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ON THE RECURRENCE OF RANDOM WALKS
SIMON LAZARUS
Abstract. We define random walks on Rd and recurrent points and demonstrate that a random walk’s recurrence to 0 implies its recurrence to each of
its possible points. We then prove two different necessary and sufficient conditions for the recurrence of random walks. Finally, we employ these results to
provide conditions under which random walks on R and on R2 are recurrent
as well as prove that when d ≥ 3, no truly d-dimensional random walk on Rd
is recurrent.
Contents
1. Preliminaries
2. Random Walks and Recurrence
3. Sufficient Conditions for Recurrence and Transience
Acknowledgments
References
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4
16
23
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1. Preliminaries
In this section, we provide some basic definitions from probability theory and
introduce some notation so that our use of various terms later in this paper shall
be clear. We also provide (without proof) several basic but important results from
probability theory that we shall employ in our discussion of random walks. For
proofs of these results, see [2]. Lastly, we present two examples which we shall later
find useful.
Below, ℘(A) denotes the collection of all subsets of A.
Definition 1.1. A probability space is a triple (Ω, F, P ), where Ω is a set, F ⊆ ℘(Ω)
is a σ-algebra, and P : F → R is a measure (a nonnegative countably additive set
function) such that P (Ω) = 1.
We refer to Ω as the sample space; its elements are outcomes. F is the set of
events, and P is the probability measure of the space. Given A ∈ F we call P (A)
the probability of the event A.
Definition 1.2. Let d ∈ N. (i) A function X : Ω → Rd is said to be a random
variable on Rd if for all Borel sets B ⊆ Rd we have X −1 (B) ∈ F. (ii) Given a
Borel set B ⊆ Rd and a random variable X on Rd , we define the event that X ∈ B
(denoted {X ∈ B}) as X −1 (B).
Date: September 15, 2013.
1
2
SIMON LAZARUS
If d = 1, we simply call X a random variable.
One important elementary result is that sums, products, supremums, and limit
superiors of random variables are themselves random variables; and in general any
Borel measurable function of random variables on Rn that maps to Rm is itself a
random variable on Rm .
Given a Borel set B ⊆ Rd and a random variable X on Rd , we can naturally
define the probability that X is in B, denoted P (X ∈ B), as P ({X ∈ B}). (This is
well defined since Definition 1.2 guarantees that {X ∈ B} ∈ F.) Letting Bd denote
the collection of Borel sets in Rd , it follows from the above definitions that if we
also define µX : Bd → R by µX (B) = P (X ∈ B), then µX is a measure on Bd such
that µX (Rd ) = 1. We call µX the probability measure of X.
In general, any measure µ on Bd such that µ(Rd ) = 1 is called a probability
measure.
Definition 1.3. Let X be a random variable.
The expectation (or expected value
R
or mean) of X is defined as E[X] = Ω XdP whenever this integral exists (this
includes the possibilities E[X] = ∞ and E[X] = −∞).
We next present a theorem that will help us compute expected values.
Theorem 1.4. Let X be a random variable on Rd and let g : Rd → R be Borel
measurable. If g ≥ 0 or if E|g(X)| < ∞, then
Z
E[g(X)] =
g dµX .
Rd
We now discuss density functions and characteristic functions of probability measures. Note that since every random variable uniquely determines a probability
measure, our coming definitions of density functions and characteristic functions of
probability measures immediately define density functions and characteristic functions of random variables.
R
In the definition that follows and throughout this paper, we shall use B f (x)dx
to denote the integral of f over B with respect to Lebesgue measure.
Definition 1.5. Let µ be a probability measure
on Bd . If f : Rd → R is measurable
R
such that for all B ∈ Bd we have µ(B) = B f (x)dx, we say f is a density function
for µ.
It is clear that any measurable fR : Rd → R can only be a density function for
some probability measure on Bd if Rd f (x)dx = 1 and f ≥ 0 almost everywhere.
It is also apparent that every f satisfying these conditions is in fact the density for
some probability measure on BdR. For given such an f , we can define a probability
measure µf on Bd by µf (B) = B f (x)dx.
In this way, we see that there may be multiple density functions corresponding
to a single probability measure (as such densities can differ on a set of measure 0),
but there is exactly one probability measure corresponding to a given density.
Definition 1.6. Let µ beR a probability measure on Bd . Then the function ϕµ :
Rd → R given by ϕµ (t) = Rd eit·y dµ(y) is called the characteristic function of µ.
R
R
R
Here, Rd eit·y dµ(y) simply means Rd cos(t · y)dµ(y) + i Rd sin(t · y)dµ(y). We
note that given any t ∈ Rd , these integrals always exist and are always finite, as
follows. For all x ∈ Rd , let f (x) = cos(t · x) and g(x) = sin(t · x). Then |f |, |g| ≤ 1
ON THE RECURRENCE OF RANDOM WALKS
3
R
R
on Rd , soRsince µ is a probability measure, Rd |f (x)|dµ(x) ≤ Rd 1dµ(x) = 1 and
similarly Rd |g(x)|dµ(x) ≤ 1.
We also note that given a random variable X on Rd , we can define the characteristic function of X, denoted ϕX , as ϕµX . In this case, by similar reasoning
to the above paragraph we can apply Theorem 1.4 to find that for all t ∈ Rd ,
ϕX (t) = E[cos(t · X)] + iE[sin(t · X)]. We write this as ϕX (t) = E[eit·X ].
We now present some fairly simple facts about characteristic functions and an
example of a characteristic function that will be relevant later.
Theorem 1.7. Let X1 , X2 be independent random variables on Rd and let ϕ1 , ϕ2
be their characteristic functions. Then ϕ1 is uniformly continuous, ϕ1 (0) = 1, and
for all t ∈ Rd we have |ϕ1 (t)| ≤ 1 and ϕ1 (−t) = ϕ1 (t). Additionally, ϕ1 · ϕ2 is the
characteristic function of X1 + X2 .
