# Problem Set 6: Linear Hypotheses II

Hypothesis testing
Ec402
PS 6
Notes on the Wald Statistic
• Want to test hypotheses of the form 𝐻0 : 𝑅𝛽 = 𝑞 (with r restrictions
= number of rows in matrix R), based on our estimate 𝛽.
• Note:
𝐸 𝑅𝛽 = 𝑅𝛽
𝑉𝑎𝑟 𝑅 𝛽 = 𝐸
𝑅𝛽 − 𝑅𝛽 𝑅 𝛽 − 𝑅𝛽
𝑉𝑎𝑟 𝑅𝛽 = 𝐸 𝑅 𝛽 − 𝛽 𝛽 − 𝛽
= 𝑅𝑉𝑎𝑟 𝛽 𝑅′
𝑉𝑎𝑟 𝑅𝛽 = 𝜎 2 𝑅 𝑋 ′ 𝑋
−1 ′
𝑅
′ ′
𝑅
′
• With A5N this implies:
𝑅𝛽~𝑁(𝑅𝛽, 𝜎 2 𝑅 𝑋 ′ 𝑋
−1 𝑅 ′ )
𝑅𝛽 − 𝑅𝛽~𝑁(0, 𝜎 2 𝑅 𝑋 ′ 𝑋
−1 𝑅 ′ )
• And so under 𝐻0 : 𝑅𝛽 = 𝑞 we have:
𝑅𝛽 − 𝑞~𝑁(0, 𝜎 2 𝑅 𝑋 ′ 𝑋
−1 ′
𝑅 ) (1)
• If 𝜎 2 is known, use a 𝜒 2 distribution.
• Why?
• 𝜒 2 (𝑟) is the sum of r squared independent N(0,1)
variables by definition.
• From (1):
[𝜎 2 𝑅 𝑋 ′ 𝑋
−1 𝑅 ′ ]−1/2 (𝑅 𝛽
− 𝑞)~𝑁(0, 𝐼)
And so from the definition of 𝜒 2 (𝑟):
𝑅𝛽 − 𝑞
′
𝜎 2𝑅 𝑋′𝑋
−1 𝑅 ′ −1
𝑅𝛽 − 𝑞 ~𝜒 2 𝑟
(2)
• If 𝜎 2 is unknown, use an F distribution.
• Why?
• F distribution is defined in terms of 2 independent 𝜒 2
distributions. Let 𝑦1 and 𝑦2 be independently
distributed 𝜒 2 variables with 𝑛1 and 𝑛2 degrees of
freedom. Then:
𝑦1 /𝑛1
𝐹=
~𝐹(𝑛1 , 𝑛2 )
𝑦2 /𝑛2
• Can show that
𝜀′𝜀
2
~𝜒
(𝑁
2
𝜎
− 𝐾) and is independent of
𝛽 (see Johnston &amp; DiNardo p.495)
• And combining this result with (2) we get:
′
𝑅𝛽 − 𝑞
𝑅 𝑋 ′ 𝑋 −1 𝑅′ −1 𝑅𝛽 − 𝑞 / 𝑟
~𝐹(𝑟, 𝑁 − 𝐾)
𝜀′𝜀/(𝑁 − 𝐾)
(Note the 𝜎 2 s cancel)
• Since
𝜀′𝜀
𝑁−𝐾
𝑅𝛽 − 𝑞
′
= 𝑠 2 this gives us
𝑠2𝑅 𝑋′𝑋
−1 𝑅 ′ −1
𝑅 𝛽 − 𝑞 /𝑟 ~ 𝐹(𝑟, 𝑁 − 𝐾)
• Note that this Wald test is equal to the
Likelihood Ratio test given by:
𝐹=
(𝑅𝑆𝑆𝑈 −𝑅𝑆𝑆𝑅 )/𝑟
𝑅𝑆𝑆𝑈 /(𝑁−𝐾)
• See lecture notes for a proof