Nonparametric Test in First Price Auction with Independent
Private Value
Serafin Grundl and Yu Zhu
February 15, 2013
Abstract
We propose a nonparametric test for first-price auction with independent private value. We
show that the model restrictions can be reduced into a continuum of moment equalities. Then test
the model assumptions is equivalent to testing the moment equalities. This test is flexible and can
allow for unobserved heterogeneity. We derive the asymptotic distribution of the test and propose
a bootstrap procedure to obtain the critical value. We apply the procedure to test the assumption
that bidders know the valuation distribution on data from US Forest Service timber auction and
reject this assumption.
[Very Preliminary. Please Do Not Circulate]
1
Introduction
Testing in auction models can be challenging, because the bidders’ valuations are not observed. A
common approach to overcome this challenge is to first estimate the bidders’ valuations from bids,
and then construct a test based on the estimated valuations. There are two drawbacks of this approach.
First, it is unclear how to test restrictions on the whole valuation distribution because the valuations
are nonparametrically estimated. Typically, the restrictions can only be tested on a finite number of
quantiles of the valuation distribution.1 Second, the approach cannot be used in applications where the
1 There
are only a handful of papers using a nonparametric test in first price auctions.Marmer, Shneyerov, and Xu (2007)
develop a nonparametric test for different entry models. Haile, Hong, and Shum (2003) test the common value model
1
valuations cannot be estimated. One important example is the model with multiplicative unobserved
heterogeneity investigated by Krasnokutskaya (2011).
In this paper, we propose a different approach. Instead of first estimating valuations from bid data,
we start from the underlying valuation distribution. The bid distribution is a function of the underlying
valuation distribution, which depends on the model. If we know the theoretical bid distributions given
any primitive under a specific model, we can test the model restrictions by comparing theoretical bid
distributions with the bid distributions we observe. If there is no primitive generating a bid distribution
similar to the data, we reject the model.
To compare theoretical bid distributions with the data, we need to have a measure of closeness of
distributions. We choose the characteristic function (hereafter, c.f.) as it can be estimated easily from
data. One difficulty is that c.f. is usually very complicated function of the primitives and impossible
to compute analytically. Following Bierens and Song (2012), we compute c.f. using simulation.
We show that we can reduce the model restriction to a continuum of moment conditions using c.f..
Therefore testing the model restrictions is equivalent to test the moment conditions. We propose a test
based on these moment conditions following Santos (2012) and Hong (2011).
The test is specially useful in first price auction with independent private values, where the bid
function can be easily computed numerically. One attractive feature of this test is its flexibility. We can
easily test models with additive or multiplicative unobserved heterogeneity nonparametrically, which
is, to the best of our knowledge, not investigated before. It can also be generalized to accommodate
affiliated value models at the cost of impose a parametric form of affiliation.
The asymptotic distribution of the test is nonstandard. One methodological contribution of this
paper is to propose a new bootstrap strategy to obtain the critical value. More generally, this procedure
can be applied to testing nonlinear moment equalities with nonparametric functions as discussed in
Hong (2011). Different from the conservative procedure proposed in Hong (2011), our procedure
leads to asymptotically exact test.
We apply the test to the U.S. Forest Service timber auction data to test the assumption that the
bidders know the valuation distribution. The assumption that bidder know the valuation distribution is
against the private value model. Both papers build tests based on pseudovalues and therefore, can only use a finite number
of quantiles.
2
a common but strong assumption in the first-price auctions which has never been assessed empirically.
The assumption is testable by exploiting variation in the number of bidders participating in an auction.
In the application, we assume Exogenous Participation conditional on unobserved heterogeneity.
Exogenous Participation is a common assumption in the structural auction literature (e.g. Haile, Hong,
and Shum (2003), Haile and Tamer (2003), Gillen (2009), Guerre, Perrigne, and Vuong (2009) and
Aradillas-López, Gandhi, and Quint (2011a) ). Aradillas-López, Gandhi, and Quint (2011b) test and
fail to reject Exogenous Participation allowing for unobserved heterogeneity using ascending USFS
timber auctions.2
The rest of the paper is organized as following. In section 2, we show how we can reduce different
model restrictions into moment conditions involving c.f.. In section 3, we propose the general test,
derive the asymptotic distribution and discuss the bootstrap procedure. In section 4, we discuss how
to adopt the general test to test that bidders know the valuation distribution assuming Exogenous
Participation given unobserved heterogeneity. In section 5, we provide the testing result on U.S.
Forest Service timber auction. Section 6 concludes.
2
General Model Restrictions
Consider auctions where there are k bidders independently draw their values from a valuation distribution. Bidders submit a bid after they observe their valuation. The number of opponents is known
when bidders submit the bids. The bidder with highest bid win the auctioned good and pay his bid.
Let Bk = (B1k , B2k , ...Bkk ) be the observed random vector of all bids from one k-bidder auction.
Suppose we observe auctions with k bidders for k ∈ K ⊂ {2, 3...} and K has K elements. Let θ be the
primitives in the model and let B̃k (θ ) = B̃1k (θ ) , B̃2k (θ ) , ...B̃kk (θ ) be the random vector of all bids
generated by θ according to the model in k bidder auction. Now let φk : Rk × T −→ C to be some
complex valued function. We first illustrate that if we choose appropriate φk , the model restrictions
can be reduced to a continuum of moment conditions involving φk which are testable, e.g. the model
2 They
refer to the model as Conditional Independent Value.
3
is correctly specified only if there exists θ such that
Eφk (Bk ,t) − Eφk B̃k (θ ) ,t = 0, for all t ∈ T, k ∈ K.
(1)
Therefore, testing the validity of a specific model is reduced to testing the moment conditions (1).
Example 2.1 (Correct Belief and Exogenous Participation). Suppose bidders know the valuation distribution and the number of opponents. They play the symmetric Bayesian Nash Equilibrium. Now let
fk denote the density of valuation distribution in k-bidder auction. Exogenous Participation means that
the valuation distribution does not change across auctions, e.g. fk = f . Therefore, with Exogenous
Participation θ = f . Following Bierens and Song (2012), T = [−κ, κ] with κ being some positive real
number and φk (Bk ,t) = 1k ∑ki=1 exp (itBik ).
Example 2.2 (Ambiguous Beliefs and Exogenous Participation). Now suppose bidders do not know
the valuation distribution, They share a common prior set of distributions which they consider as
possible candidate for valuation distribution. The bidders with Maxmin Expected Utility(MEU) play
Bayesian Nash Equilibrium according to F min , the lower contour of all distributions in the common
prior set. 3 Under Exogenous Participation, θ = f , f min the valuation distribution and the belief and
T = [−κ, κ]. The model is correctly specified if there exists an θ such that (1) holds with B̃k (θ ) being
generated by θ where bidders play BNE given f min but their valuations are drawn from f . Grundl