Example 1.8. Let r > 0. Consider the “triangular” random variable T given
if |x| ≤ r
by the density function g, where for all x ∈ R we have g(x) = r−|x|
r2
and g(x) = 0 otherwise. It can be shown that T is equivalent to the sum of
two independent random variables (call them X1 and X2 ) which are uniformly
distributed on [−r/2, r/2]. By Theorem 1.7, then, ϕT = ϕX1 · ϕX2 = ϕ2X1 . Now
X1 has density f , where for all x ∈ R, f (x) = 1/r if x ∈ [−r/2, r/2] and f (x) = 0
otherwise. Hence, for all t ∈ R \ {0} we have
Z
Z
1
1
2 sin( tr
1 r/2 itx
1 e 2 itr − e− 2 itr
2)
ϕX1 (t) =
eitx f (x)dx =
=
.
e dx = ·
r
r
it
tr
R
−r/2
So since ϕT = ϕ2X1 , this means that for all t ∈ R \ {0},
2
1−cos(2· tr )
2
4 sin2 ( tr
4
2
2)
2
ϕT (t) =
=
=
=
(1 − cos(tr)).
(tr)2
(tr)2
(tr)2
R
R r/2
We also have that ϕX1 (0) = R e0·ix f (x)dx = 1r −r/2 1dx = 1, so ϕT (0) = 12 = 1.
2 sin( tr
2)
tr
Next, we present two important theorems about characteristic functions (the
Inversion Formula and Pólya’s Criterion) and employ them in another example
that will be relevant later.
Theorem 1.9.
R Let µ be a probability measure on B1 and let ϕ be its characteristic
function. If R |ϕ(x)|dx < ∞, then µ has a density function f given by: ∀y ∈ R,
Z
1
e−ity ϕ(t)dt.
f (y) =
2π R
In this case, f is continuous and bounded on R.
Theorem 1.10. Let ϕ : R → R be even, continuous on R, and convex on (0, ∞)
and suppose ϕ(0) = 1 and limx→∞ ϕ(x) = 0. Then ϕ is the characteristic function
for some probability measure µ on B1 .
Example 1.11. Let r > 0. For all t ∈ R, let ϕ(t) = 1 − |rt| if |t| ≤ 1r and
ϕ(t) = 0 otherwise. Then ϕ clearly satisfies the hypotheses of Theorem 1.10,
so
function of some probability measure µ on B1 . Since
R ϕ is the characteristic
1
|ϕ(x)|dx
=
<
∞,
by
Theorem 1.9 we therefore know that µ has density f ,
r
R
4
SIMON LAZARUS
where for all y ∈ R we have
1
f (y) =
2π
Z
1/r
e−ity (1 − |rt|)dt.
−1/r
By considering the two cases y = 0 and y 6= 0, evaluating this integral is straight1
forward (albeit somewhat tedious), and it can be shown that f (0) = 2πr
and that
for all y ∈ R \ {0},
r(1 − cos( yr ))
.
f (y) =
πy 2
Having now discussed characteristic functions and density functions at length,
we close this section by defining three types of convergence which we shall later
employ and discussing the notion of “infinitely often.”
Below, | · | is the Euclidean norm on Rd , and A ⊆ Rd is said to be a continuity
set of X if A is a Borel set and P (X ∈ ∂A) = 0.
Definition 1.12. Let X, X1 , X2 , . . . be random variables on Rd . (i) If we have
P (limn→∞ Xn = X) = 1, then we say {Xn } converges almost surely to X and
a.s.
we write Xn → X. (ii) If for all > 0 we have limn→∞ P (|Xn − X| ≥ ) = 0,
p
then we say {Xn } converges in probability to X and we write Xn → X. (iii) If
limn→∞ P (Xn ∈ A) = P (X ∈ A) for all A ⊆ Rd that are continuity sets of X, then
d
we say {Xn } converges in distribution to X and we write Xn → X.
It is a fact that convergence almost surely implies convergence in probability,
which in turn implies convergence in distribution.
Definition 1.13. Let A1 , A2 , . . . ∈ F. Then the event An i.o. is defined as
∞
∩∞
k=1 ∪n=k An .
An i.o. is read “An infinitely often.” Note that (An i.o.) ∈ F since F is a
σ-algebra. We see that An i.o. is the collection of outcomes ω which are elements
of infinitely many of the events A1 , A2 , · · · . Intuitively, An i.o. occurs if and only
if infinitely many of the events A1 , A2 , · · · occur.
As a final note in this section, we prove the Borel-Cantelli Lemma.
P∞
Theorem 1.14. Let A1 , A2 , . . . ∈ F. If n=0 P (An ) < ∞, then P (An i.o.) = 0.
P∞
P∞
Proof. If n=0 P (An ) < ∞, then certainly inf k∈N n=k P (An ) = 0. Now note
that for any events B1 , B2 , . . . ∈ F we have P (∩∞
r=1 Br ) ≤ inf m∈N P (Bm ) since for
all m ∈ N we have ∩∞
r=1 Br ⊆ Bm . Thus,
!
!
∞ [
∞
∞
∞
\
[
X
P (An i.o.) = P
An ≤ inf P
An ≤ inf
P (An ) = 0,
k=1 n=k
k∈N
n=k
k∈N
n=k
where the second inequality follows from the countable subadditivity of P .
2. Random Walks and Recurrence
Definition 2.1. Let X1 , X2 , . . . be i.i.d. random variables on Rd . Then the sequence {Sn } given by S0 = 0 ∈ Rd and Sn = X1 + . . . + Xn is said to be a random
walk on Rd .
ON THE RECURRENCE OF RANDOM WALKS
5
In this section, we shall take up the question: what are necessary and sufficient
conditions such that with probability 1, a random walk will visit (arbitrarily close
to) a given point in Rd infinitely many times? In order to precisely state and
answer this question, we introduce some useful definitions. Below, | · | denotes the
Euclidean norm on Rd .
