Stat 5101 Lecture Notes - School of Statistics

156 Stat 5101 (Geyer) Course Notesminimum value, and where, since it is a smooth function, its derivative must bezero, that is,⎛ ⎞∂q(c)k∑=2p i c i −2p i⎝ p j c j⎠ =0∂c iNow we do not know what the quantity in parentheses is, but it does not dependon i or j, so we can write it as a single letter d with no subscripts. Thus wehave to solve2p i c i − 2dp i = 0 (5.38)for c i . This splits into two cases.Case I. If none of the p i are zero, the only solution is c i = d. Thus the onlynull eigenvectors are proportional to the vector u =(1,1,...,1). And all suchvectors determine the same hyperplane.Case II. If any of the p i are zero, we get more solutions. Equation (5.38)becomes 0 = 0 when p i = 0, and since this is the only equation containing c i ,the equations say nothing about c i , thus the solution isj=1c i = d, p i > 0c i = arbitrary, p i =0In hindsight, case II was rather obvious too. If p i = 0 then X i = 0 withprobability one, and that is another degeneracy. But our real interest is incase I. If none of the success probabilities are zero, then the only degeneracy isY 1 + ···+Y k =n with probability one.5.4.4 DensityDensity? Don’t degenerate distribution have no densities? In the continuouscase, yes. Degenerate continuous random vectors have no densities. But discreterandom vectors always have densities, as always, f(x) =P(X=x).The derivation of the density is exactly like the derivation of the binomialdensity. First we look at one particular outcome, then collect the outcomes thatlead to the same Y values. Write X i,j for the components of X i , and note thatif we know X i,m = 1, then we also know X i,j = 0 for j ≠ m, so it is enough todetermine the probability of an outcome if we simply record the X ij that areequal to one. Then by the multiplication ruleP (X 1,j1 = 1 and ···and X n,jn =1)===n∏P (X i,ji =1)i=1n∏i=1k∏j=1p jip yjj