and Zhu (2013) show that the model is just identified if auctions of two different number of bidders
are observed. Therefore, (1) have testing power only if K has more than 3 elements.
Example 2.3 (Multiplicative Unobserved Heterogeneity and Correct Belief). Krasnokutskaya (2011)
shows that auctions with multiplicative unobserved heterogeneity is identified. Suppose the auction
specific unobserved heterogeneity u with distribution fku . Notice the distributions of the unobserved
heterogeneity can differ for different number of bidders. Let T = [−κ, κ]2 and
φk (Bk ,t) =
3 We
1
exp it1 log Bik + it2 log B jk
∑
k (k − 1) i6= j
implicitly assume that F min is in the prior set.
4
where t = (t1 ,t2 ). If the model is correctly specified, there exists a θ = { fk }k∈I , fku k∈K such
that (1) holds. If in addition, Exogenous Participation conditional on the unobserved heterogeneity is
assumed, θ = ( f , f u ).
In above examples, φk are chosen such that Eφk (Bk ,t) corresponds to characteristic functions of
the bid distribution. It is well-known that there is one-to-one correspondence between characteristic
functions and the distribution functions. Therefore, the bid distribution identifies the model if and
only if corresponding characteristic function does. Bierens and Song (2011) use this observation to
generalize identification results in IPV first price auctions. Bierens and Song (2012) develop estimation procedure for standard IPV first price auction models and obtain consistent estimator of valuation
distribution. They also provide a specification test which is valid if the true valuation distribution has
certain parametric form. In next section, we propose a specification test using equation (1) which
allows for arbitrary distributions and partial identification.
3
Nonparametric Test
In this section, we introduce the test statistic that we use to test a continuum of moments with unknown
functions. But before that, we want to discuss the parameter space. The infinite dimensional parameter
involved here are the density functions. Therefore, we introduce the space of densities we work with.
3.1
Space of Densities
Let F be the collection of density functions supported on some bounded interval which is contained in
[0,C1 ] where C1 is a positive constant. h is an arbitrarily smooth density function supported on [0,C1 ]
with CDF H. Bierens (2008) finds that for any f ∈ F with distribution function F, F (x) = G [H (x)]
for some distribution function G with density g supported on [0, 1]. Notice that for each g such that
´1 2
´1
0 g (x) dx < ∞, there exists a function r (x) on [0, 1] such that 0 r (x) = 0 and
g (x) =
[1 + r (x)]2
.
´1
1 + 0 r2 (x) dx
5
√
To see this, one can verify r (x) =
g(x)
´1√
g(x)
0
− 1 satisfies all the requirements. Define
ˆ
R = r (x) |r has support lying in [0, 1]
ˆ
1
1
r (s) ds < ∞ .
2
r (s) ds = 0,
0
0
Therefore T : R → F
(Tr) (x) =
[1 + r (H (x))]2
h (x)
´1
1 + 0 r2 (x) dx
is surjective.4 Instead of considering the space of the density functions, we can work with the functions
defined on [0, 1]. It is well-known that any measurable function can be approximated by polynomials.
Bierens (2008) suggest using Legendre Polynomials defined on [0, 1] to approximate r. Let p j be the
j-th Legendre Polynomial and ϕ0 = 1 be the constant function. Therefore, r (x) = ∑∞j=1 β j p j (x) for
any r ∈ R.
In order to derive the asymptotics of the test, we have to control the bias of the approximation.
Now, let ∇λ r (x) = d λ r (x) /dxλ , 1 > α > 0 and R, C be positive integers. Define
R (C, R + α) = r|r ∈ R and ∇R r (x1 ) − ∇R r (x2 ) ≤ C |x1 − x2 |α ,
F (C,C1 , R + α) = {Tr|r ∈ R (C, R + α) , (Tr) (C1 ) = 1} .
0
Let pm (x) = (p1 (x) , p2 (x) , ..., pm (x)) and β m = (β1 , β2 , ..., βm ) ∈ Rm . Define k f k∞ = supx | f (x)|
where the sup is taken with respect to the domain of f and d∞ be the metric induced by sup norm. By
Lorentz (1966),
sup
min
d∞ (r, pm β m ) = O m−R .
m
m
r∈R(C,R+α) β ∈R
In addition, using Lemma A.1,
sup
min
d∞ ( f , T (pm β m )) = O m−R .
m
m
f ∈F (C,C1 ,R+α) β ∈R
Throughout the paper, we assume the the densities generating the data lies in F (C,C1 , R + α) so that
we can work with the space R (C, R + α) instead. Therefore it is useful to know how rich the space
4 Obviously,
for any r ∈ R, Tr is a density function.
6
F (C,C1 , R + α) is. Next lemma partly answers this question.
Lemma 3.1. Let f be a density function with support [0,C2 ] ⊂ [0,C1 ] and f (x) > 0 if x < C2 . There
exists a C > 0 such that f ∈ F (C,C1 , R + α) if
R
∇ f (x1 ) − ∇R f (x2 ) ≤ C3 |x1 − x2 |α
and ∃ε > 0, such that ∀y satisfying C2 ≥ y ≥ C2 − ε, f (y) = M (y)2 (C2 − y)2R+2α with M strictly
positive and having R times bounded derivatives.
This lemma shows that if f is bounded from 0, then we only require that its R-th derivative to
be Hölder continuous with exponent α. But if it takes value 0 on [0,C1 ], it has to vanish sufficiently
slowly.
3.2
Test
Now partition the underlying parameter θ into two parts θP and θN where θP is a d1 dimension parameter while θN = ( f1 , f2 , .., fd2 ) contains d2 density functions. By the discussion in last subsection, we can replace θN by a vector of functions (r1 , r2 , ..., rd2 ) defined on [0, 1]. With a little abuse
of notation, from now on, we refer θ2 to the corresponding functions defined on [0, 1]. Therefore,
θ ∈ Θ ⊂ Rd1 × R d2 . Let k·k∞ , k·k2 be the sup norm and L2 norm. Therefore, krk∞ = supx∈[0,1] |r (x)|,
q´
1 2
krk2 =
0 r (x) dx and
kθ k∞ = max kθN k∞ , kr1 k∞ , kr2 k∞ , ..., krd2 k∞
kθ k2 =
q
kθN k22 + kr1 k22 + kr2 k22 + ... + krd2 k22
and d∞ (θ1 , θ2 ) = kθ1 − θ2 k∞ , d2 = kθ1 − θ2 k2 for θ1 , θ2 ∈ Θ.
Because the operator T is not injective, e.g. there might be r1 and r2 ∈ R such that Tr1 = Tr2 and
r1 6= r2 , if we consider the parameter space Θ, the identifed set Θ0 is typically not a singleton. And
more generally, there are cases where the underlying parameters are not point-identified. Therefore,
we develop a test which is robust to partial identification.
7
Our test is closely related to Santos (2012) and Hong (2011). Santos (2012) proposes specification
test for linear regression model with nonparametric instrumental variables where partial identification
may be present. He derives the asymptotic distribution of the test and proposes a Bootstrap procedure
to compute the critical value. Hong (2011) extends Santos’ work to nonlinear conditional moment
equalities, but his Bootstrap critical value is conservative.
Our test can be regarded as testing infinite number of unconditional nonlinear moment equalities
which is similar to Hong (2011). But we would like to highlight two main contributions of the test.
First, because we do not have the explicit form of the moment conditions, we have to compute them
using simulation method. We derive the conditions that guarantees the test to be valid. Second, we
propose a new Bootstrap strategy which is asymptotically exact. This is very important because in
simulation we find that Hong’s test can be too conservative and therefore, not so powerful.
First define
2
Qk (θ ) = max Eφk B̃k (θ ) ,t − Eφk (Bk ,t) .
t∈T
According to (1), if the model is correctly specified, there exists a θ such that Qk (θ ) = 0 for all k.
This means
2
χ = min max max Eφk B̃k (θ ) ,t − Eφk (Bk ,t) = 0.
θ ∈Θ k∈K t∈T
Now suppose we observe bids from auctions. Let b`k denote the vector of bids from the `-th
auction with k-bidder with bi`k being its i-th element. Let Lk be the total number of k-bidder auction.
The the total number of auctions observed is L = ∑k∈K Lk with Lk /L → αk ∈ (0, 1) a.s. when L → ∞.
If we can compute the sample analog of χ, we can reject the model if the sample analog is sufficiently
big.
k
φk (b`k ,t). Because Ψk (θ ,t) =
Now we can first replace Ψk (t) = Eφk (Bk ,t) by Ψ̂k (t) = L1k ∑L`=1
Eφk B̃k (θ ) ,t usually is a very complicated function of θ , Bierens and Song (2012) suggest to com-