Definition 2.2. Let {Sn } be a random walk on Rd and let x ∈ Rd . (i) x is said to
be a possible point for {Sn } if ∀ > 0, ∃N ∈ N such that P (|SN − x| < ) > 0. (ii)
x is said to be a recurrent point for {Sn } if ∀ > 0, P (|Sn − x| < i.o.) = 1.
That is, x is a possible point for a random walk if given any desired level of
closeness , there is some time at which there is a nonzero probability of the random
walk being within distance of x. Similarly, x is a recurrent point for a random
walk if with probability one the random walk comes arbitrarily close to x infinitely
many times.
Although the notions of recurrent points and possible points may at first seem
quite disparate, we now prove that they are related. We first note that from Definition 2.2 it is obvious that all recurrent points of a random walk are possible points
of a random walk. The following theorem establishes the converse in the case where
a random walk has any recurrent points.
Theorem 2.3. Let {Sn } be a random walk on Rd and let R and P be (respectively)
the sets of its recurrent points and its possible points. If R =
6 ∅, then R = P.
Proof. Suppose R =
6 ∅. Since recurrent points are possible points, R ⊆ P. We now
show that P ⊆ R by first establishing
(1)
if r ∈ R and p ∈ P then (r − p) ∈ R
by contradiction. So let r ∈ R and p ∈ P and suppose (r − p) 6∈ R. Then by
definition there exists > 0 such that P (|Sn − (r − p)| < 2 i.o.) < 1, i.e. there
exists M ∈ N such that P (∀n ≥ M, |Sn − (r − p)| ≥ 2) > 0. Since p is a possible
point for {Sn }, by definition there exists N ∈ N such that P (|SN − p| < ) > 0.
Now note that for all n > N , Sn − SN = XN +1 + . . . + Xn is independent of the
first N steps of the random walk and has the same distribution as Sn−N (as it is
simply n − N steps starting from the origin). Thus, certainly
P (∀n ≥ N + M, |Sn − SN − (r − p)| ≥ 2) = P (∀n ≥ M, |Sn − (r − p)| ≥ 2) > 0.
Note also that the combination of the event that |SN − p| < and the event that
∀n ≥ N + M, |Sn − SN − (r − p)| ≥ 2 certainly (by the Triangle Inequality) implies
the event that ∀n ≥ N + M, |Sn − r| ≥ . Since these former two events are
independent (as established above, since M ≥ 1), we thus have
P (∀n ≥ N + M, |Sn − r| ≥ )
≥ P (∀n ≥ N + M, |Sn − SN − (r − p)| ≥ 2 and |SN − p| < )
= P (∀n ≥ N + M, |Sn − SN − (r − p)| ≥ 2) · P (|SN − p| < ) > 0,
a contradiction to the fact that r is a recurrent point for {Sn }. Thus, (1) holds.
Using (1), we now show that 0 ∈ R and then show that if x ∈ R then −x ∈ R.
Since R =
6 ∅, let x ∈ R. Then certainly x ∈ P, so applying (1) with r = p = x
gives 0 ∈ R. Similarly, applying (1) with r = 0 and p = x gives −x ∈ R.
To prove the theorem itself, we now let y ∈ P and show that y ∈ R. Applying
(1) with r = 0 and p = y gives −y ∈ R, which by the above means y ∈ R.
6
SIMON LAZARUS
By Theorem 2.3, we see that our original question “is a given possible point
for {Sn } a recurrent point for {Sn }?” reduces to the question “does {Sn } have a
recurrent point?” It thus makes sense to simply call a random walk recurrent if it
has at least one recurrent point. We henceforth adopt this definition, and we call
a random walk transient if it has no recurrent points.
Additionally, within the proof of Theorem 2.3, we found that if {Sn } has any
recurrent points, then 0 is a recurrent point of {Sn }. Thus, our original question
reduces to “is 0 a recurrent point for {Sn }?” In summary, we have:
Corollary 2.4. Let {Sn } be a random walk on Rd . Then (i) if 0 ∈ R then R = P,
and (ii) if 0 6∈ R then R = ∅.
Before continuing our discussion of general random walks on Rd , we give a more
concrete example of such a random walk. Specifically, we consider a simple random
walk on Zd . In this case, the distribution of our i.i.d. random variables Xi is
1
= P (Xi = −ej ). That is, each step of
given by ∀j ∈ {1, . . . , d}, P (Xi = ej ) = 2d
the random walk has an equal chance of moving 1 unit in any of the 2d cardinal
directions and never moves anywhere else.
In this case, given any x ∈ Zd , there is at least one possible path from the origin
to x in a finite number of steps; say this number of steps is N . There is a (2d)−N
chance of our simple random walk following this precise path from the first step to
the N th step. Thus, certainly P (SN = x) ≥ (2d)−N > 0, meaning x is a possible
point for {Sn }. In this way, we see that if a simple random walk on Zd has 0 as
a recurrent point, then its set of recurrent points is all of Zd ; and if 0 is not a
recurrent point, then it has no recurrent points.
With this example in mind, we now move on to the conditions under which 0 is
a recurrent point for any general random walk on Rd . In order to simplify some of
the proofs below, we introduce a new norm on Rd and present a useful lemma.
Definition 2.5. Let x = (x1 , . . . , xd ) ∈ Rd . We define ||x|| = max{|x1 |, . . . , |xd |}.
