Convex hulls of planar random walks with drift

0.33 MB·English

by  Andrew R. Wade

Convex hulls of planar random walks with drift
Andrew R. Wade∗ Chang Xu†
18th January 2013
Abstract
3
DenotebyL thelengthoftheperimeteroftheconvexhullofnstepsofaplanar
1 n
0 random walk whose increments have finite second moment and non-zero mean.
2 Snyder and Steele showed that n−1L converges almost surely to a deterministic
n

n limit, and proved an upper bound on the variance Var[L ] = O(n). We show that
n
a n−1Var[L ] converges and give a simple expression for the limit, which is non-zero
J n
for walks outside a certain degenerate class. This answers a question of Snyder and
7
1 Steele. Furthermore, we prove a central limit theorem for L in the non-degenerate
n

case.
]
R
P
Key words: Convex hull, random walk, variance asymptotics, central limit theorem.
.
h
t AMS Subject Classification: 60G50, 60D05 (Primary) 60J10, 60F05 (Secondary)
a
m
[
1 Introduction and main results

1
v
9 On each of n unsteady steps, a drunken gardener drops a seed. Once the flowers have
5 bloomed, what is the minimum length of fencing required to enclose the garden?
0
4 Let Z1,Z2,... be a sequence of independent, identically distributed (i.i.d.) random
. vectors on R2. Write 0 for the origin in R2. Define the random walk (S ;n ∈ Z ) by
1 n +
0 S := 0 and for n ≥ 1, S := (cid:80)n Z . Let H := hull(S ,...,S ), the convex hull of
3 0 n i=1 i n 0 n
positions of the walk up to and including the nth step, and let L := |∂H | denote the
1 n n
: length of the perimeter of H . Assume that the increments of the random walk have
v n
i finite mean: E(cid:107)Z (cid:107) < ∞.
X 1
Convex hulls of random points have received much attention over the last several
r
a decades: see [3] for an extensive survey, including more than 150 bibliographic references,
and sources of motivation more serious than our drunken gardener, such as modelling
the ‘home-range’ of animal populations. An important tool in the study of random
convex hulls is provided by a result of Cauchy in classical convex geometry. Spitzer and
Widom [5], using Cauchy’s formula, and later Baxter [1], using a combinatorial argument,
showed that
n
(cid:88) 1
E[L ] = 2 E(cid:107)S (cid:107).
n i
i
i=1
∗Department of Mathematical Sciences, Durham University, South Road, Durham DH1 3LE.
†Department of Mathematics and Statistics, University of Strathclyde, 26 Richmond Street, Glasgow
G1 1XH.
1
0
2
0
1
0
0
1
−
0
2
−
0 50 100 150 200 250
Figure 1: Example with mean drift E[Z ] of magnitude µ = 1/4 and n = 103 steps.
1
Note that E[L ] thus scales like n in the case where the one-step mean drift vector
n
E[Z ] (cid:54)= 0 but like n1/2 in the case where E[Z ] = 0. The Spitzer–Widdom–Baxter result,
1 1
in common with much of the literature, is concerned with first-order properties of L :
n
see [3] for a summary of results in this direction for various random convex hulls, with a
specific focus on (driftless) planar Brownian motion.
Much less is known about higher-order properties of L . Assuming that E[(cid:107)Z (cid:107)2] <
n 1
∞, Snyder and Steele [4] obtained an upper bound for Var[L ] using Cauchy’s formula
n
togetherwithaversionoftheEfron–Steininequality. SnyderandSteele’sresult(Theorem
2.3 of [4]) can be expressed as
π2
n−1Var[L ] ≤ (cid:0)E[(cid:107)Z (cid:107)2]−(cid:107)E[Z ](cid:107)2(cid:1), (n ∈ N := {1,2,...}). (1.1)
n 1 1
2
As far as we are aware, there are no lower bounds for Var[L ] in the literature.
n
According to the discussion in [4, §5], Snyder and Steele had “no compelling reason to
expect that O(n) is the correct order of magnitude” in their upper bound for Var[L ],
n
and they speculated that perhaps Var[L ] = o(n) (maybe with a distinction between
n
the cases of zero and non-zero drift). Our first main result settles this question under
minimal conditions, confirming that (1.1) is indeed of the correct order, apart from in
certain degenerate cases, while demonstrating that the constant on the right-hand side
of (1.1) is not, in general, sharp.