pute it by simulation. Let b̃sk (θ ) be vector of bids from the s-th simulated auction and S be the total
number of simulations and Ψ̃k (θ ,t) = 1S ∑Ss=1 φk b̃sk (θ ) ,t . Then define
2
Q̂kL (θ ,t) = Ψ̃k (θ ,t) − Ψ̂k (t) .
8
(2)
Let ΘL be a sequence of increasing spaces such that ∪∞
L=1 ΘL = Θ. Following Santos (2012), define
the test statistic as
ŜL = min max max Lk Q̂kL (θ ,t) .
(3)
θ ∈ΘL k∈K t∈TL
Here TL ∈ T is a sequence of increasing sets such that ∪∞
L=1 TL = T .
3.3
Asymptotic Distribution
First, we introduce the notion of path-wise derivative.
Definition 3.1. For any functional ϕ (·) : Θ → R, the pathwise derivative of ϕ at θ in the direction of
θ∆ is
dϕ (θ )
dϕ (θ + τθ∆ ) [θ∆ ] =
.
dθ
dτ
τ=0
The second order derivative in the direction of (θ∆1 , θ∆2 ) is
d 2 ϕ (θ )
[θ∆1 , θ∆2 ] =
dθ 2
dϕ(θ +τ∆2 )
dθ
)
[θ ∆1 ] − dϕ(θ
[θ
]
∆1 dθ
dτ
.
τ=0
Notice it is easy to show that for any nonnegative constant C,
dϕ(θ )
dθ
)
[Cθ∆ ] = C dϕ(θ
dθ [θ∆ ]. Another
thing to notice is that if θ = (θP , θN ) and θN = (r1 , r2 , ..., rd2 ), then
d2
∂ ϕ (θ )
dϕ (θ )
dϕ (θ )
· θ∆,P + ∑
[∆] =
[r∆,i ] .
dθ
∂ θP
i=1 dri
Here
∂ ϕ(θ )
∂ θP
dϕ(θ )
dri
[r∆,i ] is the directional derivative in the direction (0, 0, ..., r∆,i , 0, ..., 0) at θ .
is understood as usual partial derivative with respect to finite dimension parameter. While
Definition 3.2. Let vk (·) be a function from T to R with k ∈ K and v (t) = (vk (t))k∈K . So v (t) is a
vector valued function.
(
VLθ =
d2
∂ Ψk (θ ,t)
dΨk (θ ,t) mL
· θ∆P + ∑
[p ] βi , θ∆P ∈ Rd1 , βi ∈ Rd2
v : T → RK , vk (t) =
∂ θP
dri
i=1
Here
dΨk (θ ,t) mn
[p ] =
dri
dΨk (θ ,t)
dΨk (θ ,t)
dΨk (θ ,t)
[p1 ] ,
[p2 ] , ...,
[pmL ] .
dri
dri
dri
9
)
.
Θ0
Define VLΘ0 = ∪θ ∈Θ0 VLθ and V∞Θ0 = ∪∞
L=1VL .
In order to derive the asymptotic distribution of the test statistics, we further introduce following
assumptions.
Assumption 1. sup(t,θ )∈T ×ΘL
Assumption 2.
√ L Ψ̃k (θ ,t) − Ψk (θ ,t) = o p (1).
√ Lk Ψ̂k (t) − Ψk (t) ⇒ Gk (t) with Gk (t) being some complex valued stochastic pro-
cess indexed by t.
Assumption 3. (1) supθ ∈Θ d∞ (θ , ΘL ) = o
√1
L
. (2) supt∈T d (t, TL ) = o √1L .
Assumption 4. (1)Θ0 lies in the interior of Θ; (2) In a ε-neighborhood under d∞ of Θ0 , denoted by
Θε0 , Ψ (θ ,t) is pathwise twice differentiable with respect to θ ; (3)
d 2 Ψ(θ ,t)
dθ 2
[θ∆1 , θ∆2 ] ≤ C kθ∆1 k2 kθ∆2 k∞
for all θ ∈ Θε0 ; (4) Ψ (θ ,t) is uniformly bounded on Θε0 × T .
Assumption 5. Let θ̂L = arg minθ maxk∈K maxt∈TL Lk Q̂kL (θ ,t), d∞ θ̂L , Θ0 = O p (δL ) where δL =
o (1).
Now we briefly discuss about the assumptions. Assumption 1 requires that the simulated char√
acteristic function converges to the true characteristic function at a rate faster than L. Therefore,
the simulation errors can be negligible. Assumption 2 says that the empirical characteristic function
weakly converges to some process. Assumption 3 specifies the sieve space. Assumption 4 and 5 are
the most important assumptions there, so we want to discuss them in details. Assumption 4(1) we
require Θ0 lies in the interior of Θ, because the test is based on Santos (2012) which is, as pointed
out by Gandhi, Lu, and Shi (2012), based on pointwise asymptotics. As the asymptotic distribution is
discountinuous on the boundary of the parameter space, the bootstrap procedure might not be a good
approximation when the true parameter is on the boundary.5 Santos (2012) allows for boundary points
by requiring the projections of the boundary points onto the sieve space approximate the boundary
sufficiently slow. We get around of the problem because we can transform the parameter space to
make sure the true parameter is in the interior. The rest of this assumption is similar to Assumption
3.6 in Hong (2011). One key difference is Assumption 4(3). We require the second order pathwise
5 See
Andrews and Guggenberger (2007) for a discussion about uniformity.
10
derivative to be bounded by the product of L2 and sup norm, while Hong requires it to be bounded by
the product of L2 norm. Our assumption is weaker. We make this assumption because the characteristic function we are working with is not bounded by the product of L2 norm. To be consistent with this
bound, we require θ̂L converges in sup metric in stead of L2 metric in Assumption 5. However, these
technical differences do not affect the limiting distribution as we will see in next Theorem.
Theorem 3.1. Suppose Assumptions 1 to 5 holds with δL = o p L−1/4 , if the model is correctly
specified
χL ⇒ min max max kGk (t) − vk (t)k2 ,
Θ
v∈V∞ 0 k∈K t∈T
where Gk for k ∈ K are independent processes. If δL L1/4 → ∞,
χL ≤
min
max max Lk Q̂kL (θ ,t) ⇒ min max max kGk (t) − vk (t)k2 ,
Θ
−1/4
o p (L
) k∈K t∈TL
v∈V∞ 0 k∈K t∈T
θ ∈ΘL ∩Θ0
Under the alternative,
a.s
L−1 χL (R) −→ min max αk kΨk (θ , ·) − EΨk (·)k2∞ .
θ ∈Θ k∈K
This result follows immediately from Theorem 3.1 and 3.2 in Hong (2011). However, notice the
limiting distribution is slightly different from Santos (2012) and Hong (2011) in that the stochastic
processes Gk do not depend on θ . The restriction δL = o p L−1/4 guarantees the specified limiting
distribution to be exact. If this restriction is not met, the limiting distribution can still be used to
construct a test. But then the test is conservative.
3.4
Bootstrap Critical Value
One big challenge to construct the bootstrap critical value is to characterize the set V∞Θ0 . To characterize this set, one needs to compute the directional derivative in each direction for all θ ∈ Θ0 . Santos
(2012) shows how to estimate this set in nonparametric IV regression. He exploits the linear form
of the nonparametric IV regression. Hong (2011) extends his work to conditional moment equality
where nonlinearity is allowed. But instead of characterizing the V∞Θ0 set, he only take the minization
11
on a subset of V∞Θ0 , namely the set only includes the zero function. Therefore, the resulting test is very
conservative. In this paper, we propose a different way to approximate the set V∞Θ0 .
√
Consider the difference Lk Ψ (k (θ1 ,t) − Ψk (θ2 ,t)). If θ1 is sufficiently close to θ2 , it is close to
√ dΨ(θ2 ,t)
Lk dθ [θ1 − θ2 ]. Therefore, we can approximate the space V∞Θ0 by the space consists of functions
√
Lk [Ψ (k (θ1 ,t) − Ψk (θ0 ,t))] for all θ0 ∈ Θ0 with θ1 lying in a neighborhood of θ0 which shrinks at
a proper rate as sample size grows. Therefore, take minimization on V∞Θ0 can be replaced by taking
minimization with respect to θ1 which lies in a shrinking neighborhood of θ0 . To be precise, define
the bootstrap statistic as
IL∗ =
min
min
ε
max max Lk
θ2 ∈ΘL θ1 ∈B L (θ2 ) k∈K t∈TL
2
Φ∗ (t) − Φ̂k (t) − Φ̃k (θ1 ,t) − Φ̃k (θ2 ,t)
2
+λL Φ̃k (θ2 ,t) − Φ̂k (t)
Here Φ∗ (t) =
1
Lk
∗ ,t with b∗ to be the `-th bootstrapped auction in k-bidder auction. We
φ
b
∑`
k,`
k,`
can see that we have an additional minimization compared to Hong (2011).
Assumption 6. (1)λL → ∞ and λL = o (L/ log (log (L))).(2) There exists a sequence ηL → 0 such that
λL
(3a)
inf max max |Ψ (θ ,t) − Ψ (ΠL θ0 ,t)|2 → ∞
inf
θ ∈ΘL ,d2 (θ ,ΠL Θ0 )>ηL θ0 ∈Θ0 k∈K t∈T
√
√
√
LηL εL → 0 and LεL → ∞. (3b)εL = C/ L with C be any nonnegative constant.
Theorem 3.2. (1)Under Assumption 1 to 5 and assumption 6(1),(2),(3a),
L∗
IL∗ −→ min max kGk (·) − vk (·)k2∞ , a.s.
Θ
v∈V∞ 0 k∈K
P∗
In addition, L−1 IL∗ −→ 0 a.s..
(2)Under Assumption 1 to 5 and assumption 6(1),(2),(3b),
L∗
IL∗ −→ min max kGk (·) − vk (·)k2∞ ≥ min max kGk (·) − vk (·)k2∞ , a.s.
Θ
v∈VC 0 k∈K
Θ
v∈V∞ 0 k∈K
P∗
In addition, L−1 IL∗ −→ 0 a.s..
12
The first part of the Theorem shows that if we know the sequence ηL → 0 such that Assumption 6
holds, we can choose εL according to guarantee that the Bootstrap critical value gives asymptotically
exact level. However, in many cases, it is very difficult to find the sequence ηL as we shall see in the
application. But if we do know that there exists such sequence, we can still use the second part of
the theorem, we can still construct conservative critical value. In any case, the critical value is less