Lemma 2.6. Let > 0 and m ∈ N such that m > 1. Then
∞
∞
X
1 X
P (||Sn || < )
P
(||S
||
<
m)
≤
n
(2m)d n=0
n=0
Proof. Let A = {(x1 , . . . , xd ) ∈ Rd : ∃i ∈ {1, . . . , d} such that xi = −m} and
B = {x ∈ Rd : ||x|| < m}. We partition A ∪ B into (2m)d “half-open” cubes
as follows. Let I = {−m, −m + 1, . . . , m − 1}d . For all v = (v1 , . . . , vd ) ∈ I, let
Cv = [v1 , (v1 + 1)) × · · · × [vd , (vd + 1)). Then clearly ∪v∈I Cv = A ∪ B and the
Cv are pairwise disjoint. So since B ⊆ A ∪ B, this means
∞
∞
∞ X
X
X
X
P (Sn ∈ B) ≤
P (Sn ∈ A ∪ B) =
P (Sn ∈ Cv ).
n=0
n=0
n=0 v∈I
Since each of these probabilities is certainly nonnegative, by Fubini’s Theorem we
can exchange the order of summation to get
∞
∞
X
XX
(2)
P (Sn ∈ B) ≤
P (Sn ∈ Cv ).
n=0
v∈I n=0
Now fix v ∈ I and let Tv = inf{k ∈ {0, 1, . . .} : Sk ∈ Cv }. (Note that this infimum
may be ∞.) Note that given any n ∈ {0, 1, . . .}, the event that Sn ∈ Cv occurs if and
ON THE RECURRENCE OF RANDOM WALKS
7
only if precisely one of the events (Sn ∈ Cv and Tv = 0), (Sn ∈ Cv and Tv = 1), . . .,
(Sn ∈ Cv and Tv = n) occurs. For if the former event does not occur, certainly none
of the latter events occurred; and if the former event does occur, then the random
walk landed in Cv by time n and hence there was some earliest time k ∈ {0, 1, . . . , n}
at which it did so. We therefore have
∞
∞ X
n
X
X
P (Sn ∈ Cv ) =
P (Sn ∈ Cv and Tv = k)
n=0
n=0 k=0
Since each of these terms is nonnegative, we can again exchange the order of summation to get
∞
∞ X
∞
X
X
(3)
P (Sn ∈ Cv ) =
P (Sn ∈ Cv and Tv = k).
n=0
k=0 n=k
Now fix k ∈ {0, 1, . . .}. Note that if Tv = k, then the random walk has reached Cv
at time k, meaning in order for the walk to reach Cv again at time n we certainly
must have that the next n − k steps of the walk do not (in sum) displace the walk
by a distance greater than in any cardinal direction (for Cv is a cube with side
length ). This means
∞
X
P (Sn ∈ Cv and Tv = k) ≤
n=k
∞
X
P (||Sn − Sk || < and Tv = k)
n=k
But those next n − k steps are independent of the first k steps and have the same
distribution as the first n − k steps of a separate random walk starting at 0. Hence,
∞
∞
X
X
P (Sn ∈ Cv and Tv = k) ≤
P (||Sn − Sk || < ) · P (Tv = k)
n=k
n=k
= P (Tv = k)
∞
X
P (||Sn−k || < ) = P (Tv = k)
∞
X
P (||Si || < )
i=0
n=k
where the last equality is simply re-indexing. Since this holds for any fixed k ∈
{0, 1, . . .}, by letting k vary and summing we obtain
∞ X
∞
X
P (Sn ∈ Cv and Tv = k) ≤
k=0 n=k
∞
X
P (Tv = k)
∞
X
P (||Si || < ).
i=0
k=0
Similarly, since this holds for any fixed v ∈ I, by letting v vary we get
∞ X
∞
∞
∞
XX
XX
X
P (Sn ∈ Cv and Tv = k) ≤
P (Tv = k)
P (||Si || < ).
v∈I k=0 n=k
i=0
v∈I k=0
Applying (3) and then (2), this implies
∞
X
n=0
P (Sn ∈ B) ≤
∞
XX
v∈I k=0
P (Tv = k)
∞
X
P (||Si || < ).
i=0
We now note that by definition Sn ∈ B if and only if ||Sn || < m. We also note
that the innermost sum on the right-hand side of the above does not depend on k
or v and hence can be pulled outside of the other two sums. Thus, we have
!
∞
∞
∞
X
X
XX
P (||Sn || < m) ≤
P (||Si || < )
P (Tv = k).
n=0
i=0
v∈I k=0
8
SIMON LAZARUS
P∞
Lastly, we note that k=0 P (Tv = k) ≤ 1 (since given v ∈ I, Tv takes exactly
one value in {0, 1, . . .} ∪ {∞} and hence has probability ≤ 1 of taking a value in
{0, 1, . . .}) and we note that there are (2m)d elements in I. Thus, we have
!
∞
∞
X
X
P (||Sn || < m) ≤
P (||Si || < ) · (2m)d · 1
n=0
i=0
as desired.
With this lemma in hand, we are ready to prove an important necessary and
sufficient condition for recurrence which we shall employ in the proofs of Theorems
2.8, 3.1, and 3.3 below.
Theorem 2.7. Let {SnP
} be a random walk on Rd and let r > 0. Then {Sn } is
∞
recurrent if and only if n=0 P (||Sn || < r) = ∞.
P∞
Proof. We
P∞must show that (i) n=0 P (||Sn || < r) < ∞ implies {Sn } is transient
and (ii) n=0 P (||Sn || < r) = ∞ implies {Sn } is recurrent. To do so, we show that
in case (i) 0 is not a recurrent point for {Sn } but in case (ii) it is a recurrent point
for {Sn }.
P∞
(i) By Theorem 1.14, if n=0 P (||Sn || < r) converges, then P (||Sn || < r i.o.) =
0. Now note that the event that |Sn | < r i.o. certainly implies the event that
||Sn || < r i.o. since the Euclidean norm of Sn being less than r for some n certainly
implies that the largest coordinate of Sn is less than r. Thus, P (|Sn | < r i.o.) ≤
P (||Sn || < r i.o.). That is, P (|Sn − 0| < r i.o.) = 0, so by definition 0 is not a
recurrent point P
for {Sn }.
∞
(ii) Suppose n=0 P (||Sn || < r) = ∞. Let > 0. We wish to show P (|Sn | <
i.o.) = 1. We do so by showing P
that for every k ∈ N, P (∀n ≥ k, |Sn | ≥ ) = 0.
