A Consistent Nonparametric Test for Causality in Quantile

Using (A.10) and (A.11-4), we have the following inequality;

Pr[∖Th^ml( J - JT)| > δ I sup∖Q,(x)-Qθ(x)≤C}

≤ Pr{ max { ∖Th^m2(J -J„,)| , ∖TW'∖Jt - Jtl)| } > δ I sup∖Q_β(x)-Q(x)∖≤CT}

, for all δ > 0 .                                                             (A.12)

Invoking Lemma 1 and condition A2(iii), we have

Pr{ sup∖ Q_θ(x)-Q_θ(x)∖≤C_t} → 1 as T → ∞ .                   (A.13)

By (A.8) and (A.9), as T → ∞, we have

Pr{ max { ∖Th^m/2(J_t - J_tu)∖ , ∖Th^m/2(J_t - J_tl)∖ } > δ } → 0, for all δ > 0.

(A.14)

Therefore, as T → ∞,

the L.H.S. of the inequality (A.12) - Pr {∖ Th^m/2(J_t - J_t )∖ > δ } → 0 and

the L.H.S. of the inequality (A.12) → 0 .

In summary, we have that if both Th^m/2(J_T-J_TU) =o_p(1) and Th^m/2(J_T-J_TU) =o_p(1),
then Th^m/2(^J_t - J_t ) = op (1).                                                         □

Step 3: Asymptotic equivalence.

In the remaining proof, we focus on showing that Th^m/2(J_T-J_TU) =o_p(1), with the proof
of Th^m/2(J_T-J_TL) =o_p(1) being treated similarly. The proof of Step 3 is close in lines
with the proof in Zheng (1998). Denote

H_T(s,t,g)≡K_ts{1(y_t≤g(x_t))-,}{1(y_s≤g(x_s))-,} and             (A.15)

Then we have J_T ≡ J[Q_,] and J_TU ≡ J[Q_, -C_T]. We decompose H_T (s,t, g) into
three parts;

H_T(s,t,g)=K_ts{1(y_t≤g(x_t))-F(g(x_t)∖z_t)}{1(y_s≤g(x_s))-F(g(x_s)∖z_s)}

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