Theorem 1.1. Suppose that E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0. Then
1 1
4E[((Z −E[Z ])·E[Z ])2]
lim n−1Var[L ] = 1 1 1 =: σ2 ∈ [0,∞). (1.2)
n→∞ n (cid:107)E[Z1](cid:107)2
Remarks 1.1. (i) The assumptions E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0 ensure σ2 < ∞.
1 1
(ii) To compare the limit result (1.2) with Snyder and Steele’s upper bound (1.1),
observe that
(cid:18)E[(Z ·E[Z ])2]−(cid:107)E[Z ](cid:107)4(cid:19)
σ2 = 4 1 1 1 ≤ 4(cid:0)E[(cid:107)Z (cid:107)2]−(cid:107)E[Z ](cid:107)2(cid:1).
(cid:107)E[Z ](cid:107)2 1 1
1
(iii) The limit σ2 is zero if and only if (Z − E[Z ]) · E[Z ] = 0 with probability
1 1 1
1, i.e., if Z − E[Z ] is always orthogonal to E[Z ]. In such a degenerate case, (1.2)
1 1 1
2
says that Var[L ] = o(n). This is the case, for example, if Z takes values (1,1) and
n 1
(1,−1) each with probability 1/2. Note that the Snyder–Steele bound (1.1) applied in
this example says only that Var[L ] ≤ (π2/2)n, which is not the correct order. Here, the
n
two-dimensional trajectory can be viewed as a space-time trajectory of a one-dimensional
simple symmetric random walk. We conjecture that in fact Var[L ] = O(logn). Steele [6]
n
obtains variance results for the number of faces of the convex hull of one-dimensional
simple random walk, and comments that such results for L seem “far out of reach” [6, p.
n
242].
In the case where E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) = µ > 0, Snyder and Steele deduce from
1 1
their bound (1.1) a strong law of large numbers for L , namely lim n−1L = 2µ, a.s.
n n→∞ n
(see [4, p. 1168]). Given this and the variance asymptotics of Theorem 1.1, it is natural
to ask whether there is an accompanying central limit theorem. Our next result gives a
positive answer in the non-degenerate case, again with essentially minimal assumptions.
Theorem 1.2. Suppose that E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0. Suppose that σ2 as defined
1 1
in (1.2) satisfies σ2 > 0. Then for any x ∈ R,
(cid:20)L −E[L ] (cid:21) (cid:20)L −E[L ] (cid:21)
lim P n n ≤ x = lim P n√ n ≤ x = Φ(x), (1.3)
(cid:112)
n→∞ Var[L ] n→∞ σ2n
n
where Φ is the standard normal distribution function.
Our Theorems 1.1 and 1.2 will be deduced as consequences of the following result,
which shows, perhaps surprisingly, that L −E[L ] can be well-approximated by a sum
n n
of i.i.d. random variables.
Theorem 1.3. Suppose that E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0. Then, as n → ∞,
1 1
(cid:12) (cid:12)
(cid:12) (cid:88)n 2(Z −E[Z ])·E[Z ](cid:12)
n−1/2(cid:12)L −E[L ]− i 1 1 (cid:12) → 0, in L2.
(cid:12) n n (cid:107)E[Z ](cid:107) (cid:12)
(cid:12) 1 (cid:12)
i=1
The subsequent sections of the paper present the proofs of these theorems. The main
ingredients, which we present in turn, include a martingale difference representation,
Cauchy’s formula from convex geometry, and an analysis of the geometry of the convex
hull via extrema (the strong law of large numbers with the non-zero drift provides much
of the regularity that we need).
To finish this section we discuss some simulations. We considered a specific form
of random walk with increments Z − E[Z ] = (cosΘ ,sinΘ ), where Θ was uniformly
i i i i i
distributed on [0,2π), corresponding to a uniform distribution on a unit circle centred
at E[Z ] = (µ,0), say. We took one example with µ = 0, and two examples with
i
µ (cid:54)= 0 of different magnitudes. In these latter cases, the results above take the form:
lim n−1Var[L ] = 4E[cos2Θ ] = 2 (Theorem 1.1) and (2n)−1/2(L −E[L ]) converges
n→∞ n 1 n n
in distribution to a standard normal distribution (Theorem 1.2). The corresponding
pictures in Figures 2 and 3 show an agreement between the simulations and theory.
