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The parameter S is the same as in Example 1. The parameter R satisfies (S1) with x $^{R}$( o ) = w and (S2) by an identity analogous to that used in Example 1. Condition (S3) is satisfied by D R $_{P}$( o ) ≜ y − E$_{P}$ ( Y | W = w ). By the bounds on Y , ( P,o ) ↦→ D R $_{P}$( o ) is uniformly bounded. Here, the remainder terms are both exactly zero: Rem R P ≡ Rem S P ≡ 0. Thus, we have that L RS P ≡ 0 in this example. 

The requirement that V ar$_{P}$$_{0}$ $^{[}$D R 0 ( O ) ] > 0 in 2, and more generally that there exist a nonzero eigenvalue λ$_{j}$ for the limit distribution in Theorem 4 to be non-degenerate, may at times present an obstacle to our goal of obtaining asymptotic control of the type I error. This is the case for Example 4, which we now discuss further. Nevertheless, we show that with a little finesse the type I error can still be controlled at the desired level for the given test. In fact, the test we discuss has type I error converging to zero, suggesting it may be noticeably conservative in small to moderate samples. 

## Example 4 (Continued). 

In this example, one can take x R = w and x S = w ( − k ). Furthermore, it is easy to show that 

<img src='content_image/297309.jpg'>

The first-order approximations for R and S are exact in this example as the remainder terms Rem R P and Rem S P are both zero. However, we note that if E$_{P}$ ( Y | W ) = E$_{P}$ ( Y | W ( − k )) almost surely, it follows that D R P ≡ D S $_{P}$. This implies that Γ$_{0}$ ≡ 0 almost surely under H$_{0}$ . 

As such, under the conditions of Theorem 4, all of the eigenvalues in the limit distribution of nψ$_{n}$ in Theorem 4 are zero and nψ$_{n}$ → 0 in probability. We are then no longer able to control the type I error at level α , rendering our proposed test invalid. 

