text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
)
1 +
1
12n
(cid:6) (cid:5)
1 +
(cid:6)
0.3799
n
7/(24n) ≥ 1 +
e
1+
3n(1 − x2)
+
(1+)2
.
36n2(1 − x2)
√
At x = 0.55 the right side is less than 1 + 0.543/n, so Case I is completed since 0.543 ≤
1/12 + 0.3799 + 7/24.
CASE II. The remaining case is j < 2n. For any integer k, P (βn ≥ k) = 1−P (βn ≤ k−1). F... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
:= (j − 3)/2,
while for n even,
√
pni ≤ P ( nY /2 ∈ Ik ) with i = 2k, 1 ≤ k ≤ K := (j − 2)/2,
and
√
pn0/2 ≤ P ( nY /2 ∈ I0).
6
(1.21)
(1.22)
(1.23)
In either case,
I0 ∪ I1 ∪ · · · ∪ IK ⊂ [0, n(1 − 1 − j/n)].
(cid:4)
The intervals will be defined by
δk+1 := (k + 1)/n + k(k + 1/2)(k + 1)/n3/2 , k ≥ 0,
∆k+... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
K ≤ j/2 − 1 < n/2 − 1 and
1 − xj ). Since j < 2n, we have
√
(cid:13)
(cid:13)
δK+1 ≤ (K + 1)/n + K(K + 1/2)/(n 2) ≤ xj /2 + nxj /(4 2).
√
√
2
We have ∆K+1 = nxj /2 − 1/2 + δK+1. It will be shown next that
√
1 − 1 − x ≥ x/2 + x /8, 0 ≤ x ≤ 1.
2
(1.29)
The functions and their first derivatives agree at 0 while t... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
2).
2
.
(1.30)
(1.31)
It is easy to verify that for 0 ≤ t < 1,
g(t) = (1 + t) log(1 + t) + (1 − t) log(1 − t) =
∞
(cid:1)
2r
t /r(2r − 1).
r=1
Thus the left side of (1.31) can be expanded as
A =
. We have
(cid:15) n−1
r=2 and B =
(cid:15)
r≥n
(cid:15)
r≥2 x2r (1 − n/(2r − 1))/2r = A + B where
d2A/dx2 =... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
+ 7/288n 2 − (n − 1)i2 /2n ].
2
(1.32)
Lemma 1.6. For any 0 ≤ a < b and a standard normal variable Y ,
P (Y ∈ [a, b]) ≥
(cid:4)
1/2π(b − a) exp[−a /4 − b2/4]φ(a, b)
2
(1.33)
where φ(a, b) := [4/(b2 − a2)] sinh[(b2 − a2)/4] ≥ 1.
Proof. Since the Taylor series of sinh around 0 has all coefficients positive, and (si... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
�
Lemma 1.6 is proved.
For the intervals Ik , Lemma 1.6 yields
√
P ( nY /2 ∈ Ik) ≥
(cid:4)
2/πnφk exp[−(∆2 + ∆2
k+1
k)/n + log(∆k+1 − ∆k )]
(1.35)
where φk := φ(2∆k / n, 2∆k+1/ n). The aim is to show that the ratio of the bounds (1.35)
over (1.32) is at least 1.
√
√
First consider the case k = 0. If n is eve... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
exp[−(1 + 1/n)2/n + log(1 + 1/n)].
Using log(1 + u) ≥ u − u2/2 again, the desired inequality can be checked since n ≥ 5. This
completes the case k = 0.
Now suppose k ≥ 1. In this case, i < 2n − 2 implies n ≥ 10 for n even and n ≥ 13 for n
√
odd. Let sk := δk + δk+1 and dk := δk+1 − δk . Then for i as in the definit... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
≤ x ≤ α for each of the pairs (α, λ) = (0.207, 0.9),
(0.195, 0.913), (0.14, 0.93), (0.04, 0.98).
Proof. Since x (cid:10)→ log(1 + x) is concave, or equivalently we are proving 1 + x ≥ eλx where the
latter function is convex, it suffices to check the inequalities at the endpoints, where they hold.
�
Lemma 1.7 and (1.... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
k ≤ kn
d(n) := 1/n + 3k2
:=
n/2 − 3/2 for n odd. Let
n/n3/2 and t := 1/ n. It will be shown that d(n) is decreasing in n,
n/2 − 1 for n even or kn
√
:=
√
(cid:13)
(cid:13)
√
9
separately for n even and odd. For n even we would like to show that 3t/2 + (1 − 3 2)t2 + 3t3
is increasing for 0 ≤ t ≤ 1/ 20 and ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
So Lemma 1.8 is proved.
It will next be shown that for n ≥ 10
sk ≤ n
−1 + k/ n.
√
(1.43)
√
(cid:13)
By (1.38) this is equivalent to 2/ n + (2k2 + 1)/n ≤ 1. Since k ≤ n/2 − 1 one can check
that (1.43) holds for n ≥ 14. For n = 10, 11, 12, 13 note that k is an integer, in fact k ≤ 1, and
(1.43) holds.
After some... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
the
exact value (1.39) of d in the d/n and λd terms. We assemble together terms with factors k2 ,
k and no factor of k, getting a lower bound A for (1.44) of the form
A := α[k2/n3/2] − 2β[k/n 5/4 ] + γ[1/n]
(1.45)
where, if n is even, so i = 2k and λ = 0.9, we get
α = 0.7 − [2.5 − 2(0.98)/3]/ n − 3/n,
√
β = n
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
1/2n
,
2
γ = 0.679 − (3.625 + 7/288 − 0.98/6)/n − 1/2n .
