PROOF OFLEMMA3.1. From Lemma I.4 in [33], ψ exists and ψ (w)= e12w2 ∞
w
g(3)(s)e−12s2ds
= −e12w2 w
−∞g(3)(s)e−12s2ds.
From g(3)(s)= 3sf(s2)+ 2s3f(3)(s2), we get that (5.1) g(3)(x)≤ 3|x|f+ 2|x|3f(3). Hence, for w > 0
ψ (w)≤ 3fe12w2
∞ w
se−12s2ds+ 2f(3)e12w2
∞ w
s3e−12s2ds
= 3f+ 2w2+ 2 f(3). Similarly, for w < 0
ψ (w)≤ 3fe12w2
w
−∞(−s)e−12s2ds+ 2f(3)e12w2
w
−∞(−s)3e−12s2ds
= 3f+ 2w2+ 2 f(3),
proving the first bound. The second bound uses (3.6) to obtain xψ(x)= x2ψ (x)+ xg(3)(x)
and the bound then follows directly with (3.7) and (5.1). For the last bound, differ-entiate (3.6) to get
ψ(x)= xψ(x)+ ψ(x) + g(4)(x) and combine (3.7), (3.8) and (3.4).
PROOF OFLEMMA4.2. The derivatives of g are
∂g
∂sj(s)=1
2sjf(w),
∂2g
∂sj2(s)=1
2f(w)+ sj2f(w),
∂2g
∂sj∂sk
(s)= sjskf(w), j= k.
Note that we can write the multivariate normal Stein equation (4.7) as
Putting all the above together gives that AMVN(0,S)g(s)= wf(w)+1
where we used that 0 < θj <1 and npj ≥ 1 to obtain the first inequality. Similarly, EIj(1)ξj4≤ pj
ESj(1) 4+ 4ESj(1)3+ 6ESj(1) 2+ 4ESj(1)+ 1
< pj
4+ 4 · 43/4+ 6 + 4 + 1 <27pj
and
EIj(1)ξj6≤ pj
ESj(1) 6+ 6ESj(1)5+ 15ESj(1) 4+ 20ESj(1)3 + 15ESj(1) 2+ 6ESj(1)+ 1
< pj
42+ 6 · 425/6+ 15 · 4 + 20 · 43/4+ 15 + 6 + 1 <305pj, where we used Lemma4.1to bound the absolute moments of Sj(1). Also,
EIj(1)ξj≤ pj
ESj(1)+ 1 <2pj and
EIj(1)ξj3≤ pj
ESj(1)3+ 3ESj(1) 2+ 3ESj(1)+ 1
< pj
43/4+ 3 + 3 + 1 <14pj.
When j= k, it is clear that the same bounds still hold, as Ij(1)Ik(1)= 0. PROOF OFLEMMA 4.4. Since the covariance-matrix S is nonnegative defi-nite, the solution of the Stein equation (4.15) is well defined and is given by (see [24]):
ψi(s)= − ∞
0 Ehi
e−us+1− e−2uZ du.
By dominated convergence,
∂3ψi
∂sa∂sb∂sc
(s)= − ∞
0
e−3uE
∂3hi
∂sa∂sb∂sc
e−us+1− e−2uZ du,
and so
∂3ψi
∂sa∂sb∂sc
(s)≤ ∞
0
e−3uE ∂3hi
∂sa∂sb∂sc
e−us+1− e−2uZ du.
We now obtain bounds for the third-order partial derivatives of h1 and h2. By straightforward differentiation,
∂3h1
∂sa∂sb∂sc
(s)
= 2[δj a+ δj b+ δj c]f(3)(w)+ 4sj(sa+ sb+ sc) + sbscδj a+ sascδj b+ sasbδj c
f(4)(w)+ 8sjsasbscf(5)(w),
∂3h2
∂sa∂sb∂sc(s)
= 6δj aδj bδj cf(3)(w)
+ 12sj[saδj bδj c+ sbδj aδj c+ scδj aδj b]f(4)(w)
+ 4sj3(sa+ sb+ sc)+ 3sj2(sbscδj a+ sascδj b+ sasbδj c)f(5)(w) + 8sj3sasbscf(6)(w),
where δjj = 1 and δj k= 0 if j = k. We can bound these partial derivatives by using the inequalities δj k≤ 1 and nk=1|ak| ≤ 1nnk=1|ak|n for n≥ 1. Doing so yields the bounds
∂3h1
∂sa∂sb∂sc
(s)≤ 6f(3)+ 6f(4)sa2+ sb2+ sc2+ sj2 + 2f(5)sa4+ sb4+ sc4+ sj4 ,
∂3h2
∂sa∂sb∂sc
(s)≤ 6f(3)+ 6f(4)sa2+ sb2+ sc2+ 3sj2 +f(5)7ss4+ 7sb4+ 7sc4+ 27sj4
+4
3f(6)sa6+ sb6+ sc6+ 3sj6 .