∞
To do this, we first establish that n=0 P (|Sn | < /2) = ∞. Let M ∈ N such that
r
r
< 2√ d . Let s = M
. Then since s > 0, Lemma 2.6 gives
M > 1 and M
∞
∞
X
1 X
P
(||S
||
<
M
s)
≤
P (||Sn || < s),
n
(2M )d n=0
n=0
which by substituting s =
r
M
gives
∞
∞
∞
X
r X
1 X
√
P
(||S
||
<
r)
≤
P
||S
||
<
≤
P
||S
||
<
n
n
n
(2M )d n=0
M
2 d
n=0
n=0
r
since M
< 2√ d . Now note that given any n ∈ {0, 1, . . .}, the event that ||Sn || < 2√ d
implies the event that |Sn | < /2. For if Sn = (x1 , . . . , xd ) and ||Sn || < 2√ d , then
by definition |x1 |, . . . , |xd | < 2√ d so
s 2
q
2
2
√
|Sn | = x1 + . . . + xd < d ·
= .
2
2 d
Thus, we certainly have that
X
∞
∞
X
P ||Sn || < √
≤
P (|Sn | < /2),
2 d
n=0
n=0
ON THE RECURRENCE OF RANDOM WALKS
9
which combined with the above gives
∞
∞
X
1 X
P
(||S
||
<
r)
≤
P (|Sn | < /2),
n
(2M )d n=0
n=0
P∞
meaning necessarily n=0 P (|Sn | < /2) = ∞.
Now let k ∈ N. For all n ∈ N, let An be the event that both |Sn | < /2 and
∀m ≥ k + n, |Sm | ≥ /2. Then clearly at most k of the events An can occur. For if
j ∈ N is the smallest n for which An occurs, then by definition ∀m ≥ k + j, |Sm | ≥
/2, meaning
all of the events that can possibly occur are Aj , Aj+1 , . . . , Aj+k−1 .
P∞
Thus, n=0 P (An ) ≤ k since at most k of the events have any chance of occurring.
Now note that given any n ∈ {0, 1, . . .}, the event that both |Sn | < /2 and
∀m ≥ k + n, |Sm − Sn | ≥ certainly implies the event An . Thus,
∞
X
P (An ) ≥
n=0
∞
X
P (|Sn | < /2 and ∀m ≥ k + n, |Sm − Sn | ≥ ).
n=0
But since the event {∀m ≥ k + n, |Sm − Sn | ≥ } is independent of the event that
|Sn | < /2 and occurs with the same probability that the event {∀i ≥ k, |Si | ≥ }
occurs, this means
k≥
∞
X
n=0
P (An ) ≥
∞
X
P (|Sn | < /2) · P (∀m ≥ k + n, |Sm − Sn | ≥ )
n=0
=
∞
X
P (|Sn | < /2) · P (∀i ≥ k, |Si | ≥ ).
n=0
Now suppose ∃c > 0 such that P (∀i ≥ k, |Si | ≥ ) = c. Then the above gives
k≥c
∞
X
P (|Sn | < /2) = ∞,
n=0
a contradiction. Thus, P (∀i ≥ k, |Si | ≥ ) = 0. So since k ∈ N was arbitrary, we
have shown that for all k ∈ N, P (∀n ≥ k, |Sn | ≥ ) = 0, as desired.
Having presented in Theorem 2.7 a necessary and sufficient condition for the
recurrence of a random walk {Sn }, we now establish another such condition (originally given in [1]) that is sometimes more useful.
Theorem 2.8. Let {Sn } = {X1 + . . . + Xn } be a random walk on Rd and let r > 0.
Let ϕ be the characteristic function of X1 . Then {Sn } is recurrent if and only if
Z
1
dx = ∞.
sup
Re
1 − cϕ(x)
c∈(0,1) [−r,r]d
Remark 2.9. We note that given c ∈ (0, 1), the above integral always exists and has
value in [0, ∞]. For since 0 < c < 1 and (by Theorem 1.7) |ϕ| ≤ 1, certainly
for
all
x ∈ Rd we have cRe(ϕ(x)) < 1, so Lemma 3.7 below gives that Re
1
1−cϕ(x)
≥0
d
for all x ∈ R .
This new condition is sometimes more tractable than the one in Theorem 2.7,
for instead of requiring one to find or estimate probabilities related to {Sn }, it
requires finding ϕ and computing or estimating the above integral. For this reason,
10
SIMON LAZARUS
we employ it in proving Theorem 3.9 below. Before proceeding with the proof of
Theorem 2.8, we provide four lemmas whose uses will soon be apparent.
Lemma 2.10. For all x ∈ [−π/3, π/3] we have 1 − cos(x) ≥
x2
4 .
Proof. Since cos is an even function, it suffices to show this for x ∈ [0, π/3]. First,
note that if t ∈ [0, π/3] then clearly cos(t) ≥ 1/2. Hence, if y ∈ [0, π/3] then
Z y
Z y
1
y
sin(y) =
cos(t)dt ≥
dt = .
2
2
0
0
Thus, if x ∈ [0, π/3] then
x
Z
1 − cos(x) =
Z
sin(y)dy ≥
0
0
x
y
x2
dy =
.
2
4
Lemma 2.11. Let Y1 , . . . , Yd be random variables on R with characteristic functions ϕ1 , . . . , ϕd . Let Z = (Y1 , . . . , Yd ). Then ϕZ , the characteristic function of Z,
Qd
is given by: ∀x = (x1 , . . . , xd ) ∈ Rd , ϕZ (x) = i=1 ϕi (xi ).
Proof. Let x = (x1 , . . . , xd ) ∈ Rd . By definition,
Z
ϕZ (x) = E[eix·Z ] =
eix·(t1 ,...,td ) dµZ (t1 , . . . , td ).
Rd
Since for allR t ∈ Rd we have |eix·t | = 1 and since µZ is a probability measure,
necessarily Rd |eix·(t1 ,...,td ) |dµZ (t1 , . . . , td ) = 1. So by Fubini’s Theorem, we have
Z
Z
ϕZ (x) =
· · · eix·(t1 ,...,td ) dµY1 (t1 ) · · · dµYd (td )
R
R
Z
Z
ix1 t1
=
··· e
· · · eixd td dµY1 (t1 ) · · · dµYd (td )
R
R
Z
Z
ix1 t1
ixd td
=
e
dµY1 (t1 ) · · ·
e
dµYd (td )
R
= E[e
R
ix1 Y1
ixd Yd
] · · · E[e
] = ϕ1 (x1 ) · · · ϕd (xd ).