For the same reasons, the supremum of β2 − αγ for n ≥ 25 is now attained at n = 25 and is
negative (less than -0.015), so the conclusion (1.44) again holds.
5/4
It remains to consider the case k = 1 where n is even and n ≥ 10 or n is odd and n ≥ 13.
Here in... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
−n−3/2
√
√
(cid:13)
Proof of (1.6). For n odd, (1.6) is clear when j = 1, so we can assume j ≥ 3. For n even,
(1.6) is clear when j = 2. We next consider the case j = 0. By symmetry we need to prove
that pn0 ≤ P ( n|Y |/2 ≤ 1). This can be checked from a normal table for n = 2. For n ≥ 4
√
we have pn0 ≤
2/πn by (1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
. 175. Feller does not give a proof. For completeness, here is one:
(cid:14) ∞
(cid:14) ∞
ψ(t) = −
ψ(cid:4)(x)dx =
φ(x)(1 − 3x
−4)dx ≤ P (Y ≥ t).
t
To prove (1.6) via (1.46) for j = n ≥ 5 we need to prove
t
1/2n
≤ φ(tn)t−1
n
(1 − t−2)
n
√
where tn := (n − 2)/ n. Clearly n (cid:10)→ tn is increasing. For n ≥ ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
1
− +
4n
7
288n2
≤ −
−
+
where 3/n ≤ x ≤ 1 − 2/n. Note that 2n(1 − x2) ≥ 4. Thus it is enough to prove that
x − x 2/2 − x /4 ≥ 3/4n + 7/288n
2
2n
11
for 3/n ≤ x ≤ 1 and n ≥ 5, which holds since the function on the left is concave, and the
�
inequality holds at the endpoints. Thus (1.6) and Lemma 1.4 are pr... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
(cid:3)
√
√
Y ≤ − + 2kn = G( 2kn − n/2).
n
2
η := Φ−1(G(Z)).
n
(1.48)
This definition of η from Z is called a quantile transformation. By Theorem 1.3, G(Z) has a
U [0, 1] distribution and η a B(n, 1/2) distribution. It will be shown that
Z − 1 ≤ η ≤ Z + (Z − n/2)2 /2n + 1 if Z ≤ n/2,
(1.49)
and
Z − (Z − n/2)2/... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
(x) ≤ φ(x)/x for x > 0, e.g. Dudley (1993), Lemma 12.1.6(a). So we have
P (Z ≤ −n/2) = P
(cid:2)
n
2
+
√
n
2
Y ≤ −
(cid:3)
n
2
(cid:2) √
n
2
P
(cid:3)
Y ≤ −n = Φ(−2 n) ≤ √
√
e−2n
2 2πn
=
<
1
2n .
So G(−n/2) < G(c0) = 2−n and −n/2 < c0. Thus if Z ≤ −n/2 then η = 0. Next note that
Z + (Z − n/2)2/... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
for 1 ≤ k ≤ n/2
(1.51)
η = k ≤ (Z + n/2)2/2n + 1 = Z + (Z − n/2)2 /2n + 1.
Since P (Z ≤ n/2) = 1/2 ≤ P (η ≤ n/2), and η is a non-decreasing function of Z, Z ≤ n/2
implies η ≤ n/2. So (1.49) is proved.
It will be shown next that (η, Z) has the same joint distribution as (n − η, n − Z). It is
clear that η and n − η... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
Z| ≤ 1 + (Z − n/2)2/2n
and since Z < n/2 implies η ≤ n/2 and Z > n/2 implies η ≥ n/2,
|η − n/2| ≤ 1 + |Z − n/2|.
(1.52)
(1.53)
√
Letting Z = (n + nY )/2 and noting that then G(Z) ≡ Φ(Y ), (1.48), (1.52), and (1.53)
imply Lemma 1.2 with Cn = η − n/2.
�
1.6
Inequalities for the separate processes
We will need f... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
m
(cid:5) (cid:6)(cid:6)
x
h
1 − p m
and
P (ξ ≤ m − x) ≤ exp(−x /[2p(1 − p)]).
2
13
.
(1.55)
(1.56)
(1.57)
Proof. The first inequality in (1.55) is clear. Let E(k, n, p) denote the probability of at least
k successes in n independent trials with probability p of success on each trial, and B(k, n, p)
the prob... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
(t) := n(Fn(t) − t), 0 ≤ t ≤ 1, the corresponding empirical process. The previous lemma
extends via martingales to a bound for the empirical process on intervals.
Lemma 1.10. For any b with 0 < b ≤ 1/2 and x > 0,
(cid:5)
|αn(t)| > x/ n) ≤ 2 exp −
√
P
( sup
0
≤t≤b
nb
h
1 − b
(cid:6)(cid:6)
(cid:5)
x(1 − b)
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
:16)
(cid:16)Fn(u), u ≤ s =
1 − t
(cid:6)
E
Fn(s) − s
,
1 − s
in other words, the process (Fn(t) − t)/(1 − t), 0 ≤ t < 1 is a martingale in t (here n is
fixed). Thus, αn(t)/(1 − t), 0 ≤ t < 1, is also a martingale, and for any real s the process
exp(sαn(t)/(1 − t)) is a submartingale, e.g. Dudley (1993), 10.3.3(... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
1 − b)/(2nb))).
√
0≤t≤b
(cid:6)(cid:6)
(1.59)
It is easy to check that h(u) ≤ u2/2 for u ≥ 0, so the first inequality in Lemma 1.10 follows.
It is easily shown by derivatives that h(qy) ≥ q2h(y) for y ≥ 0 and 0 ≤ q ≤ 1. For q = 1 − b,
�
the bound in (1.58) then follows.
