Stat 5101 Lecture Notes - School of Statistics

5.1. RANDOM VECTORS 137Example 5.1.3 (A Degenerate Random Vector).Suppose U, V , and W are independent and identically distributed random variableshaving a distribution not concentrated at one point, so σ 2 = var(U) =var(V ) = var(W ) is strictly positive. Consider the random vector⎛X = ⎝ U − V ⎞V − W ⎠ (5.23)W − UBecause of the assumed independence of U, V , and W , the diagonal elementsof var(X) are all equal tovar(U − V ) = var(U) + var(V )=2σ 2and the off-diagonal elements are all equal tocov(U − V,V − W) =cov(U, V ) − cov(U, W ) − var(V )+cov(V,W) =−σ 2Thus⎛⎞2 −1 −1var(X) =σ 2 ⎝−1 2 −1⎠−1 −1 2Question Is X degenerate or non-degenerate? If degenerate, what hyperplaneor hyperplanes is it concentrated on?Answer We give two different ways of finding this out. The first uses somemathematical cleverness, the second brute force and ignorance (also called plugand chug).The first way starts with the observation that each of the variables U, V ,and W occurs twice in the components of X, once with each sign, so the sumof the components of X is zero, that is X 1 + X 2 + X 3 = 0 with probability one.But if we introduce the vector⎛ ⎞1u = ⎝1⎠1we see that X 1 + X 2 + X 3 = u ′ X. Hence X is concentrated on the hyperplanedefined byH = { x ∈ R 3 : u ′ x =0}or if you preferH = { (x 1 ,x 2 ,x 3 )∈R 3 :x 1 +x 2 +x 3 =0}.Thus we see that X is indeed degenerate (concentrated on H). Is is concentratedon any other hyperplanes? The answer is no, but our cleverness has run out.It’s hard so show that there are no more except by the brute force approach.