conservative compared to Hong (2011).
Remark 3.1. Although the bootstrap statistic is valid for point identification, it is superfluous to solve
the double minimization. Bootstrap Statistic can be constructed by
IL∗ =
min max max Lk
θ1 ∈BεL (θ̂L ) k∈K t∈TL
2
Φ∗ (t) − Φ̂k (t) − Φ̃k (θ1 ,t) − Φ̃k θ̂L ,t
.
Here εL is chosen as in the set identification case. All the results in Theorem 3.2 fall through. Therefore, computation is less intense in the point identification case.
4
Testing Whether Bidders Know the Valuation Distribution
The key assumption in the empirical analysis of first price auction is that the bidders know the valuation distribution of their opponents. However, in many auction environments bidders know very little
about the true word and therefore, they bid cautiously. Ambiguity averse preferences fit these environments. In the environment of independent private value, Grundl and Zhu (2013) develop identification
result with ambiguity averse bidders who have Maxmin Expected Utility(MEU) and find evidence of
ambiguity aversion. But there is no formal test provided.
In this section, we discuss how to apply test statistic proposed above to nonparametrically test
the assumption that bidders know the valuation distribution assuming Exogeneous Participation given
unobserved heterogeneity with multiplicative form as in Krasnokutskaya (2011).
4.1
Data Generating Processes
We provide following two alternative data generating processes.
• DGP1: Bidders’ valuations V = V ∗U has two component. An private part V ∗ which is assumed
13
to be iid distributed across bidders. Denote the distribution of V ∗ as Fk in the k-bidder auction.
It has a density function fk whose support lies in [v, M + v] with v > 0. Bidders know fk . The
observed heterogeneity U is a scalar auction specific charcteristic which is observed by the
bidders but not the econometrician. U is independent from V ∗ . It has a distribution Fu with
density fu whose support lies in [1, M + 1]. Therefore, a bidder whose V ∗ is v∗ attending a k
bidder auction with unobserved heterogeneity being u will bid us∗k (v∗ ) with
s∗k (v∗ )
1
∗
=v −
F (v)k−1
ˆ
v∗
F (s)k−1 ds.
v
The primitives in DGP1 are { f }k∈K , fu .
• DGP2: Everything are the same as in DGP1 except that now the bidders have a common prior
set about the private part of their opponents. The lower contour in of the prior set is Fkmin with
density fkmin whose support lies in [v, M + v]. This means Fkmin is in the prior set and it first order
stochastically dominate all the distribution functions in the prior set (see Lo (1998) and Grundl
and Zhu (2013) for a detailed discussion).The bidders have Maxmin Expected Utility. The bids
are generated in the same way as in DGP1 with
s∗k (v∗ )
∗
=v −
1
Fkmin (v)k−1
ˆ
v∗
Fkmin (s)k−1 ds.
(4)
v
The primitives are { f }k∈K , f min k∈K , fu .
Notice we normalize the lower bound of the unobserved heterogeneity to be 1. This normalization is
essentially the same as the normalization used in Krasnokutskaya (2011). In DGP1, because V = UV ∗
and the form, if U and V ∗ generate the bid distribution, cU and 1/cV ∗ with c > 0 also generate the
bid distribution.6 Then c is not pinned down. However, as U and V ∗ all have positive lower bound,
if we normalize the lower bound, we pin down the constant c. In DGP2, this normalization suffice as
we require the fkmin and fk have the same support. Also, notice the upper bound of the support of the
densities are not required to be known. We only require them to be smaller than M. In practice, we can
choose a very big M such that one can confidently say that the valuation is not greater than M. From
6 See
Grundl and Zhu (2013) for a discussion.
14
now on, we assume we know v as it can be easily estimated from the lower bound of the bids. We will
require fkmin and fk are bounded from below by f which is a continuous function taking positive value
at v. This condition guarantees the uniqueness of the symmetric increasing bidding strategy.
Assumption 7 (Exogenous Participation given Unobserved Heterogeneity).
The private valuation distribution and the common prior do not change across number of bidders.
In DGP1, this assumption means that Fk = F for all k ∈ K. In DGP2, this means Fk = F and
Fkmin = F min .
4.2
The Parameter Space and the Sieve Space
Now the parameters involved here are density functions. One could define the parameter space to be
the space of all density functions defined on [1, M + 1] and [v, M + v]. However, this might introduce
problems, because the true parameter might be on the boundary of the parameter space. For example,
if upper bound of the support of the density is strictly smaller than M, the density is on the boundary
of the parameter space. If the true parameter is on the boundary of the parameter space, the bootstrap
confidence interval is no longer valid. It will underestimate the critical value leading to over rejection
under the null.
Instead, we consider the space of the functions defined on [0, 1]. As discussed in Section3.1, we
can find a density h defined on [0, M] which is strictly positive and all derivatives exist and uniform
bounded by some constant. Then for any r ∈ R (C, R + α), Tr is a density function defined on [0, M].
Then the density defined on [v, M + v] and [1, M + 1] can be easily obtained by shifting the support of
Tr.
Therefore, for DGP1 the parameter space is Θ1 = R (C, R + 1)2 and for DGP2 Θ2 = R (C, R + 1)2 .
DefineRL to be a sequence of sieve space such that
(
RL =
mL
mL
i=1
i=1
)
∑ ρi ψi | ∑ ρi2 ≤ C
1
2
2
3
is increasing and ∪∞
L=1 RL = R (C, R + 1). Then ΘL = R L and ΘL = R L
15
4.3
Simulation Method
In this section, we only discuss the simulation method and convergence result of the criterion function
for DGP2. The result holds for DGP1 as it is a special case of DGP2.
For any r, rmin , ru ∈ RL , one can construct density functions f , fL , f min and fLu by
fL = (1 − 1/L) (1 − p) Tr + f + 1/ML, f = (1 − p) Tr + f ,
f min = (1 − p) Trmin + f
fLu = (1 − 1/L) Tru + 1/ML, f u = Tru
Here p =
´
f dx. Then we compute the distribution function F, F min and FuL . Then for auctions with
k bidders, we draw SL replication of random vector which has k elements iid drawn from F. To do so,
let (υ1s , υ2s , .., υks ) being the s-th iid draw from uniform distribution on [0, 1]. Then
(v∗1s , v∗2s , ..., v∗ks ) = FL−1 (υ1s ) , FL−1 (υ2s ) , ..., FL−1 (υks ) .
We can compute the vector of bids by
b∗sk = v + (s∗k (v∗1s ) , s∗k (v∗1s ) , ..., s∗k (v∗1s ))
with s∗k defined by
s∗k (v∗ )
=v −
ˆ
1
∗
F min (v)k−1
v∗
F min (s)k−1 ds.
0
Next, let µsk been s-th iid draw from uniform distribution on [0, 1].Then
bsk =
FuL
−1
(µsk ) + 1 b∗sk .
Then we can define
Ψ̃Lk (θ ,t) =
1
φk (bsk ,t) ,
SL ∑
s
ΨLk (θ ,t) = Eφk (bsk ,t) ,
16
(5)
with
φk (bsk ,t) =
1
exp it1 log bisk + it2 log b jsk
∑
k (k − 1) i6= j
And let Ψk (θ ,t) be the characteristic function of the log of two random selected bids from the same
auction generated by f , f min , fu . Now
2
Q̂kL (θ ,t) = Ψ̃Lk (θ ,t) − Ψ̂ (t)
Theorem 4.1. Suppose (1)L log N (ω/L, RL , d2 ) /SL → 0 for any ω > 0; (2)
√ −R
LmL → 0 and R ≥ 2
and
TL = {0, ±κ/NL , ±2κ/NL , .., ±2κ}2
√
with NL L → ∞, Assumption 1 to 5 hold for Ψ̃L .
√
This theorem guarantees that we can apply the test developed with εL = C/ L for some positive
constant C > 0. In this particular example, it is difficult to show the exact convergence rate of the
minimizer. Therefore, we are not sure how to construct the exact critical value. However, we can still
construct a conservative test. In next section, the Monte Carlo experiments show that the conservative
test has very good power.
5
Empirical Application: US Forest Service Timber Auctions
5.1
Data Description
USFS timber auctions have been extensively analyzed in the structural auction literature.7 The data
can be downloaded from Phil Haile’s website.
Following Haile and Tamer (2003) we construct a subsample of scaled sales with short term contracts between 1982 and 1990, for which the assumption of private values is plausible. In scaled