We next have a corresponding inequality f... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
− α. It will be shown that for any real α and y
P { sup X(t) − αt > y|X(1)} = 1{β>y} + exp(−2y(y − β))1{β≤y} .
0≤t≤1
(1.62)
Clearly, if β > y then sup0≤t≤1 X(t) − αt > y (let t = 1). Suppose β ≤ y. One can apply
a reflection argument as in the proof of Dudley (1993), Proposition 12.3.3, where details are
given on ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
B as W (t) − tW (1), 0 ≤ t ≤ 1, for a Wiener process W .
Let W1(t) := b−1/2W (bt), 0 ≤ t < ∞. Then W1 is a Wiener process. Let η := W (1) − W (b).
Then η has a normal N (0, 1√− b) distribution and is independent of W1(t), 0 ≤ t ≤ 1. Let
γ := ((1 − b)W1(1) − bη) b/x. We have
√
P ( sup B(t) > x|η, W1(1)) = P
0≤t≤b ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
x γ/b)1{γ≤1}
.
(cid:17)
2
2
0≤t≤b
From the definition of γ it has a N (0, b(1 − b)/x2) distribution. Since x is constant, the latter
integral with respect to γ can be evaluated by completing the square in the exponent and
yields (1.60).
We next need the inequality, for x ≥ 0,
1 − Φ(x) ≤
exp(−x /2).
2
1
2
(1.... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
) ≤ D exp(−2x
2),
0≤t≤1
(1.65)
16
which is the Dvoretzky-Kiefer-Wolfowitz inequality with a constant D. Massart (1990) proved
(1.65) with the sharp constant D = 2. Earlier Hu (1985) proved it with D = 4 2. D = 6
suffices for present purposes. Given D, it follows that for u ≥ n/6,
√
P ( n sup |αn(t)| ≥ u) ≤ D ex... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
2ν. Let Z be
Let N be the largest integer such that 2N ≤ n, so that ν
a ν-dimensional normal random variable with independent components, each having mean 0
:= {i + 1, · · · , m}. For any
and variance λ
two vectors a := (a1, · · · , aν ) and b := (b1, · · · , bν ) in Rν , we have the usual inner product
For any s... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
j, k.
Let Wj,k := (Z, ej,k ) and W (cid:4)
j,k ). Then since the elements of E are orthogonal it
j,k
follows that the random variables W (cid:4)
for 1 ≤ j ≤ N, 0 ≤ k < 2N −j and WN,0 are independent
j,k
normal with
:= (Z, e(cid:4)
:= {e(cid:4)
j,k
(cid:4)
(cid:4)
j,k) = λ2j−2 , Var(WN,0) = λν.
(cid:4)
(ci... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
it has a U [0, 1] distribution. It is easy to verify successively for j = N, N − 1, · · · , 0
that the random vector {Uj,k, 0 ≤ k < 2N −j } has a multinomial distribution with parameters
17
n, 2j−N , · · · , 2j−N . Let X
multinomial distribution with parameters n, 1/ν, · · · , 1/ν.
The random vector X is equal in ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
2) and (1.73) which have the joint distribution of X and Z, by the Vorob’ev-Berkes-Philipp
theorem, see Berkes and Philippp (1979), Lemma A1. Then we define a Brownian bridge by
n(Fn(t) − t), 0 ≤ t ≤ 1. By
Bn(t) := Wn(t) − tWn(1) and the empirical process αn(t) :=
our choices, we then have
√
{n(Fn(j/ν) − j/ν)}ν = ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
, · · · , ν while
the increments of the processes over the intervals between the lattice points j/ν are also not
too large.
Let C := 0.29. Let M be the least integer such that
C(x + 6 log n)
≤ λ2M +1
.
(1.77)
Since n ≥ 204 (1.67) and λ < 2 this implies M ≥ 2. We have by definition of M and (1.66)
2M ≤ λ2M ≤ C(x ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
(x + 6 log n)/ n ,
0≤t≤1
√
(cid:3)
sup |Bn(t) − Bn(πM (t))| > 0.22(x + 6 log n)/ n ,
0≤t≤1
√
(cid:2)
P2 := P
(1.78)
(1.79)
and, recalling (1.74) and (1.75),
P3 := 2N −M max P
m∈A(M )
(cid:31)(cid:5)
|D(m) −
m
ν
D(ν)| > 0.5x + 9 log n ∩ Θ
,
(1.80)
(cid:6)
where A(M ) := {k2M : k = 1, 2, · · ·} ∩ A(0... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
h(−C (cid:4)) ≥ 0.575 (note that C (cid:4) has been chosen to make 2h(C (cid:4))
and h(−C (cid:4)) approximately equal). By definition of M (1.77), λ2M +1 ≥ C(x + 6 log n), and
0.575C > 1/6, so
(1.81)
Next, to bound P1 and P2. Let b := 2M −N −1 ≤ 1/2. Since αn(t) has stationary increments,
P (Θc) ≤ 2−M exp(−x/6).
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
:8)
C
6
(cid:5)
uh 0.28 ·
(cid:6)(cid:6)
2
C
by (1.83) and (1.82) and since 1 − b > 1 − C/6, so one can calculate
P1 ≤ 2N −M +2 −u/6
e
≤ 22−M λ−1 exp(−x/6).
(1.84)
The Brownian bridge also has stationary increments, so Lemma 1.11, (1.61) and (1.82) give
P2 ≤ 2N −M +2 exp(−(0.22u)2 /(2nb))
≤ 2N −M +2 exp(−... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
single subscript j, e.g.