We now use the inequality for the third-order partial derivative of h2(s)to bound the third-order partial derivatives of ψ2(s). The random vector Z∼ MVN(0, S) can be written as (Z1, . . . , Zm), where Zj ∼ N(0, 1 − pj) and Cov(Zj, Zk)=
−√pjpk for j = k. On applying the inequality |a1+ a2|r ≤ 2r−1(|a1|r+ |a2|r), where r≥ 1, we have
∂3ψi
∂sa∂sb∂sc
(s)
≤ ∞
0 e−3uE6f(3)+ 12f(4)e−2usa2+ sb2+ sc2+ 3sj2 +1− e−2u Z2a+ Zb2+ Zc2+ 3Z2j
+ 8f(5)e−4u7sa4+ 7sb4+ 7sc4+ 27sj4
+1− e−2u 27Z4a+ 7Zb4+ 7Z4c+ 27Z4j
+128
3 f(6)e−6usa6+ sb6+ sc6+ 3sj6 +1− e−2u 3Z6a+ Zb6+ Zc6+ 3Z6j
du
≤ f(3) 2 +12
5 f(4)4+ sa2+ sb2+ sc2+ 3sj2
+ 8 This completes the proof of inequality (4.18), and inequality (4.17) follows from a similar calculation.
PROOF OF THEOREM4.3. As was the case in the proof of Theorem4.2, we require a bound for the expressionEAm−1f (W ), which is equivalent to bounding EAMVN(0,S)g(S). By using Taylor expansions in a similar manner to that used in the proof of Theorem4.2, we have that
Eh(W) − χ(m2 −1)h=EAm−1f (W )=EAMVN(0,S)g(S)≤ |R1| + |R2|, for some θj ∈ (0, 1). Carrying out a calculation similar to the one used to obtain (4.11) yields
To complete the proof, we require bounds on the expectations on the right-hand side of (5.2). Now recall that g(s)=14f (w). By a straightforward differentiation,
∂3g
∂sj3(s)= 3sjf(w)+ 2sj2f(3)(w), and so, for any j , k, l,
(5.3) ∂3g
∂sj∂sk∂sl
(s)≤|sj| + |sk| + |sl| f+2 3
|sj|3+ |sk|3+ |sl|3 f(3). To bound the expectations, we also use the inequalities E|Ij(1)ξk| < 2pj and E|Ij(1)ξk3| < 14pj, j, k= 1, . . . , m from Lemma4.3. Using (5.3) and these in-equalities yields
EIj(1) ∂3g
∂sj∂sk∂sl
(ξ )≤ 3·2pjf+2·14pjf(3)= 2pj
3f+14f(3). Hence, we obtain the bound
Eh(W) − χ(m2 −1)h≤ 1
√n
3f+ 14f(3)
m j=1
√1pj + 2m m j=1
√pj
+ m j=1
m k=j
√pk+ 2 m j=1
m k=j
m l=1
√pjpkpl
(5.4)
≤ 6
√n
3f+ 14f(3)
m j=1
√1pj
.
Using inequality (2.11) to translate bounds for the derivatives of the solution f to bounds on the derivatives of the test function h completes the proof.
PROOF OFCOROLLARY4.2. Let α > 0, and for some fixed z > 0 define
hα(x)=
⎧⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎩
1, if x≤ z,
1− 2(x − z)2/α2, if z < x≤ z + α/2, 2x− (z + α) 2/α2, if z+ α/2 < x ≤ z + α,
0, if x≥ z + α.
Then hαexists and is Lipshitz continuous with hα = 1, hα = 2/α and hα = 4/α2. Let Yd be a χ(d)2 random variable, then by (4.5),
P(W ≤ z) − P(Ym−1≤ z)
≤ Ehα(W )− Ehα(Ym−1)+ Ehα(Ym−1)− P(Ym−1≤ z) (5.5)
≤ 12
√np∗
6 hα + 46hα+ 84hα+ P(z ≤ Ym−1≤ z + α)
= 12
√np∗
6+92
α +336 α2
+ P(z ≤ Yd≤ z + α).
Now, for d= 1 (which corresponds to m = 2), P(z ≤ Y1≤ z + α) = z+α
z
e−x/2
√2π xdx≤ α
0
√1
2π xdx=
2α
π .
For d≥ 2, the mode of Yd is given by d− 2. The density of Y2is clearly bounded by 12, and, for d≥ 3, the density of Yd can be bounded by
1
2d/2(d2)xd/2−1e−x/2≤ 1
2d/2(d2)(d− 2)d/2−1e−(d−2)/2≤ 1 2√
π(d− 2), where the last inequality follows from Stirling’s inequality (x + 1) ≥
√2π xx+1/2e−x, which holds for all x > 0. Therefore,
(5.6) P(z ≤ Ym−1≤ z + α) ≤
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
2α/π , if m= 2,
α/2, if m= 3,
α 2√
π(m− 3), if m≥ 4.
Bounds for m= 2, m = 3 and m ≥ 4 now follow on substituting inequality (5.6) into (5.5) and choosing an appropriate α. For m= 2, we take α = 52.75n−1/5; for m= 3, we choose α = 25.27n−1/6; and α= 30.58(m − 3)1/6n−1/6is taken when m≥ 4. We can obtain a lower bound similarly, which is the negative of the upper bound. The proof is now complete.
Acknowledgements. The problem was first brought to our attention by Persi Diaconis, and we are thankful for that. As so many papers, this paper would not exist without him. Larry Goldstein provided many fruitful discussions. We would also like to thank an anonymous referee for their report.
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