Lemma 2.12. Let µ and ν be probability measures
on Bd R(the Borel sets in Rd )
R
with characteristic functions ϕµ and ϕν . Then Rd ϕµ dν = Rd ϕν dµ.
R
R R
Proof. By definition Rd ϕµ (t)dν(t) = Rd Rd exp(it · x)dµ(x)dν(t). Since for all
t, x ∈ Rd we
R have
R | exp(it · x)| = 1 and since µ and ν are probability measures,
necessarily Rd Rd | exp(it · x)|dµ(x)dν(t) = 1, so Fubini’s Theorem gives
Z Z
Z Z
Z
exp(it · x)dµ(x)dν(t) =
exp(it · x)dν(t)dµ(x) =
ϕν (x)dµ(x),
Rd
Rd
Rd
Rd
Rd
as desired.
Lemma 2.13. For all n ∈ {0, 1, . . .}, let an ≥ 0. Then
sup
∞
X
c∈(0,1) n=0
cn an =
∞
X
n=0
an .
ON THE RECURRENCE OF RANDOM WALKS
11
Proof. Since
c ∈ (0,
and for all n ∈ {0, 1, . . .} we have cn an ≤ an , certainly
P1)
P∞for all
∞
n
supc∈(0,1) n=0 c an ≤ n=0 an , so it suffices to show the reverse inequality. We
do so by considering two cases.
P∞
PN
First, suppose n=0 an = ∞. Let R > 0. Choose N ∈ N such that n=0 an >
2R. Then choose c ∈ (0, 1) such that for all n ∈ {0, 1, . . . , N } we have cn ≥ 12 . Then
P∞ n
PN
PN
n
≥ n=0 12 an > 12 · 2R = R. Thus, P
for all R > 0, there
n=0 c an ≥
n=0 c an P
∞
∞
exists c ∈ (0, 1) such that n=0 cn an > R, meaning supc∈(0,1) n=0 cn an = ∞.
P∞
Now suppose instead that there exists S ∈ [0, ∞) such that n=0 an = S. Let
PM
> 0. Choose M ∈ N such that n=0 an > S − 2 . Then choose c ∈ (0, 1) such
that for all n ∈ {0, 1, . . . , M } we have cn ≥ 1 − 2(S+1)
. Then
∞
X
n=0
cn an ≥
M
X
cn an ≥
n=0
M X
1−
2(S + 1)
n=0
an >
1−
2(S + 1)
S−
2
S
2
− +
> S − − + 0 = S − .
2(S + 1) 2 4(S + 1)
2 2
P∞ n
P∞
Thus, for all > 0, there exists
P∞c ∈ n(0, 1) such
P∞ that n=0 c an > n=0 an − ,
meaning necessarily supc∈(0,1) n=0 c an ≥ n=0 an , as desired.
=S−
With these tools in hand, we are ready to prove Theorem 2.8.
R
1
dx < ∞ then
Proof of 2.8. We first show that (i) if supc∈(0,1) [−r,r]d Re 1−cϕ(x)
R
P∞
1
n=0 P (||Sn || < 1/r) < ∞ and then that (ii) if supc∈(0,1) [−r,r]d Re 1−cϕ(x) dx =
P∞
∞ then n=0 P (||Sn || < 1/r) = ∞. By 2.7, showing (i) and (ii) will complete the
proof.
(i) Suppose there exists C ∈ [0, ∞) such that
Z
1
dx = C.
sup
Re
1 − cϕ(x)
c∈(0,1) [−r,r]d
Let T1 , . . . , Td be i.i.d. random variables with density g, where for all t ∈ R,
g(t) = r−|t|
if |t| ≤ r and g(t) = 0 otherwise. Then by Example 1.8, each Ti has
r2
characteristic function ϕT , where ϕT (0) = 1 and ∀t ∈ R \ {0} we have ϕT (t) =
2(1−cos(tr))
.
(tr)2
Note that if t ∈ [− 1r , 1r ], then |rt| ≤ 1 < π/3. Thus, by Lemma 2.10, for all
2
1 1
t ∈ [− 1r , 1r ] we have 1 − cos(rt) ≥ (rt)
4 , meaning for all t ∈ [− r , r ] \ {0} we have
2(1−cos(rt))
1 1
2·
≥ 1. Thus, for all t ∈ [− r , r ] we have 2ϕT (t) ≥ 1 since ϕT (0) = 1.
(rt)2
So certainly for all (x1 , . . . , xd ) ∈ [− 1r , 1r ]d we have
2d
(4)
d
Y
ϕT (xi ) =
i=1
d
Y
2ϕT (xi ) ≥ 1
i=1
since each term in the rightmost product of the above is ≥ 1.
1
1 1 d
R Now let n ∈ {0, 1, . . .}. Then we know P (||Sn || < r ) = P (Sn ∈ (− r , r ) ) =
1 dµSn (x), where µSn is the probability measure of Sn . Then by (4),
(− 1 , 1 )d
r r
P
1
||Sn || <
r
d
≤2
Z
d
Y
(− r1 , r1 )d i=1
ϕT (xi )dµSn (x),
12
SIMON LAZARUS
where dµSn (x) is shorthand for dµSn (x1 , . . . , xd ). Now for all y ∈ R \ {0}, we
know 1−cos(y)
≥ 0 since cos(y) ≤ 1. Thus, for all t ∈ R we have ϕT (t) ≥ 0 since
y2
2(1−cos(rt))
≥ 2 · 0 if t 6= 0
(rt)2
d
(x1 , . . . , xd ) ∈ R we certainly have
ϕT (t) =
and ϕT (t) = 1 ≥ 0 if t = 0. Therefore, for all
Qd
1 1 d
d
i=1 ϕT (xi ) ≥ 0. So since (− r , r ) ⊆ R , we
see
P
||Sn || <
1
r
≤ 2d
Z
d
Y
ϕT (xi )dµSn (x).