The results of this paper do not cover the case where (cid:107)E[Z ](cid:107) = 0. The simulations
1
in this case suggest that, for the example we considered, lim n−1Var[L ] exists (see
n→∞ n
the leftmost plot in Figure 2), but Figure 3 does not appear to be consistent with a
normal distribution as a limiting distribution. The method of the present paper provides
a promising approach to the zero-drift case, but a new idea will be needed to gain control
over the geometry in that case.
3
l
l ll
y1 +002e+054e+05 lllll l llllllllllllllllllly2 +002e+054e+05 lllllll l llllllllllllllllll y3 +002e+054e+05 lllllll l lllllllllllllllll
e e e
0 0 0
0 50000 150000 250000 0 50000 150000 250000 0 50000 150000 250000
x x x
Figure 2: Plots of y = Var[L ] estimates against x = n for about 25 values of n in the
n
range 102 to 2.5×105 for 3 examples with µ = (left to right) 0, 0.2, 0.36. Each point is
estimated from the sample variance of 103 repeated simulations. Also plotted are straight
lines y = 0.536x (leftmost plot) and y = 2x (other two plots).
Histogram of (len − mean(len))/Hsiqsrtto(vgarar(mle no)f) (len1 − mean(len1))/sHqirstt(ovagrr(alemn 1o)f) (len − mean(len))/sqrt(var(len))
4
0. 0.4
3
y 0.3 y 0.3 y 0.
ensit 0.2 ensit 0.2 ensit 0.2
D D D
0.1 0.1 0.1
0 0 0
0. 0. 0.
−2 0 1 2 3 4 −3 −1 0 1 2 3 −3 −1 1 2 3
Figure(l3en: −S mimeaun(llaent)e)d/sqhrti(vsatro(lgenr)a)m estim(laent1e s− fmoerant(hleen1d))i/ssqtrrti(vbaur(tleino1n))of √Ln−E([lLenn ]− wmietahn(lnen)=)/sq5rt×(va1r(0le3n))
Var[Ln]
inthethreeexamplesdescribedinFigure2. Eachhistogramiscompiledfrom103 samples.
2 Martingale difference representation
The first step in the proofs is a martingale difference argument, based on resampling
members of the sequence Z ,...,Z , to get an expression for Var[L ] amenable to ana-
1 n n
lysis. Let F denote the trivial σ-algebra, and for n ∈ N set F := σ(Z ,...,Z ), the
0 n 1 n
σ-algebra generated by the first n steps of the random walk. Then S is F -measurable,
n n
and for n ∈ N we can write L = Λ (Z ,...,Z ) for Λ : R2n → [0,∞) a measurable
n n 1 n n
function.
Let Z(cid:48),Z(cid:48),... be an independent copy of the sequence Z ,Z ,.... Fix n ∈ N. For
1 2 1 2
i ∈ {1,...,n}, we ‘resample’ the ith increment, replacing Z with Z(cid:48), as follows. Set
i i
(cid:40)
S if j < i
S(i) := j (2.1)
j S −Z +Z(cid:48) if j ≥ i;
j i i
then (S(i);0 ≤ j ≤ n) is the random walk (S ;0 ≤ j ≤ n) but with the ith step
j j
independently resampled. We let L(i) denote the perimeter length of the corresponding
n
convex hull for this modified walk, namely hull(S(i),...,S(i)), i.e.,
0 n
L(i) := Λ (Z ,...,Z ,Z(cid:48),Z ,...,Z ).
n n 1 i−1 i i+1 n
For i ∈ {1,...,n}, define
D := E[L −L(i) | F ]; (2.2)
n,i n n i
4
in words, −D is the expected change in the perimeter length of the convex hull, given
n,i
F , on replacing Z by Z(cid:48). The point of this construction is the following result.
i i i
Lemma 2.1. Let n ∈ N. Then (i) L − E[L ] = (cid:80)n D ; and (ii) Var[L ] =
n n i=1 n,i n
(cid:80)n E[D2 ], whenever the latter sum is finite.
i=1 n,i
Proof. We may rewrite (2.2) as
D = E[L | F ]−E[L | F ]. (2.3)
n,i n i n i−1
Indeed, since L(i) is independent of Z , E[L(i) | F ] = E[L(i) | F ] = E[L | F ]. Hence
n i n i n i−1 n i−1
by (2.3), we obtain the representation (cid:80)n D = E[L | F ] − E[L | F ], giving (i).