(cid:4)
ej := ej,k(j). The following orthogonal expansion holds in E:
(cid:4)
1A(0,m) =
m
ν
(cid:1)
eN,0 +
M <j≤N
(cid:4) ,
cj ej
(1.87)
(cid:4)
(cid:4)
for j ≤ M since 2M
where 0 ≤ cj ≤ 1 for m < j ≤ N . To see this, note that 1A(0,m)
⊥ e
j,k
(cid:4)
⊥ e
is a divisor of m. Also... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
cid:15)v(cid:15)2
for k (cid:14)
(cid:15)
(cid:4)
We next have
ej = 2j−N eN,0 +
(−1)s(i,j,m)2j+1−i (cid:4)
ei
(cid:1)
(1.88)
i>j
where s(i, j, m) = 0 or 1 for each i, j, m so that the corresponding factors are ±1, the signs
being immaterial in what follows. Let ∆j := (D, e(cid:4)
j ). Then from (1.87),
(ci... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
.2, (1.4),
|U (cid:4) − Uj ξj /2| ≤ 1 + ξj
j
2/8.
(1.90)
(cid:4) = (X, e(cid:4)
(cid:4)
20
Let
Lj := |Wj
(cid:4) − Ujξj/2| = |ξj|| Uj − λ2j|/2
(cid:4)
(cid:4)
√
by definition of ξj. Thus
|∆j| ≤ Lj + 1 + ξj
2/8.
(1.91)
Then we have on Θ
√
(cid:4)
| Uj − λ2j| = |Uj − λ2j|/( λ2j + Uj) ≤
√
(cid:4)
|Uj√... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
Let
C4 :=
√
1 + C (cid:4)
4(1 − 2−1/2)
=
√ √
√
2 1 + C (cid:4)( 2 + 1)
,
4
and for each M let cM := 1/(4C4 2M/2 ). Then for any real number x, we have x(1−C42M/2x) ≤
cM . It follows that
(cid:1)
(cid:1)
Lj ≤
C3ξ2
j + cM C22−j/2
M <j≤N
M <j≤N
21
≤ C2cM 2−(M +1)/2 /(1 − 2−1/2) +
(cid:1)
M <j≤N
2
C3... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
/3 1 −
C3 +
(cid:5)
2
3
(cid:6)(cid:6)(M −N )/2
1
8
≤ eN/321.513(N −M ) ≤ 22N −1.5M .
Markov’s inequality and (1.89) then yield
P3(m)
≤ e −x/6 −322N −1.5M
n
.
Thus
P3 ≤ e −x/6 −323N −2.5M ≤ 2−2.5M −x/6
.
(1.96)
Collecting (1.81), (1.84), (1.85) and (1.96) we get that P0 ≤ (23−M λ−1 + 2−M + 2−2.5M )e
−x/6
. ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
L on H is linear, that is, for
any f, g ∈ H and constant c, L(cf + g) = cL(f ) + L(g) almost surely.
Proof. The variable L(cf + g) − cL(f ) − L(g) clearly has mean 0 and by a short calculation
�
one can show that its variance is also 0, so it is 0 almost surely.
The Wiener process (Brownian motion) is a Gaussian s... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
a Gaussian
stochastic process Bt defined for 0 ≤ t ≤ 1 with mean 0 and covariance EBtBu = t(1 − u) for
0 ≤ t ≤ u ≤ 1. Given a Wiener process Wt, it is easy to see that Bt = Wt − tW1 for 0 ≤ t ≤ 1
defines a Brownian bridge.
For j = 0, 1, 2, ..., and k = 1, ..., 2j let Ij,k be the open interval ((k − 1)/2j , k/2j ). Le... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
constant or if f (t) ≡ t, then use linearity of the
�
operation Wj,k on functions for each j and k.
Lemma 1.14. For any f :
[0, 1] (cid:10)→ R and r = 0, 1, ..., for 0 ≤ t ≤ 1
[f ]r (t) = f (0) + t[f (1) − f (0)] +
fj,kTj,k(t),
r−1 2j
(cid:1) (cid:1)
where the sum is defined as 0 for r = 0.
j=0 k=1
Proof. For ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
] is uniformly continuous:
Lemma 1.15. If f is continuous on [0, 1] then [f ]r converges to f uniformly as r → ∞.
It follows that for any f ∈ C[0, 1],
f (t) = f (0) + t[f (1) − f (0)] +
fj,kTj,k(t),
∞ 2j
(cid:1) (cid:1)
where the sum converges uniformly on [0, 1]. Thus, the sequence of functions
j=0 k=1
1, t, ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
Lemma 1.13.
Lemma 1.17. The random variables Wj,k(B.) for j = 0, 1, ... and k = 1, ..., 2j are independent
with distribution N (0, 2−j−2).
Proof. We have by the previous lemma
Wj,k(B.) = Wj,k(W.) = W(2k−1)/2j+1 −
#
W(k−1)/2j + Wk/2j
$
1
2
= L(1[0,(2k−1)/2j+1]) −
$
#
L(1[0,(k−1)/2j ]) + L(1[0,k/2j ])
1
2 ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
gj,k are orthogonal for different k since they are supported on
non-overlapping intervals Ij,k. If j (cid:14)= j(cid:4), say j(cid:4) < j, then gj,k is 0 outside of Ij,k, equal to 1/2
on the left half of it and −1/2 on the right half, while gj(cid:1) ,k(cid:1) is constant on the interval, so the
�
functions are orth... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
. However, the Schauder functions are indefinite integrals of constant multiples
of the orthogonal functions gj,k or equivalently constant multiples of Haar functions, and it
turns out that the indefinite integral fits well with the processes we are considering, as in the
above proof. In a sense, the Wiener process Wt ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
,1 has
law bin(n, 1/2) by Lemma 1.3(b). We will define empirical distribution functions Un for the
U [0, 1] distribution recursively over dyadic rationals, beginning with Un(0) = 0, Un(1) = 1,
and Un(1/2) = V1,1/n. These values have their correct distributions so far. Now given Vj−1,k
for some j ≥ 2 and all k = 1, .... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
which has all the properties
of an empirical distribution function for U [0, 1].
