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Assume that we have a random vector $\mathbf{Y}=(Y_1,...,Y_n)$ with unknown joint distribution. We know however that the conditional distribution of an entry $Y_i$ given the rest of the vector, without the entry $Y_i$ ($Y_{-i}$) is normal: $$ Y_i|Y_{-i}\sim N(\mu+\beta Y_kY_j,\sigma^2) $$ with $k,j\neq i$ standing in some relation with $i$. For example $k=i-1, j=k+1$ and $\mu$ some constant. Is it possible to find the joint distribution function? I.e. if for example we would have $$ Y_i|Y_{-i}\sim N(\mu+\beta Y_k,\sigma^2) $$ then the joint probability function would be a multivariate normal (with some resitriction on the parameter space (i.e. $\beta<1$), but what about the case above? I do understand that the joint distribution cannot be normal but isn't it possible that the joint distribution is just another multivariate distribution?

A short Gibbs sampling code in R shows that it is at least plausible that such a distribution exists as the sampler does not diverge (depending on the parameter space):

Y<-1
Z<-1
col<-c()
for (i in 1:10000){
X<-rnorm(1,1+0.1*Y*Z,1)
Y<-rnorm(1,1+0.1*X*Z,1)
Z<-rnorm(1,1+0.1*X*Y,1)
col<-cbind(col,c(X,Y,Z))
}

scatterplot3d::scatterplot3d(x=col[1,],y=col[2,],z=col[3,])
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To determine whether or not conditional distributions are compatible with one another, an approach is to apply the Hammersley-Clifford Theorem:

Under the condition that the supports of the full conditionals $g_{j}$ are the entire sets $\mathcal{Y}_j$, the joint distribution $g$ satisfies $$ g(y_1,\ldots,y_p) \propto \prod_{j=1}^p \; {g_{\ell_j}(y_{\ell_j}|y_{\ell_1}, \ldots,y_{\ell_{j-1}},y_{\ell_{j+1}}^\prime,\ldots,y_{\ell_p}^\prime) \over g_{\ell_j}(y_{\ell_j}^\prime|y_{\ell_1},\ldots,y_{\ell_{j-1}}, y_{\ell_{j+1}}^\prime,\ldots,y_{\ell_p}^\prime)} $$ for every permutation $\ell$ on $\{1,2,\ldots,p\}$ and every $y,y' \in \mathcal{Y}=\prod_{j=1}^p \mathcal{Y}_j$.

In the current setting, it however is impossible to find a joint distribution: the products $y_i y_j y_k$ appearing in the joint mean that the joint cannot be Normal. But since all conditionals are Normal, there is no other joint distribution possible.

Take as an example \begin{align*}X|Y,Z&\sim\mathcal{N}(\beta YZ,1)\\ Y|X,Z&\sim\mathcal{N}(\beta XZ,1)\\ Z|Y,X&\sim\mathcal{N}(\beta XY,1) \end{align*} and apply Hammersley-Clifford's formula with $x'=y'=z'=0$. The joint does not depend on the choice of $(x',y',z')$ and this choice makes a lot of terms vanish. The log joint density should be (up to an additive constant) $$-\{x^2+y^2+(z-\beta xy)^2-\beta^2 x^2 y^2\}/2$$ since all other terms cancel. Which means that the joint density is proportional to $$\exp(-\{x^2+y^2+(z-\beta xy)^2-\beta^2 x^2 y^2\}/2)$$ If I integrate this density in $z$, the term $(z-\beta xy)^2$ vanishes and integrate to a constant in $(x,y)$. I am thus left with $$\exp(-\{x^2+y^2-\beta^2 x^2 y^2\}/2)$$ which does not integrate in $x$ conditional on $y$ when $y^2>\beta^{-2}$. Therefore there cannot exist a joint probability density.

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  • $\begingroup$ I think I am understanding your comment as well as the hint on the HC-Theorem. But the most important point for me is the last sentence. Is it true that normal conditionals must imply a joint normal? Clearly with the setting above the conditionals cannot imply a joint normal but why is it impossible that they are forming another joint distribution? $\endgroup$ – Michael L. Nov 16 '17 at 15:00
  • $\begingroup$ As I am pretty sure that gaussian conditionals can end up to form a non-gaussian joint distribution. I have updated my question. $\endgroup$ – Michael L. Nov 17 '17 at 8:04
  • $\begingroup$ Start your Gibbs sampler at $X=Y=Z=9$ and then you will watch it diverge very quickly! $\endgroup$ – Xi'an Nov 17 '17 at 8:13
  • $\begingroup$ Yes thats correct but only depending on the parameter space. I.e. with beta=-0.1 it will not diverge. Again I do not say that for all combinations of parameters a joint distribution exists but only for some cases. Furthermore I am still not convinced that gaussian conditionals must end up in a joint normal. $\endgroup$ – Michael L. Nov 17 '17 at 8:33
  • $\begingroup$ With other starting values, Gibbs also diverges for $\beta=-1$. See my update explaining why there cannot be a joint. $\endgroup$ – Xi'an Nov 17 '17 at 8:56

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