STAT 3690 Lecture 11
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STAT 3690 Lecture 11
2022
Testing on Au (J&W pp. 279)
● A is of q x p and rk(A) = q , i.e., AΣAT > 0
● Model: iid AXi ~ MVNq (Au, AΣAT ).
● LRT for H0 : Au = p0 v.s. H1 : Au p0
– Test statistic: n(A _ p0 )T (ASAT )- 1 (A _ p0 ) ~ T2 (q, n _ 1) = F (q, n _ q) under H0
– Rejction region at level α : R = {z1 , . . . , zn : (A _ p0 )T (ASAT )- 1 (A _ p0 ) > F1 -α,q,n -q }
– p-value: p(z1 , . . . , zn ) = 1 _ FF (q,n -q){ (A _ p0 )T (ASAT )- 1 (A _ p0 )}
● Multiple comparison
– Interested in H0 : µ 1 = . . . = µp v.s. H1 : Not all entries of u are equal.
* µk : the kth entry of u
– p = 2 (i.e., A = [1, _1]): A/B testing
(1 _ Ⅲ ) x 100% confidence region (CR) for u (J&W Sec. 5.4)
● Pr((1 _ α) x 100%CR covers u) = 1 _ α
– CR is a set made of observations and is hence random
– u is fixed
– (1 _ α) x 100% CR covers u with probability (1 _ α) x 100%
● Dual problem of testing H0 : u = u0 v.s. H1 : u u0 at the α level
– Translated from rejection region. Steps:
1. Take R as a function of u0 ;
2. Replace u0 with u;
3. Take the complement.
– (1 _ α) x 100% CR = {u : n( _ u)T Σ- 1 ( _ u) < χ1(2) -α,p } if Σ is known
– (1 _ α) x 100% CR = {u : ( _ u)T S- 1 ( _ u) < F1 -α,p,n -p } if Σ is not known
(1 _ Ⅲ ) x 100% CR for p = Au
● X1 , . . . , Xn MVNp (u, Σ)
– Unknown Σ
– n > p
● A is of q x p and rk(A) = q , i.e., AΣAT > 0
● Then iid AXi ~ MVNq (p , AΣAT )
● (1 _ α) x 100% CR for p is {p : (A _ p)T (ASAT )- 1 (A _ p) < F1-α,q,n -q }
● Special case: A = a e 毖p
– (1 _ α) x 100% confidence interval (CI) for scalar ν = aT u is
{ν : n(aT _ν)2 (aT Sa)- 1 < F1-α,1,n - 1 } = │aT _t1-α/2,n - 1 |aT Sa/n, aT +t1-α/2,n - 1 |aT Sa/n\
Simultaneous confidence intervals
● Construct (1 _ αk ) CI for scalars a k(T)u, say CIk , k = 1, . . . , m, simultaneously
● Make sure Pr(|k {ak(T)u e CIk }) > 1 _ α
● Bonferroni correction
– Bonferroni inequality:
m m m m
Pr( | {ak(T)u e CIk }) = 1 _ Pr( | {ak(T)u CIk }) > 1 _ L Pr(ak(T)u CIk ) = 1 _ L αk
k=1 k=1 k=1 k=1
– Taking αk such that α = | αk , e.g., αk = α/m, i.e.,
(ak(T) _ t1-α/(2m),n - 1 |a k(T)Sak /n, a k(T) + t1-α/(2m),n - 1 |a k(T)Sak /n)
– Working for small m
● Scheffé’s method
– Let CI出 = (wT _ c |wT Sw/n, wT + c |wT Sw/n) for all w e 毖p
– Derive that c = |p(n _ 1)(n _ p)- 1 F1 -α,p,n -p
* By Cauchy–Schwarz: {wT ( _ u)}2 = [(S1/2w)T {S- 1/2( _ u)}]2 s {(wT Sw)T /n}{n( _ u)T S- 1 ( _ u)} ÷
m
Pr( | {ak(T)u e CIk }) > Pr( | {wT u e CI出 }) = 1 _ Pr( | {wT u CI出 })
k=1 出 egψ 出 egψ
= 1 _ Pr( | [{wT ( _ u)}2 /{(wT Sw)T /n} > c2])
出 egψ
> 1 _ Pr({n( _ u)T S- 1 ( _ u) > c2 })
* Pr({n( _ u)T S- 1 ( _ u) > c2 }) = α ÷ c = |p(n _ 1)(n _ p)- 1 F1 -α,p,n -p
– Working for large even infinite m
Comparing two multivariate means (J&W Sec. 6.3)
● Two independent samples of (potentially) different sizes from two distributions with equal covariance
– X11 , . . . , X1n1 MVNp (u1 , Σ)
– X21 , . . . , X2n2 MVNp (u2 , Σ)
● Let i and Si be the sample mean and sample covariance for the ith sample
● Hypotheses H0 : u1 = u2 v.s. H1 : u1 u2
● Test statistic following LRT
(1 _ 2 )T {(n1(-)1 + n2(-)1 )Spool }- 1 (1 _ 2 ) ~ F (p, n1 + n2 _ p _ 1)
● Rejection region at level α
,x11 , . . . , x1n1 , x21 , . . . , x2n2 : (1 _2 )T {(n1(-)1 +n2(-)1 )Spool }- 1 (1 _2 ) > F1-α,p,n 1 +n2 -p - 1 、
– Spool
● p-value
1 _ FF1 − 玄」ψ」念 1+念2 −ψ − 1 ┌ (1 _ 2 )T {(n1(-)1 + n2(-)1 )Spool }- 1 (1 _ 2 )┐
2022-03-17