With the help of Lemma 1.2, one can show that the Schauder coefficients of the empirical
process αn := n1/2(Un − U ), where U is the U [0, 1] distribution function, are close to those
of B(n). Lemma 1.2 has to be applied not only for the... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
´ev´esz, P. (1981). Strong Approximations in Probability and Statistics.
org˝
Academic, New York.
Donsker, Monroe D. (1952). Justification and extension of Doob’s heuristic approach to
the Kolmogorov-Smirnov theorems. Ann. Math. Statist. 23, 277–281.
Dudley, Richard M. (1984). A Course on Empirical Processes. Ecole ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
Gebiete 32,
111–131.
Mason, D. M. (1998). Notes on the the KMT Brownian bridge approximation to the
uniform empirical process. Preprint.
Mason, D. M., and van Zwet, W. (1987). A refinement of the KMT inequality for the
uniform empirical process. Ann. Probab. 15, 871-884.
Massart, P. (1990). The tight constant in t... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
., and Watson, G. N. (1927). Modern Analysis, 4th ed., Cambridge Univ.
Press, Repr. 1962.
26 | https://ocw.mit.edu/courses/18-465-topics-in-statistics-nonparametrics-and-robustness-spring-2005/0a472bc75921bd9a7a12c37bb261d572_bretagn_massart.pdf |
15.093 Optimization Methods
Lecture 3: The Simplex Method
1 Outline
� Reduced Costs
� Optimality conditions
� Improving the cost
� Unboundness
� The Simplex algorithm
Slide 1
� The Simplex algorithm on degenerate problems
2 Matrix View
Slide 2
min c x
0
s:t: Ax � b
x � 0
x � (x
; x
)
x
basic variables
B
... | https://ocw.mit.edu/courses/15-093j-optimization-methods-fall-2009/0a506ec77f46e9366688254eb6026cb9_MIT15_093J_F09_lec03.pdf |
7
�1
�1
0
0
0
y � x + �d � (2 � �; 0; 2 + �; 0; �; 1 � �; 4 � �)
What happens
as � increases�
�
� � min
�
�
fi�1;:::;mjd
g
B(i)<0
d
i
�
�
�
�
x
B(i)
min �
; �
; �
� 1:
(�1)
(�1)
(�1)
2
1
4
l � 6 (A
exits the basis).
6
New solution
0
y � (1; 0; 3; 0; 1; 0; 3)
Slide 16
New
basis B � (A
; A
; A
; A
)
Slide 17
... | https://ocw.mit.edu/courses/15-093j-optimization-methods-fall-2009/0a506ec77f46e9366688254eb6026cb9_MIT15_093J_F09_lec03.pdf |
(i)
B(i)<0
Theorem
� B � fA
; A
g basis
B
;i�6
l
j
(i)
� y � x + � d is a BFS associated with basis B .
�
7 The Simplex Algorithm
1. Start with basis B � [A
; : : : ; A
]
B(1)
B(m)
and a BFS x.
2. Compute c
� c
� c
B A
j
j
j
B
0
�1
� If c
� 0; x optimal; stop.
j
� Else select j : c
� 0.
j
3. Compute u � �d � ... | https://ocw.mit.edu/courses/15-093j-optimization-methods-fall-2009/0a506ec77f46e9366688254eb6026cb9_MIT15_093J_F09_lec03.pdf |
7.2 Degenerate problems
�
� � can equal zero (why�) ) y � x, although B 6� B .
Slide 22
� Even if � � 0, there might be a tie
�
x
B(i)
min
)
1�i�m;u
>0
i
u
i
next BFS degenerate.
� Finite termination not guaranteed; cycling is possible.
7.3 Avoiding Cycling
Slide 23
� Cycling can be avoided by carefully selec... | https://ocw.mit.edu/courses/15-093j-optimization-methods-fall-2009/0a506ec77f46e9366688254eb6026cb9_MIT15_093J_F09_lec03.pdf |
3. The rational Cherednik algebra
3.1. Definition and examples. Above we have made essential use of the commutation
relations between operators x ∈ h∗, g ∈ G, and Da, a ∈ h. This makes it natural to consider
the algebra generated by these operators.
Definition 3.1. The rational Cherednik algebra associated to (G, h) i... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
of h∗. Then it is clear from the relations of Hc
spanned over C[�] by the elements
� that Hc
(3.1)
g
r
r
� �
mi
yi
ni .
xi
Thus it remains to show that the images of the elements (3.1) under the map φ, i.e. the
i=1
i=1
elements
g
r
�
Dyi (c, �)mi
r
�
ni .
xi
i=1
i=1
are linearly independent. But this fol... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
called the Dunkl operator embedding.
Example 3.4. 1. Let G = Z2, h = C. In this case c reduces to one parameter, and the
algebra Ht,c is generated by elements x, y, s with defining relations
s 2 = 1, sx = −xs, sy = −ys, [y, x] = t − 2cs.
14
2. Let G = Sn, h = Cn . In this case there is also only one complex parame... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
gebras). The maps ξ and ξt
are isomorphisms.