i=1 n,i n n n 0
Here (D ;1 ≤ i ≤ n) is a martingale difference sequence, since D is F -measurable
n,i n,i i
and E[D | F ] = 0. By orthogonality of martingale differences (see e.g. [2, p. 218]),
n,i i−1
n n n
(cid:88) (cid:88) (cid:88)
Var[L ] = Var D = Var[D ] = E[D2 ],
n n,i n,i n,i
i=1 i=1 i=1
since E[D ] = 0, which gives (ii).
n,i
Remark 2.1. Lemma 2.1 with the conditional Jensen’s inequality gives the upper bound
n
(cid:88)
Var[L ] ≤ E[(L(i) −L )2],
n n n
i=1
which is a factor of 2 larger than the upper bound obtained from the Efron–Stein in-
equality: see equation (2.3) in [4].
3 Cauchy formula
Let e = (cosθ,sinθ) be the unit vector in direction θ ∈ (−π,π]. For θ ∈ [0,π], define
θ
M (θ) := max(S ·e ), and m (θ) := min (S ·e ).
n j θ n j θ
0≤j≤n 0≤j≤n
Note that since S = 0, we have M (θ) ≥ 0 and m (θ) ≤ 0, a.s. In the present setting
0 n n
(see [4], formula (2.1)), Cauchy’s formula for convex sets yields
(cid:90) π (cid:90) π
L = (M (θ)−m (θ))dθ = R (θ)dθ,
n n n n
0 0
where R (θ) := M (θ)−m (θ) ≥ 0 is the parametrized range function. Similarly, when
n n n
the ith increment is resampled as described in Section 2,
(cid:90) π (cid:90) π
(cid:0) (cid:1)
L(i) = M(i)(θ)−m(i)(θ) dθ = R(i)(θ)dθ,
n n n n
0 0
where R(i)(θ) = M(i)(θ)−m(i)(θ), defining
n n n
M(i)(θ) := max(S(i) ·e ), and m(i)(θ) := min (S(i) ·e ).
n j θ n j θ
0≤j≤n 0≤j≤n
5
Thus to study D = E[L −L(i) | F ] we will consider
n,i n n i
(cid:90) π (cid:90) π
(cid:0) (cid:1)
L −L(i) = R (θ)−R(i)(θ) dθ = ∆(i)(θ)dθ, (3.1)
n n n n n
0 0
where ∆(i)(θ) := R (θ)−R(i)(θ). For θ ∈ [0,π], let
n n n
¯
J (θ) := argmin(S ·e ), and J (θ) := argmax(S ·e ),
n j θ n j θ
0≤j≤n 0≤j≤n
so m (θ) = S ·e and M (θ) = S ·e . Similarly, recalling (2.1), define
n Jn(θ) θ n J¯n(θ) θ
J(i)(θ) := argmin(S(i) ·e ), and J¯(i)(θ) := argmax(S(i) ·e ).
n j θ n j θ
0≤j≤n 0≤j≤n
We will use the following simple bound repeatedly in the arguments that follow. In
fact, with a little more work one can reduce the bound on the right-hand side of (3.2) by
a factor of 2 (cf [4], Lemma 2.1), but the form given here is good enough for us.
Lemma 3.1. Almost surely, for any θ ∈ [0,π] and any i ∈ {1,2,...,n},
|∆(i)(θ)| ≤ 2(cid:107)Z (cid:107)+2(cid:107)Z(cid:48)(cid:107). (3.2)
n i i
Proof. The triangle inequality implies that
|∆(i)(θ)| ≤ |M(i)(θ)−M (θ)|+|m(i)(θ)−m (θ)|.
n n n n n
For some J¯ (θ) ∈ {0,1,...,n}, we have M (θ) = S ·e and, by definition, M(i)(θ) ≥
n n J¯n(θ) θ n
S(i) ·e . If J¯ (θ) < i, then, by (2.1), S(i) = S and so M(i)(θ) ≥ M (θ). Otherwise,
J¯n(θ) θ n J¯n(θ) J¯n(θ) n n
if J¯ (θ) ≥ i, then, by (2.1), S(i) = S −Z +Z(cid:48) and so
n J¯n(θ) J¯n(θ) i i
M(i)(θ) ≥ S ·e −Z ·e +Z(cid:48) ·e
n J¯n(θ) θ i θ i θ
≥ M (θ)−(cid:107)Z (cid:107)−(cid:107)Z(cid:48)(cid:107).
