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Dependence in dimension 2 is difficult. But one has to admit that dimension 2 is way more simple than dimension 3 ! I recently rediscovered a nice paper, Langford, Schwertman &#38; Owens (2001), on transitivity of the property of being positively correlated (which inspired the odd title of this post). And more recently, Castro Sotos, Vanhoof, Van Den Noortgate &#38; Onghena (2001) conducted a study which confirmed that there are strong misconceptions of correlation (and I guess, not only because probabilistic reasoning is extremely weak, as mentioned in Stock &#38; Gross (1989)) and association, or correlation (as already stated in Estapa &#38; Bataneor (1996), or Batanero, Estepa, Godino and Green (1996)). My understanding is that is it possible to have almost anything... even counterintuitive results. For instance, if we want to mix independence and comonotonicity (i.e. perfect positive dependence), all the theorems you might think of should probably be incorrect. Consider the following result (based on some old examples I have been using in my courses 5 or 6 years ago, see e.g. here)"If X and Y are comontonic, and if Y and Z are comonotonic, then X and Z are comonotonic"Well, this result seems to be intuitive, and probably valid. But it is not. Consider the following triplet,Projections on bivariate planes of the three dimensional vector areHere, X and Y are comonotonic, so are Y and Z, but X and Z are independent... Weird, isn't it ? Another one ?"If X and Y are comontonic, and if Y and Z are independent, then X and Z are independent"Again, even if it is intuitive, it is not correct... Consider for instance the following 3 dimensional distribution,Here, X and Y are comonotonic, while Y and Z are independent, but X here and Z are countercomonotonic (perfect negative dependence). It is also possible to consider the following distribution,that can be visualized below,In that case, X and Y are comonotonic, while Y and Z are independent, but X here and Z are comonotonic (perfect positive dependence). So obviously, we should be able to construct any kind of counterexample, on any kind of result we might think as intuitive.To be honest, the problem with intuition is that is usually comes from the Gaussian case, and from the perception that dependence is related to correlation. Pearson's linear correlation. Consider the case of a 3 dimensional random vector, with correlation matrixGiven two pairs of correlations,  and , what could we say about  ? For instance, the intuition is that if  and  are positive, then  is likely to be positive too (perhaps). The only property (at least the most important) we have on that correlation matrix is that it should be positive-semidefinite. So if we play on eigenvalues, it should be possible to derive inequalities satisfied by  .Langford, Schwertman &#38; Owens (2001) claim (in Theorem 3) that correlations have to satisfy some property, likewhich is simply the fact that the determinant of the correlation matrix has to be positive, that property was already mentioned in Kendall (1948), as an exercise,But is that a sufficient and necessary condition ? Since I am extremely lazy, let us run some numerical calculation to visualize possible values for , as function of  and . Consider the following codeU=seq(-1,1,by=.1)V=seq(-1,1,by=.001)FSUP=function(a,b){DF=function(c){min(eigen(matrix(c(1,a,b,a,1,c,b,c,1),3,3))$values)};V[max(which(Vectorize(DF)(V)&#62;0))]}FINF=function(a,b){DF=function(c){min(eigen(matrix(c(1,a,b,a,1,c,b,c,1),3,3))$values)};V[min(which(Vectorize(DF)(V)&#62;0))]}MSUP=outer(U,U,Vectorize(FSUP))MINF=outer(U,U,Vectorize(FINF))library(RColorBrewer)clr=rev(brewer.pal(6, "RdBu"))U=U[2:20]MSUP=MSUP[2:20,2:20]MINF=MINF[2:20,2:20]persp(U,U,MSUP,col="green",shade=TRUE)image(U,U,MSUP,breaks=((-3):3)/3,col=clr)persp(U,U,MINF,col="green",shade=TRUE)image(U,U,MINF,breaks=((-3):3)/3,col=clr)Here, we can derive the lower and the upper bound for , as function of  and .In the dark blue area, the bound for the correlation can be really low, while in the dark red, the bound is very high (either the lower bound on the left, or the upper bound on the right). Since it might be hard to read, it is possible to fix for instance , and to derive bonds for , as function of .V=seq(-1,1,by=.001)U=seq(-1,1,by=.1)U=U[2:(length(U)-1)]V=V[2:(length(V)-1)]U=c(-.9999,U,.9999)V=c(-.99999,V,.99999)FSUP=function(a){DF=function(c){min(eigen(matrix(c(1,a,-.7,a,1,c,-.7,c,1),3,3))$values)};V[max(which(Vectorize(DF)(V)&#62;0))]}FINF=function(a){DF=function(c){min(eigen(matrix(c(1,a,-.7,a,1,c,-.7,c,1),3,3))$values)};V[min(which(Vectorize(DF)(V)&#62;0))]}VS=Vectorize(FSUP)(U)VI=Vectorize(FINF)(U)plot(c(U,U),c(VS,VI),col="white")polygon(c(U,rev(U)),c(VS,rev(VI)),col="yellow",border=NA)lines(U,VS,lwd=2,col="red")lines(U,VI,lwd=2,col="red")On the graph below, we have bound for a negative correlation for  (on the left, with -0.7) and a positive correlation for  (on the right, here +0.7),We do observe here extremely nice ellipses... Consider the case of a null correlation  then the region for possible values for  and  is the unit circle.The interpretation is that if  is null, and so is  then  might take any value between -1 and 1 (under the assumption that marginal distribution allow such values, e.g. marginal Gaussian distributions). On the other hand if  is either -1 or +1 (perfect negative/positive correlation) then  has to be null...