Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

6.3. Maximum Likelihood Tests 385

i.e.,V={v:v=a 1 ,for somea∈R}. Then in vector notation we can write the
model as
X=μ+e, μ∈V. (6.3.26)

Then we can summarize the model by saying, “Except for the random error vector
e,Xwould reside inV.” Hence, it makes sense intuitively to estimateμby a vector
inVthat is “closest” toX. That is, given a norm‖·‖inRn,choose

μ̂=Argmin‖X−v‖, v∈V. (6.3.27)

(a)If the error pdf is the Laplace, (2.2.4), show that the minimization in (6.3.27)

is equivalent to maximizing the likelihood when the norm is thel 1 norm given

by

‖v‖ 1 =

∑n

i=1

|vi|. (6.3.28)

(b)If the error pdf is theN(0,1), show that the minimization in (6.3.27) is equiv-

alent to maximizing the likelihood when the norm is given by the square of

thel 2 norm

‖v‖^22 =

∑n

i=1

vi^2. (6.3.29)

6.3.15. Continuing with Exercise 6.3.14, besides estimation there is also a nice
geometric interpretation for testing. For the model (6.3.26), consider the hypotheses

H 0 :θ=θ 0 versusH 1 : θ =θ 0 , (6.3.30)

whereθ 0 is specified. Given a norm‖·‖onRn,denotebyd(X,V) the distance
betweenXand the subspaceV; i.e.,d(X,V)=‖X−μ̂‖,whereμ̂is defined in
equation (6.3.27). IfH 0 is true, thenμ̂should be close toμ=θ 01 and, hence,
‖X−θ 01 ‖should be close tod(X,V). Denote the difference by

RD=‖X−θ 01 ‖−‖X−̂μ‖. (6.3.31)

Small values ofRDindicate that the null hypothesis is true, while large values
indicateH 1. So our rejection rule when usingRDis

RejectH 0 in favor ofH 1 ifRD > c. (6.3.32)

(a)If the error pdf is the Laplace, (6.1.6), show that expression (6.3.31) is equiv-

alent to the likelihood ratio test when the norm is given by (6.3.28).

(b)If the error pdf is theN(0,1), show that expression (6.3.31) is equivalent to

the likelihood ratio test when the norm is given by the square of thel 2 norm,

(6.3.29).

6.3.16.LetX 1 ,X 2 ,...,Xnbe a random sample from a distribution with pmf

p(x;θ)=θx(1−θ)^1 −x,x=0,1, where 0<θ<1. We wish to testH 0 : θ=1/ 3

versusH 1 :θ =1/3.