Stat 5101 Lecture Notes - School of Statistics

140 Stat 5101 (Geyer) Course NotesWhen a random vector X is degenerate, it is always possible in theory (notnecessarily in practice) to eliminate one of the variables. For example, if X isconcentrated on the hyperplane H defined by (5.22), then, since c is nonzero,it has at least one nonzero component, say c j . Then rewriting c ′ x = a with anexplicit sum we getn∑c i X i = a,i=1which can be solved for X j⎛⎞X j = 1 ⎜n∑⎟⎝a − c i X i ⎠c jThus we can eliminate X j and work with the remaining variables. If the randomvectorX ′ =(X 1 ,...,X j−1 ,X j+1 ,...,X n )of the remaining variables is non-degenerate, then it has a density. If X ′ is stilldegenerate, then there is another variable we can eliminate. Eventually, unlessX is a constant random vector, we get to some subset of variables that havea non-degenerate joint distribution and hence a density. Since the rest of thevariables are a function of this subset, that indirectly describes all the variables.i=1i≠jExample 5.1.5 (Example 5.1.3 Continued).In Example 5.1.3 we considered the random vector⎛ ⎞ ⎛ ⎞X 1 U − VX = ⎝X 2⎠ = ⎝V − W ⎠X 3 W − Uwhere U, V , and W are independent and identically distributed random variables.Now suppose they are independent standard normal.In Example 5.1.3 we saw that X was degenerate because X 1 + X 2 + X 3 =0with probability one. We can eliminate X 3 , sinceX 3 = −(X 1 + X 2 )and consider the distribution of the vector (X 1 ,X 2 ), which we will see (in Section5.2 below) has a non-degenerate multivariate normal distribution.5.1.10 Correlation MatricesIf X =(X 1 ,...,X n ) is a random vector having no constant components,that is, var(X i ) > 0 for all i, the correlation matrix of X is the n × n matrix Cwith elementscov(X i ,X j )c ij = √var(Xi ) var(X i ) = cor(X i,X j )