sales bidders pay only for the timber that is actually harvested. This insures the bidders against the
risk of overestimating the volume of timber and reduces the common value component in the val7 Baldwin,
Marshall, and Richard (1997, Appendix A) contains a detailed description of the auction procedure and
background on the timber industry.
17
Variable
Mean
Median
Std. Dev.
Min
Max
Winning Bid ($/mbf)
80.01
72.45
63.40
6.25
1112.75
Appraisal Value ($/mbf)
52.43
49.72
31.27
3.71
194.63
# Bidders
3.32
3
1.39
2
7
# Potential Bidders
14.27
11
10.81
0
48
Total Volume (mbf)
349.22
106
909.62
19
15000
Species Concentration
0.76
0.75
0.20
0.28
1
Inventory (1000 mbf)
700.44
697.04
97.49
484.73
940.69
Timber Density (mbf/acres)
5.70
2.72
6.77
0.001
58.75
Harvesting Cost ($/mbf)
31.05
30.07
17.09
0
93.23
Table 1: Summary Statistics. Sample size: 995 auctions.
uations. Short term contracts with contract length less than one year limit resale opportunities and
thereby reduce the common value component generated by the resale market. In 1981 the Forest Service introduced new policies designed to limit subcontracting and speculative bidding (Haile (2001)).
Therefore, only auctions after 1981 are included. Geographically, we focus on timber tracts from
the Southern Region, ranging from Texas and Oklahoma to Florida and Virginia, where most of the
first-price auctions take place.
Following Haile, Hong, and Shum (2003), we assume that auction characteristics changes valuation distribution in a multiplicative way. This implies bi` = exp (X` β1 + β0 ) bni` . 8 Here bni` = u` b∗i` is
the normalized bid independent of the auction characteristics X` . Then
log bi` = X` β1 + β0 + log bni`
(6)
We only consider one covariates, the log appraisal value.
Because, log bni` are independent of X` , one can estimate 6 using linear regression and take the
residuals as log bni` . Instead, we use a different approach. Notice β > 0,
log bi` = X ` β1 + β0 + log bni`
where bi` , X ` and bni` are lower bounds of bi` ,X` and bni` . Then if X ` is different for auctions with different number of bidders, we can estimate β1 ,β0 use the lower bound. One advantage of this procedure
8 The
test can allow for more general form. But our sample is not big enough to allow for more flexible setup.
18
is that then β1 ,β0 can be estimated with rate 1/n. Then we can treat estimated bni` as the true values.9
To estimate β1 and β0 , we take the lower bound of bids, appraisal values from auctions with 2 to 6
bidders. And we regress the lower bound of bids on the lower bound of appraisal values. Then we get
log bi` = 0.9045X` + 0.4513 + log bni`
Then we construct
log b̂ni` = log bi` − (0.9045X` + 0.4513) .
To check the validity of this procedure, we regress log b̂ni` on X` and find no significant correlation.
Then we treat b̂ni` as true bids and normalize the lower bound of unobserved heterogeneity to 1 and
v is estimated using the lower bound of b̂ni` . In the test, we only use the auctions with 2 to 4 bidders
because we do not get much data in auctions with more than 4 bidders. We delete three auctions
where there are bids far more than other bids in the sample. Therefore, we end up with 353 2 - bidder
auctions, 257 3- bidder auctions and 163 4- bidder auctions.
5.2
Testing Results
We apply the test described in last section to test whether the data is generated by DGP 1. We choose
the upper bound of the valuation distribution and unobserved heterogeneity M to be 15 which is more
than four times the largest b̂ni` . The H function is chosen to be uniform on the interval [0, M]. f , the
lower bound of f and f u is set to be (1 − v)8 1 (0 ≤ v ≤ 1) /1010 . We use a sieve space with mL = 8.
4/3
The numbers of simulation draws for auction with k bidders are Lk , which satisfies Theorem 4.1.
Because the rate of the minimizer is hard to derive, we use the second part of Theorem 3.2 to construct
the bootstrap critical value. C is set to be 5.TL is taken to be RL × RL with RL being 100 equally spaced
points on [−1, 1]
The test statistic is 6.776. And the bootstrap 95% critical value is .4227 which is based on 150
repetitions. Therefore, we reject that the DGP 1 generates the data.
[Test for DGP 2 is Under Construction]
9 This treatment is not necessary. One can always test the model based on the joint distribution of bids and auction
characteristics. We do this because of computational considerations.
19
6
Conclusion
20
A
Lemmas
Proof to Lemma 3.1. First suppose C2 = C1 and f (C1 ) > 0. Then f is bounded from below on [0,C1 ].
As h (x) is positive and continuous, it is also bounded from below on [0,C1 ]. Therefore, f /h is positive
p
on [0,C1 ]. For any positive function g, g (x) has R times continuous derivative if g has R times
√
continuous derivative. In addition, x has infinite continuous derivatives if x > ε > 0.
∇R
p
p
g (x1 ) − ∇R g (x2 )
If f (C2 ) = 0, there exists ε > 0 such that ∀y satisfying |y −C2 | < ε, f (y) = M 2 (y) |y − x|2R+2α .
Here the R-th derivative of M is H/"older continuous with exponent α. For any x1 , x2 ∈ [0,C2 − ε),
f /h is bounded away from zero. The same argument as above applies.
If x1 , x2 ∈ [C2 − ε,C2 ],
∇R
p
R
f /h (x1 ) =
n
d R−n
R d M
∑ n dyn √h (x1 ) dyR−n (C2 − x1 )R+α .
n=0
The same thing holds for x2 . Then
p
p
R
f /h (x1 ) − ∇R f /h (x2 )
∇
R n
dn M
d R−n
d R−n
R d M
R+α
R+α √
√
(C
−
x
)
−
(C
−
x
)
(x
)
(x
)
≤ ∑
2
1
2
2
1
2
dyn h
dyR−n
dyn h
dyR−n
n=0 n
R−n
R n
d
R
d M
d R−n
R+α
R+α √ (x1 ) R−n (C2 − x1 )
≤ ∑
− R−n (C2 − x2 )
n
dy h
dy
dy
n=0 n
R−n
n
d
d M
dn M
+ R−n (C2 − x2 )R+α n √ (x1 ) − n √ (x2 )
dy
dy h
dy h
R R
≤ ∑
(c1n + c2n ) |x1 − x2 |α
n
n=0
21
If x1 ∈ [0,C2 − ε) and x2 ∈ [C2 − ε,C2 ],
p
p
R
R
∇
f
/h
(x
)
−
∇
f
/h
(x
)
1
2 p
p
p
p
≤ ∇R f /h (x1 ) − ∇R f /h (C2 − ε) + ∇R f /h (x2 ) − ∇R f /h (C2 − ε)
≤ M1 |C2 − ε − x1 |α + M2 |C2 − ε − x2 |α
≤ M3 |C2 − ε − x1 |α + |x2 − (C2 − ε)|α
≤ M4 |x1 − x2 |α
Lemma A.1. There exists a constant C > 0 such that for any r1 , r2 ∈ R and kr1 k∞ , kr2 k∞ ≤ M, we
haved∞ (Tr1 Tr2 ) ≤ Cd∞ (r1 r2 ) .
Proof.
[1 + r (H (x))]2
2
[1
+
r
(H
(x))]
1
2
|(Tr1 ) (x) − (Tr2 ) (x)| = h (x) −
h (x)
´1 2
´1 2
1 + r (x) dx
1 + 0 r2 (x) dx
0 1
´
[1 + r (H (x))]2 1 + 1 r2 (x) dx − [1 + r (H (x))]2 1 + ´ 1 r2 (x) dx
2
1
1
2
0
0
= h
(x)
´1 2
´1 2
1 + 0 r1 (x) dx 1 + 0 r2 (x) dx
[1 + r1 (H (x))]2 − [1 + r2 (H (x))]2 ≤ khk∞ ´1
1 + 0 r12 (x) dx
[1 + r (H (x))]2 ´ 1 r2 (x) dx − ´ 1 r2 (x) dx 2
0 1
0 1
+ khk∞ ´1 2
´1 2
1 + 0 r1 (x) dx 1 + 0 r2 (x) dx
≤ khk∞ (M + 2) sup |r1 (x) − r2 (x)| + khk∞ (M + 1)2 M sup |r1 (x) − r2 (x)|
r∈[0,1]
r∈[0,1]
h
i
= khk∞ (M + 1)2 M + M + 2 sup |r1 (x) − r2 (x)|
r∈[0,1]
22
B
Proof to Theorem 3.2
Proof. First note that
2
λL Φk (θ2 ,t) − Φ̂k (t) = λL kΦk (θ2 ,t) − Φk (t)k2 + o p (1)
√ sup Ψ̃ (θ , ·) − Ψ (θ , ·)∞ = o p 1/ L
θ ∈ΘL
. Therefore,
IL∗ =
In addition,
min
mine
2
max max Lk Φ∗k (t) − Φ̂k (t) − (Φk (θ1 ,t) − Φk (θ2 ,t))
θ2 ∈Θn θ1 ∈B(θ2 ) L ∩ΘL k∈K t∈TL
+λL kΦk (θ2 ,t) − Φk (t)k2 + o p (1) .
√
Lk Φ∗k (t) − Φ̂k (t) ⇒ Gk (t) almost surely.
For the first part of the theorem, by assumption, there exists ηL → 0 such that
infd∞ (θ ,Θ0 )≥ηL λL kΦk (θ2 ,t) − Φk (t)k2 → ∞.
IL∗ =
min
η
minε
2
max max Lk Φ∗k (t) − Φ̂k (t) − (Φk (θ1 ,t) − Φk (θ2 ,t))
θ2 ∈Θ0 L ∩ΘL θ1 ∈B(θ2 ) L ∩ΘL k∈K t∈T
+λL kΦk (θ2 ,t) − Φk (t)k2
=
min
η
mine
2
p
max max G∗k,L (t) − Lk (Φk (θ1 ,t) − Φk (θ2 ,t))
θ2 ∈Θ0 L ∩ΘL θ1 ∈B(θ2 ) L ∩ΘL k∈K t∈T
+λL kΦk (θ2 ,t) − Φk (t)k2 + o p (1)
2
p
∗
≥
min
min
max
max
G
(t)
−
L
(Φ
(θ
,t)
−
Φ
(θ
,t))
k
k 1
k 2
k,L
e
η
θ2 ∈Θ0 L ∩ΘL θ1 ∈B(θ2 ) L ∩ΘL k∈K t∈T
2
p dΦk (θ2 ,t)
∗
=
min
min
max max Gk,L (t) − Lk
[θ1 − θ2 ]
+ o p (1)
ηL
ε
dθ
θ2 ∈Θ0 ∩ΘL θ1 ∈B L (θ2 )∩ΘL k∈K t∈T
Here B (θ2 )εL is the εL -neighborhood of θ2 under L2 metric and Θη0 L is the ηL -neighborhood under sup
23
0
metric. By assumption, when θ , θ ∈ B (θ0 )ε under L2 metric for some θ0 ∈ Θ0 ,
0 dΦ
θ
,t
k
dΦk (θ ,t)
≤ Cd∞ θ , θ 0 × d2 (θ1 , θ2 )
[θ
−
θ
]
−
[θ
−
θ
]
1
2
1