Proof. The statement is equivalent to the claim that the elements (3.1) are a basis of Ht,c,
�
which follows from the proof of Proposition 3.2.
Remark 3.6. 1. We have
H0,0 = CG � C[h ⊕ h∗] and H1,0 = CG � D(h).
2. For any λ ∈ C∗, the algebra Ht,c is naturally isomorph... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
G , e (CG � C[hreg × h∗]) e = C[hreg × h∗]G .
Therefore, the restriction gives the embeddings: Θ1,c : B1,c
C[h∗ × hreg]G . In particular, we have
Proposition 3.8. The spherical subalgebra B0,c is commutative and does not have zero
divisors. Also B0,c is finitely generated.
)G, and Θ0,c : B0,c �
�→ D(hreg
→
15
�
... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
the Calogero-Moser space of G, h with
parameter c.
3.4. The localization lemma. Let H loc = Ht,c[δ−1] be the localization of Ht,c as a module
over C[h] with respect to the discriminant δ (a polynomial vanishing to the first order on
each reflection plane). Define also Bloc = eH loc e.
t,c
t,c
t,c
Proposition 3.12. ... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
The Poisson structure on Mc is obtained by extension of the symplectic
Poisson structure on (hreg × h∗)/G.
h/G, such that βc
→
Example 3.14. Let W = Z2, h = C. Then B0,c = �x2, xp, p2 − c2/x2�. Let X := x2, Z := xp
and Y := p2 − c2/x2 . Then Z 2 − XY = c2 . So Mc is isomorphic to the quadric Z 2 − XY = c2
in the 3-... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
GMλ,
where
Mλ = {v ∈ M |∀p ∈ (Sh)G , ∃N s.t. (p − λ(p))N v = 0},
(notice that h∗/G = Specm(Sh)G).
Proposition 3.16. Mλ are H1,c-submodules.
Proof. Note first that we have an isomorphism µ : H1,c(G, h) ∼ H1,c
by xa
Suppose P = P (x1, . . . , xr) ∈ (Sh∗)G . Then we have
�→ ya, yb �→ −xb, g �→ g.
(G, h∗), which is ... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
(G, h)λ,
�
λ∈h∗/G
where Oc(G, h)λ is the subcategory of modules where (Sh)G acts with generalized eigenvalue
λ.
Proof. Directly from the definition and the proposition.
�
Note that Oc(G, h)λ is an abelian category closed under taking subquotients and exten
sions.
3.6. The grading element. Let
�
(3.2)
Proposi... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
i
i
2 , F =
xi
1 �
2
i
2 .
yi
Then
�
(i) h =
(ii) h, E, F form an sl2-triple.
i(xiyi + yixi)/2;
Proof. A direct calculation.
Theorem 3.20. Let M be a module over H1,c(G, h).
�
�
(i) If h acts locally nilpotently on M , then h acts locally finitely on M .
(ii) If M is finitely generated over Sh∗, then M ∈ O... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
/�U �.
·
·
·
(ii) We need to show that if h acts locally finitely on M , then h acts locally nilpotently
on M . Assume h acts locally finitely on M . Then M = ⊕β∈B M [β], where B ⊂ C. Since M
is finitely generated over Sh∗, B is a finite union of sets of the form z + Z≥0, z ∈ C. So Sh
�
must act locally nilpotently... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
τ , where Sh
acts on τ by λ. These modules are called the Whittaker modules.
Let τ be irreducible, and let hc(τ ) be the number given by the formula
hc(τ ) =
dim h � 2cs
−
2
1 − λs
s∈S
s|τ .
Then we see that h acts on τ ⊗ Smh∗ by the scalar hc(τ ) + m.
Definition 3.23. A vector v in an H1,c-module M is singular... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
�
which is impossible since by Proposition 3.24, β0 must equal hc(τ ) for some τ .
Corollary 3.26. Any M ∈ Oc(G, h)0 has finite length.
Proof. Directly from the proposition.
�
3.9. Characters. For M ∈ Oc(G, h)0, define the character of M as the following formal
series in t:
ch M (g, t) =
tβTr M [β](g) = Tr M (gth... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
of h, with eigenvalues β such that β − hc(τ ) > 0.
We define Lc(τ ) = Mc(τ )/Jc(τ ), which is an irreducible module.
Proposition 3.30. Any irreducible object of Oc(G, h)0 has the form Lc(τ ) for an unique τ .
Proof. Let L ∈ Oc(G, h)0 be irreducible, with lowest eigenspace of h containing an irreducible
G-module τ . ... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
(G, h∗)0.
Proof. Clearly, if L is irreducible, then so is L†. Then L† is generated by its lowest h
eigenspace over H1,c¯(G, h∗), hence over Sh∗. Thus, L† ∈ Oc¯(G, h∗)0. Now, let M ∈ Oc(G, h)0
be any object. Since M has finite length, so does M †. Moreover, M † has a finite filtration
with successive quotients of the f... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
is a symmetric form on Mc(τ ).
Proposition 3.34. The maximal proper submodule Jc(τ ) is the kernel of φ (or, equivalently,
of the contravariant form βc).
Proof. Let K be the kernel of the contravariant form. It suffices to show that Mc(τ )/K is
irreducible. We have a diagram:
= c, and τ ∼ τ ∗
via a
=
=
¯
Mc(G, h, ... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
), Mc(τ �)) =
0, since for any extension
0 Mc(τ �) N
→
→ → Mc(τ ) → 0,
�
by Proposition 3.24 we have a splitting Mc(τ ) N .
Remark 3.36. In fact, our argument shows that if Ext1(Mc(τ ), Mc(τ �)) �= 0, then hc(τ ) −
hc(τ �) ∈ N.
