Suppose x1 is a single observation from a distribution with density f over [0,1]. It is desired to test H0:f(x)=1 against H1:f(x)=2x.
Let δ:[0,1]→{0,1} define a test by δ(x1)=i⟺ 'accept Hi '. Let αi(δ)=P(δ(x1)=1−i∣Hi). State the Neyman-Pearson lemma using this notation.
Let δ be the best test of size 0.10. Find δ and α1(δ).
Consider now δ:[0,1]→{0,1,⋆} where δ(x1)=⋆ means 'declare the test to be inconclusive'. Let γi(δ)=P(δ(x)=⋆∣Hi). Given prior probabilities π0 for H0 and π1=1−π0 for H1, and some w0,w1, let
cost(δ)=π0(w0α0(δ)+γ0(δ))+π1(w1α1(δ)+γ1(δ))
Let δ∗(x1)=i⟺x1∈Ai, where A0=[0,0.5),A⋆=[0.5,0.6),A1=[0.6,1]. Prove that for each value of π0∈(0,1) there exist w0,w1 (depending on π0) such that cost(δ∗)=minδcost(δ).[ Hint :w0=1+2(0.6)(π1/π0).]
Hence prove that if δ is any test for which
αi(δ)⩽αi(δ∗),i=0,1
then γ0(δ)⩾γ0(δ∗) and γ1(δ)⩾γ1(δ∗).