2
dθ
dθ
∞
.As long as
IL∗
√
LεL ηL → 0,
p dΦk (θ2 ,t) m 2
∗
L
min max max [p ] β ≥
min
+ o p (1)
Gk,L (t) − Lk
η
dθ
θ2 ∈Θ0 L ∩ΘL kβ k2 ≤εL k∈K t∈T
√
p dΦk (θ2 ,t) m 2
∗
L
= min min max max Gk,L (t) − Lk
[p ] β +
o
Lε
η
+ o p (1)
L
L
θ2 ∈Θ0 kβ k2 ≤εL k∈K t∈T
dθ
2
= min max max G∗k,L (t) − v (t) + o p (1)
v∈V BL k∈K t∈T
√ with BL = O εL L and
V
BL
dΦk (θ2 ,t) mL
= v|vk =
[p ] β , kβ k2 ≤ BL and θ ∈ Θ0 .
dθ
On the other hand, we can find η̃L such that λL kΦk (θ2 ,t) − Φk (t)k2 → 0 for all θ ∈ Bη̃ (Θ0 ), then
IL∗ ≤
=
=
min
min
eL
θ2 ∈Θ0 L ∩Θn θ1 ∈B(θ2 )
η̃
min
min
2
p
max max Lk G∗k,L (t) − Lk (Φk (θ1 ,t) − Φk (θ2 ,t)) + o p (1)
k∈K t∈T
2
p
∗
max max Lk Gk,L (t) − Lk (Φk (θ1 ,t) − Φk (θ2 ,t)) + o p (1)
θ2 ∈Θ0 θ1 ∈B(θ2 )eL k∈K t∈T
2
min max max G∗k,L (t) − v (t) + o (1)
v∈V BL k∈K t∈T
For the first part of the theorem, V BL is increasing as BL → ∞. By the fact G∗k,L →d Gk
2
min max max G∗k,L (t) − v (t) ⇒ min max max kGk (t) − v (t)k2 .
B
v∈VL L k∈K t∈T
Θ
v∈V∞ 0 k∈K t∈T
∗
If the null hypothesis does not hold, since λL = o p (log (log L) /L), I ∗ /L → p 0 a.s.
√
For the second part, BL = B. Obviously, LεL ηL = BηL → 0.
2
min max max G∗k,L (t) − v (t) ⇒ min max max kGk (t) − v (t)k2 .
v∈VLB k∈K t∈T
Θ
v∈VB 0 k∈K t∈T
24
The second claim follows for the same reason as in the first part.
Lemma B.1. For any distribution function F1 with density f1 and F2 with density f2 supported on Ω
and in addition M > f1 , f2 > f for some constant M and f is continuous with f (v) > ε for some ε > 0.
Let the bidding function given F to be
ˆ
v
s (v| f , I) = v −
v
F I−1 (s)
ds.
F I−1 (v)
Then there exist a constant C > 0 which depends only on M, f ,I and Ωsuch that
ks (·| f1 , I) − s (·| f2 , I)k∞ ≤ C k f1 − f2 k∞ .
In addition, if f > ε and k f1 − f2 k∞ < ε/2, C = O ε −1 .
Proof. To simplify notation, let s1 (v) = s (v| f1 , I) and s2 (v) = s (v| f2 , I).
ˆ
I−1 ´ v I−1
´
ˆ v I−1
v F I−1 (s)
F (v) F (s) ds − F I−1 (v) v F I−1 (s) ds F
(s)
1
1
2
2
v
v
1
2
|s1 (v) − s2 (v)| ≤ ds −
ds = I−1
I−1
I−1
I−1
v F1 (v)
(v) F1 (v) F2 (v)
v F2
´
´
´
v
v
v
F2I−1 (v) v F1I−1 (s) ds − v F2I−1 (s) ds + F1I−1 (v) − F2I−1 (v) v F2I−1 (s) ds
≤
F1I−1 (v) F2I−1 (v)
´
´v
´
v I−1
v F1 (s) ds − v F2I−1 (s) ds F1I−1 (v) − F2I−1 (v) vv F2I−1 (s) ds
+
≤
F1I−1 (v)
F1I−1 (v) F2I−1 (v)
´
´v
v I−1
v F1 (s) ds − v F2I−1 (s) ds F I−1 (v) − F I−1 (v) (v − v)
1
2
≤
+
F1I−1 (v)
F I−1 (v)
1
I−2− j
kF1 − F2 k∞ (v − v) ∑I−2
(v) F2j (v)
j=0 F1
≤2
F1I−1 (v)
≤2 (I − 1)
kF1 − F2 k∞ (v − v)I−1 M I−2
.
´
I−1
v
v f (s) ds
Now since for any v, by mean value theorem,
´v
v
f (s) ds/ (v − v) = f (ṽ) with ṽ ∈ (v, v). Since f
is continuous and f (v) > ε, ∃δ > 0 such that f (s) > ε/2 on [v, v + δ ]. Then for all v ≤ v + δ ,
´v
´v
εδ
εδ
f
(s)
ds/
(v
−
v)
>
ε/2
and
for
all
v
>
v+δ
,
f
(s)
ds/
(v
−
v)
>
/
(v
−
v).
Let
C
=
min
/
(v
−
v)
,
ε/2
.
1
v
v
2
2
25
Then
´v
v
´
f (s) ds/ (v − v) > C1 . In addition, kF1 − F2 k∞ ≤ | f1 − f2 | dx ≤ d2 ( f1 , f2 ). Therefore,
|s1 (v) − s2 (v)| ≤ 2 (I − 1)C11−I M I−2 d2 ( f1 , f2 ) = Cd2 ( f1 , f2 ) .
As above inequality holds for every v ∈ Ω, the first claim follows.
For the second claim, just note
I−2− j
j
kF1 − F2 k∞ (v − v) ∑I−2
F
(v)
F
(v)
j=0 1
2
F1I−1 (v)
=
j
kF1 − F2 k∞ (v − v) I−2 F2 (v)
∑ j .
F11 (v)
j=0 F1 (v)
Here because
F2j (v)
F1j (v)
´v
=
f2 (s) ds
´vv
v f 1 (s) ds
´v
!j
≤
j
F (v)
v
´v
!j
f1 (s) + ε2 ds
´v
≤
v f 1 (s) ds
2
∑I−2
j=0 F j (v) is some finite constant. While
1
j
3
−1 .
C = Cε1 ∑I−2
j=0 2 (v − v) is O ε
kF1 −F2 k∞ (v−v)
F11 (v)
≤
v
C1
ε
!j j
ε + ε2 ds
3
´v
≤
(v − v) ,
2
v εds
k f1 − f2 k2 for some constant C1 , then
Lemma B.2. Let F be a distribution with density f with support Ω and M ≥ f ≥ ε for some ε > 0,
then
∂
∂ v s (v| f , I)
< C for some C > 0 depending on only ε, M and I.
Proof. Just notice
∂
∂ v s (v| f , I)
=
(I−1) f (v) ´ v
I−1
ds
v F (s)
F(v)I
f (v)
≤ (I − 1) (v−v)
≤ (I − 1) Mε .
F(v)
Lemma B.3. For any distribution function F1 with density f1 and F2 with density f2 supported on Ω
and in addition f1 , f2 > ε for some constant ε > 0 . F1−1 (·) − F2−1 (·)∞ ≤ ε −1 (v − v) k f1 − f2 k∞ .
´x
´x
Proof. Let x1 = F1−1 (α) and x2 = F2−1 (α). Then v 1 f1 (s) ds = α and v 2 f1 (s) ds = α, which imply
´ x1
´ x2
´ x2
v f 1 (s) ds = v f 2 (s) ds. Without loss of generality, assume x1 > x2 . Then v f 1 (s) − f 2 (s) ds =
´
´ x1
x2
f
(s)
ds.
Since
f
(s)
−
f
(s)
ds
< k f1 − f2 k2 and f2 > ε, |x1 − x2 | ≤ ε −1 k f1 − f2 k2 . Because
2
1
2
x2
v
the inequality holds for every α, the lemma holds.
Proof of Theorem4.1 . (1) We begin with showing Assumption 1 holds for Ψ̃Lk . The proof proceeds
√ in two steps. We show that sup(t,θ )∈T ×ΘL L Ψ̃Lk (θ ,t) − ΨLk (θ ,t) = o p (1). Then we conclude with
26
showing ΨLk = Ψk + o (1) uniformly. Notice
P
sup
(t,θ )∈T ×ΘL
√ L
L Ψ̃k (θ ,t) − ΨLk (θ ,t) > 2ε
!
≤ P
sup
(t,θ )∈T ×ΘL
+P
√ L ReΨ̃Lk (θ ,t) − ReΨLk (θ ,t) > ε
sup
(t,θ )∈T ×ΘL
√ L ImΨ̃Lk (θ ,t) − ImΨLk (θ ,t) > ε
= P1 + P2 .
For P1 , because
√ √ 1
L
L
P
L ReΨ̃k (θ ,t) − ReΨk (θ ,t) > ε = P
L ∑ Reφk (bsk ,t) − EReφk (bsk ,t) > ε .
SL s
Obviously, Reφk (bsk ,t) is bounded by 1. Applying Bernstein Inequality, we have
√ 1
ε2
1
√
P
L ∑ Reφk (bsk ,t) − EReφk (bsk ,t) > ε ≤ 2 exp −
.
SL s
2 L/SL + ε L/3SL
There exists a constant C such that
0 o
0 0 n
0
φk (b,t) − φk b ,t ≤ C · max b − b , t − t .
∞
(7)
∞
In addition, for any f1min , f2min and f1 , f2
max s∗k F1−1 (u) | f1min − s∗k F2−1 (u) | f2min ≤ C1 max Ld2 ( f1 , f2 ) , d2 f1min , f2min .
u∈[0,1]
To see this, notice
∗ −1
sk F (u) | f1min − s∗k F −1 (u) | f2min ≤ s∗k F −1 (u) | f1min − s∗k F −1 (u) | f2min 1
2
1
1
+ s∗k F1−1 (u) | f2min − s∗k F2−1 (u) | f2min .
27
!
(8)
!
Lemma B.3 shows
−1
−1
(υ) − F2L
(υ) ≤ C2 L × d2 f1L , f2L
sup F1L
(9)
υ∈[0,1]
Applying Lemma B.1, B.2 and inequality (9) yields (8).
L and F uL
Similarly for F1,u
2,u
L −1
L L
L
sup F1,u
(υ) − F2,u
, f2,u
≤ C2 L × d2 f1,u
(10)
υ∈[0,1]
Combing (7), (8) and (10),
L
L
L
Ψ̃k (θ1 ,t1 ) − Ψ̃Lk (θ2 ,t2 ) ≤ C3 · max Ld2 ( f1 , f2 ) , d2 f1min , f2min , Ld2 f1,u
, f2,u
The last inequality holds because of Lemma (A.1). Therefore, let Tω × ΘωL being an ω covering of
T × ΘL and
d (θ1 , θ2 ) = max d2 (r1 , r2 ) , d2 r1min , r2min , d2 (r1u , f2u ) .
If ω = ε/2C3 , for any ε > 0, when L is large enough
!
√ 1
P
sup
L ∑ Reφk (bsk ,t) − EReφk (bsk ,t) > 2ε
SL s
(t,θ )∈T ×ΘL
!
√ 1
L ∑ Reφk (bsk ,t) − EReφk (bsk ,t) > ε
≤ P
sup
SL s
(t,θ )∈Tω ×ΘωL
1
ε2
2
√
≤ N (ω, T, d2 ) × N (ω, RL , d2 ) × N (ω/L, RL , d2 ) exp −
2 L/SL + ε L/3SL
This terms converges to 0 for any ε > 0 if for any ω,
L log N(ω/L,RL ,d2 )
SL
→ 0.
A similar argument show that if the same condition holds
!
√ 1
P
sup
L ∑ Imφk (bsk ,t) − EImφk (bsk ,t) > 2ε → 0.
SL s
(t,θ )∈T ×ΘL
√ Consequently, P sup(t,θ )∈T ×ΘL L Ψ̃Lk (θ ,t) − ΨLk (θ ,t) > 2ε → 0.
28
In addition, it is easy to see that supT ×ΘL ΨLk (θ ,t) − Ψk (θ ,t) = O (1/L). Therefore,
√ L
L Ψ̃k (θ ,t) − Ψk (θ ,t) = o p (1) .
(2) Next, we show Assumption 4. We only need to show (2) to (4). Note because φk us∗k v|T rmin + τr∆min
T (r + τr∆ ) and T (ru + τr∆u ) are all differentiable,
dΨk (θ ,t)
dθ
[θ∆ ] can be computed as follows,
ˆ ˆ
d
k
∗
min
min
u
u
φk usk v|T r + τr∆
,t [T (r + τr∆ ) (v)] T (r + τr∆ ) (u) dudv
dτ
τ=0
ˆ ˆ
d
=
φk us∗k v|T rmin + τr∆min ,t [T (r) (v)]k T (ru ) (u) dudv
dτ
τ=0
ˆ ˆ
d
k
∗
min
[T (r + τr∆ ) (v)] T (ru ) (u) dudv
+
φk usk v|T r
,t
dτ
τ=0
ˆ ˆ
d
k
∗
min
u
u
+
φk usk v|T r
,t [T (r) (v)]
T (r + τr∆ ) (u) dudv
dτ
τ=0
dΨk (θ ,t)
[θ∆ ] =
dθ
= D1 (θ ,t) + D2 (θ ,t) + D3 (θ ,t) .
We first investigate D1 . Define Biτ (v, u) = us∗k v|T rimin + τr∆min
and denote s∗k v|T rimin + τr∆min
as s∗kiτ (v) for i = 1, 2. Then
|D1 (θ1 ,t) − D1 (θ2 ,t)|
ˆ ˆ dφk (B1τ (v, u) ,t) dφk (B2τ (v, u) ,t) k
u
≤
−
[T (r1 ) (v)] T (r1 ) (u) dudv
dτ
dτ
τ=0
τ=0
ˆ ˆ
dφk (B2τ (v, u) ,t) k
k
u
u
[T
(r
)
(v)]
T
(r
)
(u)
−
[T
(r
)
(v)]
T
(r
)
(u)
+
dudv
1
2
1
2
dτ
τ=0
= A1 + A2
φk B v, u, r1min + τr∆min ,t
=
2
exp (i (t1 + t2 ) log (1 + u) + it1 log (v + s∗k1τ (vi )) + it2 log (v + s∗k1τ (v j )))
k (k − 1) i6∑
=j
29
,t ,
Then
dφk (B1τ (v, u) ,t) dτ
τ=0