→
3.13. The matrix of multiplicities. For τ, σ ∈ IrrepG, write τ < σ if
Re hc(σ) − ... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
Compute the multiplicities nσ,τ or, equivalently, ch Lc(τ ) for all τ .
In general, this problem is open.
3.14. Example: the rank 1 case. Let G = Z/mZ and λ be an m-th primitive root of 1.
Then the algebra H1,c(G, h) is generated by x, y, s with relations
[y, x] = 1 − 2
cj sj , sxs−1 = λx, sys−1 = λ−1 y.
m−1
�
j... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
any n ≥ 1.
Remark 3.41. According to Remark 3.31, this proposition in fact describes all the irre
ducible lowest weight modules.
Example 3.42. Consider the case m = 2. The Mc(C) is irreducible unless c ∈ 1/2 + Z≥0.
If c = (2n + 1)/2 ∈ 1/2 + Z, n ≥ 0, then Lc(C) has dimension 2n + 1. A similar answer is
obtained fo... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
irreducible H1,c-module, i.e., A = Lc(C). By Proposition 3.19, A
is naturally a finite dimensional sl2-module (in particular, it integrates to the group SL2(C)).
Hence, by the representation theory of sl2, the top degree of A is 1-dimensional. Let φ ∈ A∗
denote a nonzero linear function on the top component. Let βc b... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
general.
22
�
Now consider the Frobenius property of Lc(C) for any G ⊂ GL(h).
Theorem 3.45. Let U ⊂ Mc(C) = C[h] be a G-subrepresentation of dimension l = dim h,
sitting in degree r, which consists of singular vectors. Let J = �U �. Assume that A = Mc/J
is finite dimensional. Then
(i) A is Frobenius.
(ii) A admi... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
module is rl . Consider the Koszul complex attached to this module. Since the module is
free, the Koszul complex is exact (i.e., it is a resolution of the zero fiber). At the level of
SU -modules, it looks exactly like we want in (3.45).
So we only need to show that the maps of the resolution are morphisms over H1,c.... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
paper [CE].
Let G = Sn, and h be its reflection representation. In this case the function c reduces
to one number. We will denote the rational Cherednik algebra H1,c(Sn) by Hc(n). It is
generated by x1, . . . , xn, y1, . . . , yn and CSn with the following relations:
�
yi = 0,
�
xi = 0,
[yi, xj] = −
23
1
n
+ c... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
The space spanned by fi is (n − 1)-dimensional, since
is the residue of an exact differential).
�
i
fi = 0 (this sum
Proof. This proposition can be proved by a straightforward computation. The functions fi
�
are a special case of Jack polynomials.
Let Ic be the submodule of Mc(C) generated by fi. Consider the Hc(... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
is finite dimensional. The rest follows from Theorem 3.45.
Proof of Theorem 3.49. The support of Vc is the zero set of Ic, i.e. the common zero set of
fi. Fix x1, . . . , xn ∈ C. Then fi(x1, . . . , xn) = 0 for all i iff
λifi = 0 for all λi, i.e.
�
n
�
n
�
(z − xj )
r
n
n
�
Res∞
i=1
�
dz = 0.
λi
z − xi
i=1
... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
(z) is a polynomial.
Proof. Let g(z) be a polynomial. Then
0 = Res∞d(g(z) a(z)) = Res∞(g�(z)a(z) + a�(z)g(z))dz,
·
and hence
∞
�
Res
�
�
g�(z) +
Let g(z) = z l
�
µj g(z) a(z)dz = 0.
z − yj
�
µj
z −
�
with highest coefficient l + p +
µj > −p). This means that for every l ≥ 0,
Res∞z l+p−1 a(z)dz is a linea... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
For c > 0, the above representations are the only irreducible finite di
mensional representations of H1,c(Sn). Namely, it is proved in [BEG] that the only finite
dimensional representations of H1,c(Sn) are multiples of Lc(C) for c = r/n, and of Lc(C−)
(where C− is the sign representation) for c = −r/n, where r is a po... | https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/0a639a83911449ff554a28b7772cb49f_MIT18_735F09_ch03.pdf |
Undamped DC motor system: complete response
ω(t) =
V0
Kv
³
1 − cos (ωnt)
.
ve(t) = Kvω(t).
´
Electro-mechanical
equations of motion
(time domain)
L
di
dt
+ Ri + Kvω = vs
J
dω
dt
+ bω = Kmi
Step—function source
vs(t) = V0u(t).
L = 0.1H, J = 2kg · m2,
Kv = 6V · sec, Km = 6N · m/A,
V0 = 30V.
i(t) =
J
Km
dω(t)
dt
.
vL(t) =... | https://ocw.mit.edu/courses/2-004-systems-modeling-and-control-ii-fall-2007/0ab1a19f48f33bc89a7e81f160663a8e_lecture08.pdf |
6.864: Lecture 10 (October 13th, 2005)
Tagging and History-Based Models
Overview
• The Tagging Problem
• Hidden Markov Model (HMM) taggers
• Log-linear taggers
• Log-linear models for parsing and other problems
Tagging Problems
• Mapping strings to Tagged Sequences
a b e e a f h j � a/C b/D e/C e/C a/D f/C h/D j... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
/NA as/NA
their/NA CEO/NA Alan/SP Mulally/CP announced/NA first/NA
quarter/NA results/NA ./NA
NA
SC
CC
SL
CL
. . .
= No entity
= Start Company
= Continue Company
= Start Location
= Continue Location
Extracting Glossary Entries from the Web
Input:
Images removed for copyright reasons.