Rd i=1
It should be noted that in the steps that follow (up to and including (8) below),
every manipulation we perform only substitutes for the right-hand side of the above
something that is equivalent to it. In particular, it is clear (since ϕT is a real
function on R) that the right-hand side of the above inequality is real; this does
not (and cannot) change in any of the steps below.
Let Z = (T1 , . . . , Td ). Then by Lemma 2.11, the above inequality gives
Z
1
≤ 2d
ϕZ (x)dµSn (x),
P ||Sn || <
r
Rd
which by Lemma 2.12 means
Z
1
≤ 2d
ϕSn (x)dµZ (x).
P ||Sn || <
r
Rd
Since Sn = X1 + . . . + Xn and each of the Xj has characteristic function ϕ, by
Theorem 1.7 we know ϕSn = ϕn . (If n = 0, we must use other reasoning to show
that ϕSn = ϕn on Rd . First, note that since S0 = 0, we know ϕS0 = 1 on Rd ,
meaning if x ∈ Rd such that ϕ(x) 6= 0, then ϕ0 (x) = 1 = ϕS0 (x). Therefore, we
do have that ϕS0 (x) = ϕ0 (x) for all x ∈ Rd such that ϕ(x) 6= 0. However, we may
run into trouble if x ∈ Rd such that ϕ(x) = 0, as in this case ϕ0 (x) = 00 is not
conventionally defined. Therefore, we now define ϕ0 (x) to be 1 whenever x ∈ Rd
such that ϕ(x) = 0. In this way, ϕS0 = ϕ0 holds on all of Rd .) Thus,
Z
1
≤ 2d
ϕn (x)dµZ (x).
P ||Sn || <
r
d
R
Now note that if I1 , . . . , Id ⊆ R are intervals, then since T1 , . . . , Td are independent,
Qd
we know P (Z ∈ (I1 × · · · × Id )) = P (T1 ∈ I1 , . . . , Td ∈ Id ) = i=1 P (Ti ∈ Ii ). So
since each of the Ti has the density g given above, we see that Z has the density h,
Qd
where for all x = (x1 , . . . , xd ) ∈ Rd , h(x) = i=1 g(xi ). Therefore, we have
Z
Z
d
Y
1
P ||Sn || <
≤ 2d
ϕn (x)h(x)dx = 2d
ϕn (x)
g(xi )dx.
r
Rd
Rd
i=1
Since g is given by ∀t ∈ R, g(t) = r−|t|
r 2 if |t| ≤ r and g(t) = 0 otherwise, the above
clearly implies
Z
d Y
1
r − |xi |
d
n
P ||Sn || <
≤2
ϕ (x)
dx.
r
r2
[−r,r]d
i=1
Now let c ∈ (0, 1). Then we have
Z
d Y
1
r − |xi |
n
d n
n
c P ||Sn || <
≤2 c
ϕ (x)
dx.
r
r2
[−r,r]d
i=1
ON THE RECURRENCE OF RANDOM WALKS
13
This inequality holds for any given n ∈ {0, 1, . . .}. So letting n vary and summing,
we see
X
Z
∞
∞
d X
Y
1
r − |xi |
cn P ||Sn || <
dx
≤
2 d cn
ϕn (x)
r
r2
[−r,r]d
n=0
n=0
i=1
(5)
d X
∞ Z
d Y
2
r − |xi |
n n
=
dx.
c ϕ (x)
r
r
d
n=0 [−r,r]
i=1
We now seek to swap the order of summation and integration on the right-hand
side of (5). By Fubini’s Theorem, we can do so as long as we can show that both
d ∞ Z
Y
X
r − |xi | n n
(6)
dx < ∞
c ϕ (x)
r
[−r,r]d n=0
i=1
and
d ∞ Y
X
r − |xi | n n
c ϕ (x)
dx < ∞.
r
[−r,r]d
Z
(7)
n=0
i=1
To prove that (6) and (7) hold, recall that |ϕ| ≤ 1 on Rd and that for all t ∈ [−r, r],
≤ 1. Thus, for all x = (x1 , . . . , xd ) ∈ [−r, r]d we have
0 ≤ r−|t|
r
d Y
r − |xi | n n
c ϕ (x)
≤ |cn | · 1n · 1d = cn .
r
i=1
This means
∞ Z
X
n=0
d ∞
∞ Z
Y
X
X
r − |xi | n n
(2r)d cn < ∞
cn dx =
c ϕ (x)
dx ≤
r
[−r,r]d
[−r,r]d i=1
n=0
n=0
(since the right-hand side of the above is a geometric series and |c| < 1), thus
proving (6); and (7) follows similarly.
This means we can switch the order of summation and integration in (5) and
then move the product outside of the sum to obtain
d Z
! X
∞
d ∞
X
Y
1
2
r − |xi |
n
c P ||Sn || <
≤
cn ϕn (x)dx.
r
r
r
d
[−r,r]
n=0
n=0
i=1
Now note that given x ∈ Rd , the sum on the right-hand side of the above is a
1
convergent (complex) geometric series with value 1−cϕ(x)
. (If ϕ(x) 6= 0, this is
1
obvious since 0 < |cϕ(x)| < 1; and if ϕ(x) = 0 then 1−cϕ(x)
= 1 = 1 + 0 + 0 + ···,
0 0
1 1
which we see equals c ϕ (x)+c ϕ (x)+· · · when we recall that we defined ϕ0 (x) = 1
when ϕ(x) = 0.) Thus,
d Z
!
∞
d X
Y
1
2
r
−
|x
|
1
i
cn P ||Sn || <
≤
dx.
r
r
r
1
−
cϕ(x)
d
[−r,r]
n=0
i=1
As noted above, the right-hand side of this inequality is (and has always been) real,
so we can take its real part without affecting its value. Hence,
d Z
!