=
ds∗k1τ (vi )  it1 dτ τ=0

 v + s∗ (vi )
k10
i6= j
2
k (k − 1) ∑
+
ds∗ (v j ) it2 k1τdτ 
τ=0 
∗
v + sk10 (v j ) 

× exp (i (t1 + t2 ) log (1 + u) + it1 log (v + s∗k10 (vi )) + it2 log (v + s∗k10 (v j )))
Obviously,
dφk (B1τ (v,u),t) dτ
τ=0
≤ C r∆min 2 by Lemma B.4. And
dφk (B1τ (v, u) ,t) dφ
(B
(v,
u)
,t)
2τ
k
−
dτ
dτ
τ=0
τ=0 

∗
ds
v
(
)
j
it ds∗k1τ (vi ) it2 k1τdτ 1


dτ
2
τ=0
τ=0 

≤
+
∑
k (k − 1) i6= j  v + s∗k10 (vi )
v + s∗k10 (v j ) 
× exp (it1 log (v + s∗k10 (vi )) − it1 log (v + s∗k20 (vi )) + it2 log (v + s∗k10 (v j )) − it2 log (v + s∗k20 (v j ))) − 1
 


ds∗k1τ (v j ) ds∗k2τ (v j ) ∗ (v ) ∗ (v ) ds
ds
i
i
it k1τ k2τ
it2 dτ it2 dτ it
1
1




dτ
dτ
2
τ=0
τ=0  
τ=0
τ=0 

+
+
−
+
∑
k (k − 1) i6= j  v + s∗k10 (vi )
v + s∗k10 (v j )   v + s∗k20 (vi )
v + s∗k20 (v j ) 
= B1 + B2 .
For B1 , by Lemma B.4,


ds∗k1τ (v j ) ∗
it dsk1τ (vi ) it1 dτ 2KMT min  1 dτ τ=0

τ=0 

r∆ .
 v + s∗ (vi ) + v + s∗ (v j )  ≤ v
2
k1τ
k1τ
In addition, use the inequality |exp (ix) − 1| ≤ |x| to conclude
B12 = exp (it1 log (v + s∗k10 (vi )) − it1 log (v + s∗k20 (vi )) + it2 log (v + s∗k10 (v j )) − it2 log (v + s∗k20 (v j ))) − 1
≤ t1 [log (v + s∗k10 (vi )) − log (v + s∗k20 (vi ))] + t2 [log (v + s∗k10 (v j )) − log (v + s∗k20 (v j ))]
≤ |t1 | /v |s∗k10 (vi ) − s∗k20 (vi )| + |t2 | /v s∗k10 (v j ) − s∗k20 (v j ) .
30
The last inequality uses the concavity of the log function. Combining above inequality with Lemma
B.1 and ..., we have
2MT
2MT
d2 r1min , r2min ≤
d∞ r1min , r2min .
v
v
r1min , r2min for some C1 .
B12 ≤
Therefore, A1 ≤ C1 r∆min 2 d∞
For B2 , notice
∗
ds∗ (v ) dsk1τ (v j ) ds∗k2τ (v j ) ∗ (v ) ds
k1τ i dτ k2τ i
dτ
dτ τ=0
dτ
2
τ=0 τ=0
τ=0 |t1 | |t
|
B2 ≤
−
−
+
2
∑
∗ (v ) .
v + s∗ (v j )
k (k − 1) i6= j v + s∗k10 (vi ) v + s∗k20 (vi ) v
+
s
j k10
k20
We can only focus on the terms inside the summation. For the first term,
ds∗ (v ) ds∗k2τ (vi ) k1τ i dτ τ=0
dτ τ=0 v + s∗ (vi ) − v + s∗ (vi ) k10
k20
ds∗ (v ) ∗
k1τ i − ds∗k2τ (vi ) ∗
sk10 (vi ) − s∗k20 (vi ) dτ τ=0
dτ
τ=0 dsk2τ (vi ) ≤ ∗ (v )
+
2
v
+
s
dτ
v
i
τ=0
k10
0 ≤ C2 r∆min 2 d∞ r1min , r2min
0
for some constant C2 by Lemma B.1 and .... The same thing holds for the second term in the summa
tion. Therefore B2 ≤ C2 r∆min 2 d∞ r1min , r2min for some C2 > 0. Consequently, A1 ≤ (C1 +C2 ) r∆min 2 d∞ r1min , r2min .
For A2 ,
ˆ ˆ
A2 =
≤
≤
≤
≤
dφk (B2τ (v, u) ,t) k
k
u
u
[(Tr
)
(v)]
T
(r
)
(u)
−
[(Tr
)
(v)]
T
(r
)
(u)
dudv
1
2
1
2
dτ
τ=0
ˆ ˆ k
k
C r∆min 2
[(Tr1 ) (v)] T (r1u ) (u) − [(Tr2 ) (v)] T (r2u ) (u) dudv
ˆ ˆ n
o
min k
k
C r∆ 2
[(Tr1 ) (v)] − [(Tr2 ) (v)] T (r1u ) (u) dudv
ˆ ˆ min k
+C r∆ 2
[(Tr2 ) (v)] [Tr1u (u) − Tr2u (u)] dudv
C1 r∆min 2 d2 (r1 , r2 ) +C2 r∆min 2 d2 (r1u , r2u )
(C1 +C2 ) r∆min 2 d2 (θ1 , θ2 ) .
31
Consequently, we can conclude
|D1 (θ1 ,t) − D1 (θ2 ,t)| ≤ C r∆min 2 d∞ (θ1 , θ2 ) ≤ C kθ∆ k2 d∞ (θ1 , θ2 ) .
We can apply similar argument for the rest two terms. Therefore,
dΨk (θ1 ,t)
dΨ
(θ
,t)
2
k
[θ∆ ] −
[θ∆ ] ≤ C kθ∆ k2 d∞ (θ1 , θ2 ) .
dθ
dθ
(11)
Notice C does not depend on θ1 , θ2 , θ∆ and t.
2
∂ Ψk (θ ,t)
Now to see ∂ τ 2 [θ∆1 , θ∆2 ] ≤ C kθ∆1 k2 kθ∆2 k∞ for all θ ∈ Θε0 with some ε > 0, notice
2
∂ Ψk (θ ,t)
= lim
[θ
,
θ
]
∆1
∆2
τ→0
∂ τ2
dΨk (θ +τθ∆2 ,t)
dθ
k (θ ,t)
[θ∆1 ] − dΨdθ
[θ∆1 ] dτ
≤ C kθ∆1 k2 kθ∆2 k∞ .
The last inequality holds as a consequence of (11).
(3) Now we show Assumption 5. First notice Ψ̂ (t) converges to Ψ (t) a.s. uniformly on T . By
Lemma XXX,
sup |Ψk (θ ,t1 ) − Ψk (θ ,t1 )| ≤ Cd2 (t1 ,t2 )
θ ∈Θ
. In addition, if we fix t, Ψk is continuous in θ . These two condition implies.
lim sup |Ψk (θ1 ,t) − Ψk (θ2 ,t)| = 0.
d∞ (θ1 ,θ2 ) t∈T
32
Therefore
P d θ̂L , Θ0 ≥ ε
≤ P max max αk Qk θ̂L ,t ≥ δ
k t∈T
= P max max αk Qk θ̂L ,t − max max αk Qk θ̂L ,t + max max αk Qk θ̂L ,t − max max Lk /LQ̂k θ̂L ,t
k
t∈T
t∈TL
k
k
t∈TL
k
t∈TL
+ max max Lk /LQ̂k θ̂L ,t − max max Lk /LQ̂k (θ0 ,t) + max max Lk /LQ̂k (θ0 ,t) − max max αk Qk (θ0 ,t) ≥ δ
k
t∈TL
k
t∈TL
k
t∈TL
k
t∈TL
≤ P (o (1) + o p (1) + o p (1) ≥ δ ) → 0.
Lemma B.4. There exists a K > 0 such that for any r1min , r2min and r∆min . Here r∆min = r̄ − r̃ where
ds∗k (v|Trimin ) min r̄, r̃ ∈ R (C, R + 1), supv∈[0,M] drmin
r∆ ≤ K r∆min 2 for i = 1 and 2. And in addition,
i
ds∗ v|Trmin ∗ v|Tr min ds
k
2
1
k
min
min r
−
r
≤ K r∆min 2 d∞ r1min , r2min .
∆
∆
min
min
dr1
dr1
min = pT r min + τr min + f and f min = pT r min + τr min + f . Because
Proof. Denote f1τ
1
2τ
2
∆
∆
ds∗k v|Tr1min min r∆
(v) =
dr1min
=
=
=
d ∗
min
min s v|T r1 + τr∆
dτ k
τ=0
ˆ v min
k−1
F1τ (s)
ds d
−
min k−1 dτ 0
F1τ (v)
τ=0
ˆ v
min
k−2 dF1τ
k−1
min
ds
F
(s)
(s)
−
10
min (v) k−1 0
dτ
F10
τ=0
ˆ v
min
min k−1
(k − 1) dF1τ (v) ds +
F10 (s)
min (v) k
dτ
0
F10
τ=0
min
min
D1 r1 (v) + D2 r1 (v) .
33
A similar equation holds for
ds∗k (v|Tr2min )
dr2min
min r∆ . We first show the boundedness. Notice
ˆ
min s d pT rmin + τrmin + f
dF1τ
1
∆
(x) dx dτ (s) = dτ
0
τ=0
τ=0
ˆ
s d 1 + rmin (H (x)) + τrmin (H (x))2 h (x) 1
∆
= p
dx
´ min
2
min
0 dτ
dx
1 + r1 + τr∆
τ=0
 