n i i
Hence we conclude that, a.s., M(i)(θ) ≥ M (θ)−(cid:107)Z (cid:107)−(cid:107)Z(cid:48)(cid:107). The analogous argument
n n i i
in the other direction shows that |M(i)(θ)−M (θ)| ≤ (cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107). Moreover, a similar
n n i i
argument shows that the same bound holds for |m(i)(θ)−m (θ)|, and (3.2) follows.
n n
4 Control of extrema
For the remainder of the paper, without loss of generality, we suppose that E[Z ] = µe
1 π/2
with µ ∈ (0,∞). Observe that (S · e ;0 ≤ j ≤ n) is a one-dimensional random walk:
j θ
indeed, S ·e = (cid:80)j Z ·e . The mean drift of this one-dimensional random walk is
j θ k=1 k θ
E[Z ·e ] = E[Z ]·e = µsinθ. (4.1)
1 θ 1 θ
Note that the drift µsinθ is positive if θ ∈ (0,π). This crucial fact gives us control
over the behaviour of the extrema such as M (θ) and m (θ) that contribute to (3.1), and
n n
this will allow us to estimate the conditional expectation of the final term in (3.1) (see
Lemma 5.1 below).
For γ ∈ (0,1/2) and δ ∈ (0,π/2) (two constants that will be chosen to be suitably
small later in our arguments), we denote by E (δ,γ) the event that the following occur:
n,i
6
¯
• for all θ ∈ [δ,π −δ], J (θ) < γn and J (θ) > (1−γ)n;
n n
• for all θ ∈ [δ,π −δ], J(i)(θ) < γn and J¯(i)(θ) > (1−γ)n.
n n
We write Ec (δ,γ) for the complement of E (δ,γ). The idea is that E (δ,γ) will occur
n,i n,i n,i
withhighprobability, andonthiseventwehavegoodcontrolover∆(i)(θ). Thenextresult
n
formalizes these assertions. For γ ∈ (0,1/2), define I := {1,...,n}∩[γn,(1−γ)n].
n,γ
Lemma 4.1. For any γ ∈ (0,1/2) and any δ ∈ (0,π/2), the following hold.
(i) If i ∈ I , then, a.s., for any θ ∈ [δ,π −δ],
n,γ
∆(i)(θ)1(E (δ,γ)) = (Z −Z(cid:48))·e 1(E (δ,γ)). (4.2)
n n,i i i θ n,i
(ii) If E(cid:107)Z (cid:107) < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0, then min P[E (δ,γ)] → 1 as n → ∞.
1 1 1≤i≤n n,i
Proof. First we prove part (i). Suppose that i ∈ I , so γn ≤ i ≤ (1−γ)n. Suppose that
n,γ
θ ∈ [δ,π −δ]. Then on E (δ,γ), we have J (θ) < i < J¯ (θ) and J(i)(θ) < i < J¯(i)(θ).
n,i n n n n
Then from (2.1) it follows that in fact J (θ) = J(i)(θ) and J¯ (θ) = J¯(i)(θ). Hence
n n n n
m (θ) = m(i)(θ) and M(i)(θ) = S(i) ·e = M (θ)+ (Z(cid:48) −Z )·e , by (2.1). Equation
n n n J¯n(θ) θ n i i θ
(4.2) follows.
Next we prove part (ii). Suppose that µ = (cid:107)E[Z ](cid:107) > 0. Since E(cid:107)Z (cid:107) < ∞, the
1 1
strong law of large numbers implies that (cid:107)n−1S − E[Z ](cid:107) → 0, a.s., as n → ∞. In
n 1
other words, for any ε > 0, there exists N := N(ε ) such that P[N < ∞] = 1 and
1 1
(cid:107)n−1S −E[Z ](cid:107) < ε for all n ≥ N. In particular, for n ≥ N, by (4.1),
n 1 1
(cid:12)(cid:12)n−1S ·e −µsinθ(cid:12)(cid:12) = (cid:12)(cid:12)n−1S ·e −E[Z ]·e (cid:12)(cid:12) ≤ (cid:13)(cid:13)n−1S −E[Z ](cid:13)(cid:13) < ε , (4.3)
n θ n θ 1 θ n 1 1
for all θ ∈ [0,2π).
