Page 385 - Contributed Paper Session (CPS) - Volume 2

CPS1891 Tzee-Ming Huang
the knot removed. Therefore, we can construct a test for testing the null
ℓ
hypothesis:
= is not needed (5)
0
ℓ

based on estimated difference between ( 1,1 , … , 1, ) and ( 2,1 , … , 2, ).
To construct a test for testing (5), we take > 0 and perform least square
estimation for ( 1,1 , … , 1, ) and ( 2,1 , … , 2, ) based on ( , )s such that

∈ ( − , ) and ( , )s such that ∈ ( , + ) respectively. Let
ℓ

ℓ
ℓ
ℓ

(̂ 1,1 , … , ̂ 1, ) and (̂ 2,1 , … , ̂ 2, ) be the least square estimators for
( 1,1 , … , 1, ) and ( 2,1 , … , 2, ) respectively. Let

∆= (̂ 1,1 , … , ̂ 1, ) − (̂ 2,1 , … , ̂ 2, ) ,

then the in (5) should be rejected if
0

1
−1
≝ ∆ Σ Δ (6)
̂ 2

is large, where ̂ is an estimator for given by

1
̂ = ( ) . median of {| − 2−1 |: = 1, … , },
2
√2 0.25

and 0.25 is the 0.75 quantile of the standard normal distribution. If we assume
that s are IID N(0, ) and
2

≤ min( − ℓ−1 , ℓ+1 − ), (7)
ℓ
ℓ

−1
2
2
then under the in (5), the distribution of ∆ Σ Δ/ is (), the chi-
0
squared distribution with degrees of freedom. Thus the distribution of the
statistic in (6) is approximately () under the in (5) if ̂ is close to .
2
0
Let be the test that rejects the in (5) at level if exceeds the (1 −
,, ℓ 0
2
) quantile of (), then is an approximate size test.
,, ℓ
To apply for knot selection, the following algorithm is proposed.
,, ℓ

Algorithm 1.
• For = 1, … , , take = 1/2 .

(i) For = 1, … , , conduct the test with = and = 2. If
0.05,, ℓ ℓ
the −value is less than 0.05, then is considered as a potential

knot.
(ii) Let be the set of potential knots from (i). Take = Ø.

(a) Let be the knot in with the smallest p-value when
conducting the test 0.05,,. Add to .

(b) Remove knots in T that are in the range [ − , + ].
(c) Repeat (a)(b) until is empty.
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