ˆ 
2
´ min min 
 s min
2r
r
dx
1
+
r
(H
(x))
h
(x)
1
1
∆
dx
2 1 + r1min (H (x)) h (x) r∆min (H (x)) −
= p
h
´ min 2 i2

0 
 dx
1 + r1


 ˆ 
´
2
s  min r min dx 1 + r min (x) 
2r
1
1
∆
= p
2 1 + r1min (x) r∆min (x) −
dx
h
´ min 2 i2
0 

 1 + r1
dx

2
ˆ s ´ min min ˆ s
2 r∆ r1 dx 1 + r1min (x)
min
min
dx
≤ p
1 + r1 (x) r∆ (x) dx + p
h
2 i2
´
0
0
1 + r1min dx
Because
ˆ
0
sˆ
s 1 + r1min (x) r∆min (x) dx ≤
0
s
2
1 + r1min (x) dx
sˆ
0
s
2
r∆min (x) dx
≤ 1 + r1min 2 r∆min (x)2 ≤ K1 r∆min (x)2
and
2
ˆ s ´ min min min min r∆ r1 dx 1 + r1min (x)
r∆ (x) r1 (x) ≤ K2 r∆min (x) .
dx
≤
h
i
2
2
2
2
2
´
0
1 + r1min dx
min dF
Therefore, dτ1τ (s)
τ=0
≤ (K1 + K2 ) r∆min (x)2 = K r∆min (x)2 .
For the second claim, we have
ds∗ v|Trmin ∗ v|Tr min ds
k
1
2
k
min
min r
−
r
∆
∆
dr1min
dr1min
≤ D1 r1min (v) − D1 r2min (v) + D2 r1min (v) − D2 r2min (v) .
34
We first investigate the first term.
D1 r1min (v) − D1 r2min (v)
ˆ v
ˆ v
min
min
dF
dF
k
−
1
k
−
1
k−2
k−2
min
min
2τ
1τ
= F
(s)
(s)
ds
−
F
(s)
(s)
ds
min k−1
10
F min (v) k−1 0 20
dτ
dτ
0
F
(v)
τ=0
τ=0
10
20
ˆ
v min
k−2 dF2τ
k−1
k−1
min
≤ F
−
(s)
(s)
ds
20
min (v) k−1 0
F min (v) k−1
dτ
F
τ=0
20
10
ˆ v min
min k−2 min k−2 dF2τ
k−1
F20 (s)
− F10 (s)
+
(s) ds
k−1
min (v)
dτ
0
F10
τ=0
ˆ v
min min
min k−2 dF2τ
dF
k−1
+
F10 (s)
(s) − 1τ (s)
ds
k−1
min (v)
dτ
dτ
0
F10
τ=0
τ=0
= A1 (v) + A2 (v) + A3 (v) .
For A1 , we have
F min (v) − F min (v) ∑k−2 F min (v)k−2−i F min (v)i k−1
k−1
20
10
20
10
i=0
−
k−1 = (k − 1) min k−1 min k−1
min
F min (v) k−1
F10 (v)
F20 (v)
F10 (v)
20
k−2 min k−2−i F min (v)i 0 ∑i=0 F20 (v)
10
≤ C1 d∞ f1min , f2min v.
k−1
k−1
min (v)
F min (v)
F10
20
∑k−2 F min (v) k−2−i F min (v) i i=0 20
10
min min
≤ C1 d
r
,
r
v.
∞
1
2
F min (v) k−1 F min (v) k−1 20
10
Here C1 is some generic constant and last inequality holds by Lemma... Therefore, for any v
min k−2−i min i ˆ v
min ∑k−2
min k−2
F10 (v) i=0 F20 (v)
min min
A1 (v) ≤ C1 r∆ 2 F20 (s)
ds.
k−1 min k−1 d∞ r1 , r2 v
min
F (v)
0
F10 (v)
20
(12)
Because f1min and f2min are bounded from below by f . There exists an δ > 0 such that for any s < δ ,
f (s) > η for some η > 0. Then for any v ≥ δ ,
min k−2−i min i ˆ v
min ∑k−2
min k−2
F10 (v) i=0
min min
r∆ F20 (v)
d
r
,
r
v
F20 (s)
ds
∞
1
2
2
min (v) k−1 F min (v) k−1 0
F20
10
k−1
≤
C M r∆min 2 d∞ r1min , r2min .
2k−2 3
(ηδ )
35
If v < δ ,
≤
≤
≤
≤
min k−2−i min i ˆ v
min ∑k−2
min k−2
F10 (v) i=0
min min
r∆ F20 (v)
r
,
r
v
d
F20 (s)
ds
∞ 1
2
2
k−1
k−1
min (v)
min (v)
0
F20
F10
∑k−2 F min (v)k−2−i F min (v)i min 10
r∆ d∞ r1min , r2min v2 i=0 20
2
k−1
min (v) F min (v)
F20
10
min (v ) k−2−i f min (v ) i vk−2 min 2 ∑k−2
f
2
1
20
10
i=0
min
min
r∆ d∞ r1 , r2 v min
k−1
2
min
k
f20 (v2 ) f10 (v1 )
v
k−2−i min i min k−2 min
F10 (v) r∆ d∞ r1min , r2min v2 ∑i=0 F20 (v)
2
k
k
η v
C4 r∆min d∞ r1min , r2min .
2
k
η
We used mean value theorem in the inequality. Combine two cases, we have (12). Therefore,
(
A1 ≤ max
)
C4
k−1
,
C3 M r∆min 2 d∞ r1min , r2min = C1 r∆min 2 d∞ r1min , r2min .
2k−2
k
η (ηδ )
.
A similar argument for A2 shows that A2 (v) ≤ C5 r∆min 2 d∞ r1min , r2min .
For A3 ,
´ min min ´ s ˆ s
min (x) 2 dx
min
2r
r
dx
1
+
r
dF2τ
2
2
∆
0
(s)
=p
2 1 + r2min (x) r∆min (x) dx − p
.
h
´ min 2 i2
dτ
0
τ=0
1 + r2
dx
´ min min ´ s ˆ s
min (x) 2 dx
min
min
2r
r
dx
1
+
r
dF1τ
1
1
∆
min
0
(s)
=p
2 1 + r1 (x) r∆ (x) dx − p
.
h
2 i2
´
dτ
0
τ=0
1 + r1min dx
First notice
ˆ s
ˆ s
min
min
min
min
2 1 + r2 (x) r∆ (x) dx −
2 1 + r1 (x) r∆ (x) dx
0
ˆ 0s
min
2 r2 (x) − r1min (x) r∆min (x) dx
≤
0
≤ 2d2 (r1 , r2 ) r∆min 2 .
36
In addition,
≤
≤
≤
≤
≤
≤
´
2rmin rmin dx ´ s 1 + rmin (x)2 dx ´ 2rmin rmin dx ´ s 1 + rmin (x)2 dx 1
1
2
2
∆
∆
0
0
−
i2
i2
h
h
´
´
2
2
1 + r2min dx
1 + r1min dx
2 2 ´ min min ´ s 2r∆ r2 dx 0 1 + r2min (x) − 1 + r1min (x) dx
h
´ min 2 i2
dx
1 + r2
´
´
ˆ s
min r min dx
min r min dx
2r
2r
2
1
2
∆
∆
+
1 + r1min (x) dx h
−
i
h
i
2
2
´
´
2
2
0
1 + r1min dx 1 + r2min dx
´
´ min min
min
min
min r
dx
2r
r
dx
2r
min
min
2
1
∆
∆
C7 r∆ 2 d2 r1 , r2 +C8 h
−
´ min 2 i2 h
´ min 2 i2 dx
1 + r1
dx 1 + r2
´
´
min r min dx
min r min dx
min 2r
2r
1
2
∆
∆
−
C7 r∆ 2 d2 r1min , r2min +C8 h
i
h
i
2
2
´
´
2
2
1 + r2min dx 1 + r2min dx
´
´
2r∆min r1min dx 2r∆min r1min dx
+C8 h
2 i2 ´ min 2 i2 − h
´
dx
1 + r1min dx 1 + r2
ˆ
min min
2C8
r∆ r2 − r1min dx
C7 r∆min 2 d2 r1min , r2min + h
i
2
´
2
1 + r2min dx
ˆ
´ rmin 2 dx − ´ rmin 2 dx 2 + ´ rmin 2 dx + ´ rmin 2 dx
2
1
1
2
+C8 2r∆min r1min dx
h
i
h
i
2
2
2
2
´
´
1 + r2min dx 1 + r1min dx
C7 r∆min 2 d2 r1min , r2min + 2C8 r∆min 2 d2 r2min , r1min
+C8 r∆min 2 r1min 2 2 + r1min 2 + r2min 2 r1min + r2min 2 d2 r1min , r2min
C9 d2 r1min , r2min r∆min 2 .
Therefore, A3 ≤ pC9 r∆min 2 d2 r1min , r2min ≤ pC9 r∆min 2 d∞ r1min , r2min .
Combining results for A1 , A2 and A3 , we proved the lemma.
37
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