Take ε < µsinδ. If n ≥ N, then, by (4.3),
1
S ·e > (µsinθ−ε )n ≥ (µsinδ −ε )n,
n θ 1 1
provided θ ∈ [δ,π −δ]. By choice of ε , the last term in the previous display is strictly
1
positive. Hence, for n ≥ N, for any θ ∈ [δ,π −δ], S ·e > 0. But, S ·e = 0. So
n θ 0 θ
P(cid:2)∩ {J (θ) < γn}(cid:3) ≥ P[N < γn] → 1,
θ∈[δ,π−δ] n
as n → ∞, since N < ∞ a.s.
Now,
(cid:26) (cid:27)
max S ·e ≤ max max S ·e , max S ·e . (4.4)
j θ j θ j θ
0≤j≤(1−γ)n 0≤j≤N N≤j≤(1−γ)n
For the final term on the right-hand side of (4.4), (4.3) implies that
max S ·e ≤ max (µsinθ+ε )j ≤ (µsinθ+ε )(1−γ)n.
j θ 1 1
N≤j≤(1−γ)n 0≤j≤(1−γ)n
On the other hand, if n ≥ N, then (4.3) implies that S · e ≥ (µsinθ − ε )n. Here
n θ 1
µsinθ −ε ≥ (µsinθ +ε )(1−γ) if ε < γµsinθ. Now we choose ε < γµsinδ. Then, for
1 1 1 2−γ 1 2
any θ ∈ [δ,π −δ], we have that, for n ≥ N,
S ·e > max S ·e .
n θ j θ
N≤j≤(1−γ)n
7
Hence, by (4.4),
(cid:20) (cid:26) (cid:27)(cid:21)
P(cid:2)∩ {J¯ (θ) > (1−γ)n}(cid:3) ≥ P ∩ S ·e > max S ·e
θ∈[δ,π−δ] n θ∈[δ,π−δ] n θ j θ
0≤j≤(1−γ)n
(cid:20) (cid:26) (cid:27)(cid:21)
≥ P N ≤ n, ∩ S ·e > max S ·e .
θ∈[δ,π−δ] n θ j θ
0≤j≤N
Also, for n ≥ N, S ·e > (1− γ)µnsinδ, so we obtain
n θ 2
(cid:20) (cid:21)
(cid:16) γ(cid:17)
P(cid:2)∩ {J¯ (θ) > (1−γ)n}(cid:3) ≥ P N ≤ n, max (cid:107)S (cid:107) ≤ 1− µnsinδ ,
θ∈[δ,π−δ] n j
0≤j≤N 2
using the fact that max S ·e ≤ max (cid:107)S (cid:107) for all θ.
0≤j≤N j θ 0≤j≤N j
Now, as n → ∞, P[N > n] → 0, and
(cid:20) (cid:21)
(cid:16) γ(cid:17)
P max (cid:107)S (cid:107) > 1− µnsinδ → 0,
j
0≤j≤N 2
since N < ∞ a.s. So we conclude that
P(cid:2)∩ {J (θ) < γn, J¯ (θ) > (1−γ)n}(cid:3) → 1,
θ∈[δ,π−δ] n n
as n → ∞, and the same result holds for J(i)(θ) and J¯(i)(θ), uniformly in i ∈ {1,...,n},
n n
since resampling Z does not change the distribution of the trajectory.
i
5 Approximation lemma
The following result is a key component to our proof. Recall that D = E[L −L(i) | F ].