Set of webpages from ... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
P Hugo/NNP
victims/NNS ,/, and/CC sending/VBG them/PRP to/TO San/NNP Francisco/NNP
instead/RB ./.
• From the training set, induce a function or “program” that
maps new sentences to their tag sequences.
Our Goal (continued)
• A test data sentence:
Influential members of the House Ways and Means Committee introduce... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
a determiner
• Sometimes these preferences are in conflict:
The trash can is in the garage
A Naive Approach
• Use a machine learning method to build a “classifier” that
maps each word individually to its tag
• A problem: does not take contextual constraints into account
Hidden Markov Models
• We have an input sen... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
assumption: each word only depends on underlying tag
P (wj |tj )
An Example
• S = the boy laughed
• T = DT NN VBD
P (T, S) = P (END|NN, VBD)×
P (DT|START, START)×
P (NN|START, DT)×
P (VBD|DT, NN)×
P (the|DT)×
P (boy|NN)×
P (laughed|VBD)
Why the Name?
P (T, S) = P (END|tn−1, tn)
P (tj | tj−2, tj−1) ×
P (wj... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
Dealing with Low-Frequency Words: An Example
[Bikel et. al 1999] An Algorithm that Learns What’s in a Name
Word class
Example
Intuition
twoDigitNum
fourDigitNum
containsDigitAndAlpha
containsDigitAndDash
containsDigitAndSlash
containsDigitAndComma
containsDigitAndPeriod
othernum
allCaps
capPeriod
firstWord
... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
iterbi Algorithm
• Question: how do we calculate the following?:
T � = argmaxT log P (T, S)
• Define n to be the length of the sentence
• Define a dynamic programming table
�[i, t−2, t−1] = maximum log probability of a tag sequence ending
in tags t−2, t−1 at position i
• Our goal is to calculate maxt−2,t−1�T �[n, t... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
all i, t,
t−2, t−1.
• n|T |2 entries in � to be filled in.
• O(T ) time to fill in one entry
(assuming O(1) time to look up Score(S, i, t, t−2, t−1))
• ∞ O(n|T |3) time
Pros and Cons
• Hidden markov model taggers are very simple to train
(compile counts from the training corpus)
• Perform relatively well (over 9... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
Independence assumptions
�
• We take t0 = t−1 = START
• Assumption: each tag only depends on previous two tags
P (tj |tj−1, tj−2, w1, . . . , wn)
An Example
Hispaniola/NNP
important/JJ base/??
its empire into the rest of the Western Hemisphere .
an/DT
became/VB
from which Spain expanded
quickly/RB
• There a... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
functions f : X × Y � {0, 1}).
• Say we have m features �k for k = 1 . . . m
∞ A feature vector �(x, y) ≤ Rm for any x ≤ X and y ≤ Y.
An Example (continued)
• X is the set of all possible histories of form ∈t−1, t−2, w[1:n], i→
• Y = {NN, NNS, Vt, Vi, IN, DT, . . . }
• We have m features �k : X × Y ≥ R for k = 1 ... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
(h, t) =
�104(h, t) =
�105(h, t) =
�106(h, t) =
�107(h, t) =
�
�
�
�
�
if ∈t−2, t−1, t→ = ∈DT, JJ, Vt→
1
0 otherwise
if ∈t−1, t→ = ∈JJ, Vt→
1
0 otherwise
if ∈t→ = ∈Vt→
1
0 otherwise
if previous word wi−1 = the and t = Vt
1
0 otherwise
if next word wi+1 = the and t = Vt
1
0 otherwise
The F... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
| x) for any x ≤ X
and y ≤ Y.
• A feature is a function f : X × Y ≥ R
(Often binary features or indicator functions f : X × Y � {0, 1}).
• Say we have m features �k for k = 1 . . . m
∞ A feature vector �(x, y) ≤ Rm for any x ≤ X and y ≤ Y.
• We also have a parameter vector W ≤ Rm
• We define
P (y | x, W) =
W·�(... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
finitions
• Base case:
�[0, �, �] =
�[0, t−2, t−1] =
log 1 = 0
log 0 = −∗ for all other t−2, t−1
here � is a special tag padding the beginning of the sentence.
• Recursive case: for i = 1 . . . n, for all t−2, t−1,
�[i, t−2, t−1] = max {�[i − 1, t, t−2] + Score(S, i, t, t−2, t−1)}
t�T �{�}
Backpointers allow us t... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
<answer>programs to work properly. They are as far as I know t
<answer>agreed upon by commercial comms software developers fo
<answer>
<answer> Pins 1, 4, and 8 must be connected together inside
<answer>is to avoid the well known serial port chip bugs. The
FAQ Segmentation: Line Features
begins-with-number
begins-with... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
P ( 2.6) | question)×
P (W hat | question)×
P (conf iguration | question)×
P (of | question)×
P (serial | question)×
. . .
• i.e. have a language model for each tag
FAQ Segmentation: McCallum et. al
• Second solution: first map each sentence to string of features:
<question>2.6) What configuration of serial cable sho... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
j, tj−2, tj−1)
�
Independence
assumptions
• Estimate P (tj
| w1 . . . wn, j, tj−2, tj−1) using log-linear
models
• Use the Viterbi algorithm to compute
argmaxT �T n log P (T | S)
A General Approach: (Conditional) History-Based Models
• We’ve shown how to define P (T
| S) where T is a tag
sequence
• How do we... | https://ocw.mit.edu/courses/6-864-advanced-natural-language-processing-fall-2005/0abc5d7ab6458ec55a14c9f7c300438b_lec10.pdf |
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