,X_2,X_3)$

道格拉斯·霍夫施塔特（Douglas Hofstadter）的戈德尔（Escher），巴赫（Bach）的封面暗示了可能性。

这些实体中的每一个在坐标平面上的三个正交投影（阴影）都相同，但是这些实体明显不同。尽管阴影与边缘分布不完全相同，但是阴影的作用方式非常相似，以限制但不能完全确定投射阴影的3D对象。

— ub
source

Y 1, Y 2, Y 3

$Y1,Y2,Y3$

本着求助者的精神，

$U, V, W$

\begin{aligned} (1) & f_{U, V, W} (u, v, w) = {\begin{cases} 2 ϕ (u) ϕ (v) ϕ (w) & if u \geq 0, v \geq 0, w \geq 0, \\ or if u < 0, v < 0, w \geq 0, \\ or if u < 0, v \geq 0, w < 0, \\ or if u \geq 0, v < 0, w < 0, \\ 0 & otherwise \end{cases} \end{aligned}

$\begin{align} f_{U,V,W}(u,v,w) = \begin{cases} 2\phi(u)\phi(v)\phi(w) & ~~~~\text{if}~ u \geq 0, v\geq 0, w \geq 0,\\ & \text{or if}~ u < 0, v < 0, w \geq 0,\\ & \text{or if}~ u < 0, v\geq 0, w < 0,\\ & \text{or if}~ u \geq 0, v< 0, w < 0,\\ & \\ 0 & \text{otherwise} \end{cases}\tag{1} \end{align}$ where

ϕ (\cdot)

$\phi(\cdot)$ denotes the standard normal density function.

It is clear that $U, V$ , and $W$ are dependent random variables. It is also clear that they are not jointly normal random variables. However, all three pairs $(U,V), (U,W), (V,W)$ are pairwise independent random variables: in fact, independent standard normal random variables (and thus pairwise jointly normal random variables). In short, $U,V,W$ are an example of pairwise independent but not mutually independent standard normal random variables. See this answer of mine for more details.

In contrast, if $X,Y,Z$ are mutually independent standard normal random variables, then they are also pairwise independent random variables but their joint density is

\begin{matrix} (2) & f_{X, Y, Z} (u, v, w) = ϕ (u) ϕ (v) ϕ (w), u, v, w \in R \end{matrix}

$f_{X,Y,Z}(u,v,w) = \phi(u)\phi(v)\phi(w), ~~u,v,w \in \mathbb R \tag{2}$ which is not the same as the joint density in

(1)

$(1)$ . So, NO, we cannot deduce the trivariate joint pdf from the bivariate pdfs even in the case when the marginal univariate distributions are standard normal and the random variables are pairwise independent.

— Dilip Sarwate
source

You're basically asking if CAT reconstruction is possible using only images along the 3 main axes.

It is not... otherwise that's what they would do. :-) See the Radon transform for more literature.

— user541686
source

I like the analogy. Two aspects are troubling, though. One is the logic: just because the Radon transform (or some other technique) uses more data than the three marginals does not logically imply it really needs all those data. Another problem is that CT scans are inherently two-dimensional: they reconstruct a solid body slice by slice. (It's true that the Radon transform is defined in three and higher dimensions.) Thus they don't really get to the heart of the matter: we already know the univariate marginals aren't enough to reconstruct a 2D distribution.

— whuber

@whuber: I think you misunderstood what I was saying... and the 2D vs 3D is a red herring. I was trying to say that the inverse of the Radon transform requires the full integral for its inversion (i.e. if you literally just look at the inversion formula, you see the inversion requires an integral over all angles, not a sum over d angles). The CAT scan was just to help the OP see it's the same problem as CT.

— user541686

That's where the logic breaks down: it's not the same problem as the CT. Your argument sounds like an analog of "every vehicle I see on the road uses at least four wheels. Therefore ground transportation with fewer than four wheels is impossible, for if it were possible, then people would be using fewer wheels to save tire costs. If you doubt this, just look at the blueprints for a car." Incidentally, the transform as implemented in a CT scanner does not integrate over all angles--the measure of the set of angles it uses is zero!

— whuber

@whuber: Forget the CT thing for a moment. Do you agree with the rest of the logic?

— user541686 '18