n,i n n i
Lemma 5.1. Suppose that E(cid:107)Z (cid:107) < ∞, γ ∈ (0,1/2), and δ ∈ (0,π/2). For any i ∈ I ,
1 n,γ
(cid:12)(cid:12)(cid:12)D − 2(Zi −E[Z1])·E[Z1](cid:12)(cid:12)(cid:12) ≤ 6δ(cid:107)Z (cid:107)+6δE(cid:107)Z (cid:107)+3π(cid:107)Z (cid:107)P[Ec (δ,γ) | F ]
(cid:12) n,i (cid:107)E[Z ](cid:107) (cid:12) i 1 i n,i i
1
+3πE[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ], a.s. (5.1)
i n,i i
Proof. Taking (conditional) expectations in (3.1), we obtain
(cid:90) π (cid:90) π
D = E[∆(i)(θ)1(E (δ,γ)) | F ]dθ+ E[∆(i)(θ)1(Ec (δ,γ)) | F ]dθ. (5.2)
n,i n n,i i n n,i i
0 0
For the second term on the right-hand side of (5.2), we have
(cid:12)(cid:90) π (cid:12) (cid:90) π
(cid:12)(cid:12) E[∆(i)(θ)1(Ec (δ,γ)) | F ]dθ(cid:12)(cid:12) ≤ E[|∆(i)(θ)|1(Ec (δ,γ)) | F ]dθ. (5.3)
(cid:12) n n,i i (cid:12) n n,i i
0 0
Applying the bound (3.2), we obtain
(cid:90) π
E[|∆(i)(θ)|1(Ec (δ,γ)) | F ]dθ ≤ 2πE[((cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107))1(Ec (δ,γ)) | F ]
n n,i i i i n,i i
0
= 2π(cid:107)Z (cid:107)P[Ec (δ,γ) | F ]+2πE[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ], (5.4)
i n,i i i n,i i
8
since Z is F -measurable with E(cid:107)Z (cid:107) < ∞.
i i i
We decompose the first integral on the right-hand side of (5.2) as I +I +I , where
1 2 3
(cid:90) δ
I := E[∆(i)(θ)1(E (δ,γ)) | F ]dθ,
1 n n,i i
0
(cid:90) π−δ
I := E[∆(i)(θ)1(E (δ,γ)) | F ]dθ,
2 n n,i i
δ
(cid:90) π
I := E[∆(i)(θ)1(E (δ,γ)) | F ]dθ.
3 n n,i i
π−δ
First we deal with I and I . We have
1 3
(cid:90) δ
|I | ≤ E[|∆(i)(θ)| | F ]dθ ≤ 2δE[(cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107) | F ], a.s.,
1 n i i i i
0
by another application of (3.2). Here E[(cid:107)Z (cid:107) | F ] = (cid:107)Z (cid:107), since Z is F -measurable,
i i i i i
and, since Z(cid:48) is independent of F , E[(cid:107)Z(cid:48)(cid:107) | F ] = E(cid:107)Z(cid:48)(cid:107) = E(cid:107)Z (cid:107). A similar argument
i i i i i 1
applies to I , so that
3
|I +I | ≤ 4δ(cid:107)Z (cid:107)+4δE(cid:107)Z (cid:107), a.s. (5.5)
1 3 i 1
We now consider I . From (4.2), since i ∈ I , we have
2 n,γ
(cid:90) π−δ
I = E[(Z −Z(cid:48))·e 1(E (δ,γ)) | F ]dθ
2 i i θ n,i i
δ
(cid:90) π−δ (cid:90) π−δ
= E[(Z −Z(cid:48))·e | F ]dθ− E[(Z −Z(cid:48))·e 1(Ec (δ,γ)) | F ]dθ.
i i θ i i i θ n,i i
δ δ
Here, by the triangle inequality,
(cid:12)(cid:90) π−δ (cid:12) (cid:90) π
(cid:12)(cid:12) E[(Z −Z(cid:48))·e 1(Ec (δ,γ)) | F ]dθ(cid:12)(cid:12) ≤ E[((cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107))1(Ec (δ,γ)) | F ]dθ
(cid:12) i i θ n,i i (cid:12) i i n,i i
δ 0
= π(cid:107)Z (cid:107)P[Ec (δ,γ) | F ]+πE[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ], (5.6)
i n,i i i n,i i
similarly to (5.4). Finally, similarly to (5.5),
(cid:12)(cid:90) π−δ (cid:90) π (cid:12)
(cid:12)(cid:12) E[(Z −Z(cid:48))·e | F ]dθ− E[(Z −Z(cid:48))·e | F ]dθ(cid:12)(cid:12) ≤ 2δE[(cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107) | F ]
(cid:12) i i θ i i i θ i (cid:12) i i i
δ 0
= 2δ((cid:107)Z (cid:107)+E(cid:107)Z (cid:107)). (5.7)
i 1
We combine (5.2) with (5.3) and the bounds in (5.4), (5.5), (5.6) and (5.7) to give
(cid:12) (cid:90) π (cid:12)
(cid:12)(cid:12)D − E[(Z −Z(cid:48))·e | F ]dθ(cid:12)(cid:12) ≤ 6δ(cid:107)Z (cid:107)+6δE(cid:107)Z (cid:107)+3π(cid:107)Z (cid:107)P[Ec (δ,γ) | F ]
(cid:12) n,i i i θ i (cid:12) i 1 i n,i i
0
+3πE[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ], a.s. (5.8)
i n,i i
To complete the proof of the lemma, we compute the integral on the left-hand side of
(5.8). First note that E[(Z −Z(cid:48))·e | F ] = (Z −E[Z(cid:48)])·e , since Z is F -measurable
i i θ i i i θ i i
and Z(cid:48) is independent of F , so that
i i
(cid:90) π (cid:90) π
E[(Z −Z(cid:48))·e | F ]dθ = (Z −E[Z ])·e dθ.
i i θ i i i θ
0 0
9
To evaluate the last integral, it is convenient to introduce the notation Z −E[Z ] = R e
i i i Θi
where R = (cid:107)Z −E[Z ](cid:107) ≥ 0 and Θ ∈ [0,2π). Then
i i i i
(cid:90) π (cid:90) π (cid:90) π
(Z −E[Z ])·e dθ = R e ·e dθ = R cos(θ−Θ )dθ
i i θ i Θi θ i i
0 0 0
= 2R sinΘ = 2R e ·e .
i i i Θi π/2
Now (5.1) follows from (5.8), and the proof is complete.
6 Completing the proofs of the theorems
For ease of notation, we write Y := 2(cid:107)E[Z ](cid:107)−1(Z −E[Z ])·E[Z ], and define
i 1 i 1 1
W := D −Y .
n,i n,i i
The upper bound for |W | in Lemma 5.1 together with Lemma 4.1(ii) will enable us to
n,i
prove the following result, which will be the basis of our proof of Theorem 1.3.
Lemma 6.1. Suppose that E[(cid:107)Z (cid:107)2] < ∞ and (cid:107)E[Z ](cid:107) =(cid:54) 0. Then
1 1
n
(cid:88)
lim n−1 E[W2 ] = 0.
n,i
n→∞
i=1
Proof. Fix ε > 0. We take γ ∈ (0,1/2) and δ ∈ (0,π/2), to be specified later. We divide
the sum of interest into two parts, namely i ∈ I and i ∈/ I . Now from (3.1) with
n,γ n,γ
(3.2) we have |L(i) −L | ≤ 2π((cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107)), a.s., so that
n n i i
|D | ≤ 2πE[(cid:107)Z (cid:107)+(cid:107)Z(cid:48)(cid:107) | F ] = 2π((cid:107)Z (cid:107)+E(cid:107)Z (cid:107)).
n,i i i i i i
It then follows from the triangle inequality that
|W | ≤ |D |+2(cid:107)Z −E[Z ](cid:107) ≤ (2π +2)((cid:107)Z (cid:107)+E(cid:107)Z (cid:107)).
n,i n,i i i i i
So provided E[(cid:107)Z (cid:107)2] < ∞, we have E[W2 ] ≤ C for all n and all i, for some constant
1 n,i 0
C < ∞, depending only on the distribution of Z . Hence
0 1
1 (cid:88) 1
E[W2 ] ≤ 2γnC = 2γC ,
n n,i n 0 0
i∈/In,γ
using the fact that there are at most 2γn terms in the sum. From now on, choose γ > 0
small enough so that 2γC < ε.
0
Now consider i ∈ I . For such i, (5.1) shows that, for some constant C < ∞,
n,γ 1
|W | ≤ C (1+(cid:107)Z (cid:107))δ+C (cid:107)Z (cid:107)P[Ec (δ,γ) | F ])+C E[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ], a.s. (6.1)
n,i 1 i 1 i n,i i 1 i n,i i
Here, for any B ∈ (0,∞), a.s.,
1
E[(cid:107)Z(cid:48)(cid:107)1(Ec (δ,γ)) | F ] ≤ E[(cid:107)Z(cid:48)(cid:107)1{(cid:107)Z(cid:48)(cid:107) > B } | F ]+B P[Ec (δ,γ) | F ]
i n,i i i i 1 i 1 n,i i
= E[(cid:107)Z(cid:48)(cid:107)1{(cid:107)Z(cid:48)(cid:107) > B }]+B P[Ec (δ,γ) | F ],
i i 1 1 n,i i
10