∞
d X
Y
1
2
r − |xi |
1
n
(8)
c P ||Sn || <
≤
Re
dx.
r
r
r
1 − cϕ(x)
[−r,r]d
n=0
i=1
14
SIMON LAZARUS
1
By Remark 2.9, we know that for all x ∈ Rd , Re 1−cϕ(x)
≥ 0. So since we also
Qd r−|xi | know that i=1
≤ 1 whenever (x1 , . . . , xd ) ∈ [−r, r]d (as 0 ≤ r−|t|
≤ 1
r
r
whenever t ∈ [−r, r]), (8) implies that
d Z
d
∞
X
1
2
1
2
n
c P ||Sn || <
C.
≤
Re
dx ≤
r
r
1
−
cϕ(x)
r
d
[−r,r]
n=0
The above inequality holds for any c ∈ (0, 1). Therefore, taking the supremum
of both sides
using the fact (given
by Lemma 2.13) that
P∞ over all such c andP
∞
supc∈(0,1) n=0 cn P ||Sn || < 1r = n=0 P ||Sn || < 1r , we have
d
∞
X
2
1
≤
C < ∞,
P ||Sn || <
r
r
n=0
as desired.
(ii) Now suppose instead that
Z
sup
c∈(0,1)
[−r,r]d
Re
1
1 − cϕ(x)
dx = ∞.
1
and for
Let Y1 , . . . , Yd be i.i.d. random variables with density g, where g(0) = 2πr
r(1−cos(t/r))
all t ∈ R\{0}, g(t) =
. Then by Example 1.11, each Yi has characteristic
πt2
function ϕY , where for all t ∈ R we have ϕY (t) = 1 − |rt| if |t| ≤ 1r and ϕY (t) = 0
otherwise.
Let n ∈ {0, 1, . . .}. Note that if t ∈ [− 1r , 1r ], then 0 ≤ 1−|rt| ≤ 1. Thus, certainly
Qd
for all (x1 , . . . , xd ) ∈ [− 1r , 1r ]d we have 0 ≤ i=1 (1 − |rxi |) ≤ 1. Therefore,
P
Z
Z
d
Y
1
(1 − |rxi |)dµSn (x).
=
1dµSn (x) ≥
||Sn || <
r
(− r1 , r1 )d
(− r1 , r1 )d i=1
As before, it should be noted that the manipulations that follow (up to and including (11) below) merely substitute the right-hand side of the above inequality
with something equivalent and hence do not change the fact that that side of the
inequality is real.
Since ϕY (t) = 1 − |rt| if t ∈ (− 1r , 1r ) and ϕY (t) = 0 if t 6∈ (− 1r , 1r ), the above
clearly gives
Z
Z Y
d
d
Y
1
P ||Sn || <
≥
ϕY (xi )dµSn (x) =
ϕY (xi )dµSn (x).
r
(− r1 , r1 )d i=1
Rd i=1
So letting Z = (Y1 , . . . , Yd ), Lemma 2.11 and then Lemma 2.12 imply
Z
Z
1
P ||Sn || <
≥
ϕZ (x)dµSn (x) =
ϕSn (x)dµZ (x).
r
Rd
Rd
Since (defining ϕ0 as before) ϕSn = ϕn on Rd and since (as before) Z has the
Qd
density h given by ∀x = (x1 , . . . , xd ) ∈ Rd , h(x) = i=1 g(xi ), the above gives
Z
Z
Z
d
Y
1
n
n
P ||Sn || <
≥
ϕ (x)dµZ (x) =
ϕ (x)h(x)dx =
ϕn (x)
g(xi )dx.
r
Rd
Rd
Rd
i=1
ON THE RECURRENCE OF RANDOM WALKS
15
Let c ∈ (0, 1). Then the above gives
Z
d
Y
1
n
c P ||Sn || <
≥ cn
ϕn (x)
g(xi )dx.
r
Rd
i=1
This inequality holds for any n ∈ {0, 1, . . .}. So letting n vary and summing, we see
X
∞
∞ Z
d
X
Y
1
n
n n
(9)
≥
c P ||Sn || <
c ϕ (x)
g(xi )dx.
r
d
n=0
n=0 R
i=1
As before, we wish to apply Fubini’s Theorem to switch the order of summation
and integration on the right-hand side of the above, and to do so we must show
Z X
d
∞ d
∞ Z
Y
Y
X
n n
n n
g(xi ) dx < ∞.
g(xi ) dx < ∞ and
(10)
c ϕ (x)
c ϕ (x)
Rd
Rd n=0
n=0
i=1
i=1
We show the first of these inequalities; the second follows similarly.
1
and for all t ∈ R \ {0},
Let n ∈ {0, 1, . . . , }. First, note that since g(0) = 2πr
r(1−cos(t/r))
g(t) =
,
we
see
that
g
≥
0
on
R
since
|
cos
|
≤
1
on R. That is |g| = g
πt2
on R. So since |ϕ| ≤ 1 on Rd , we see
Z
Z Y
Z d
d
Y
n n
n
g(x1 ) · · · g(xd )dx1 · · · dxd .
g(xi ) dx ≤ c
g(xi ) dx = cn
c ϕ (x)
Rd
Rd Rd i=1
i=1
Since g ≥ 0 on R, Fubini’s Theorem thus gives
Z d
d Z
d
Y
Y
Y
n n
n
g(xi ) dx ≤ c
g(t)dt = cn
1 = cn
c ϕ (x)
R
Rd i=1
i=1
i=1
since g is a probability density function. The above inequality holds for any n ∈
{0, 1, . . . , }. So letting n vary and summing, we see
∞
d
∞ Z
X
Y
X
n n
cn < ∞
g(xi ) dx ≤
c ϕ (x)
Rd n=0
i=1
n=0
since c ∈ (0, 1). Thus, the first half of (10) holds; the second half follows similarly.
Therefore, we can exchange the order of summation and integration in (9) and
move the product outside of the sum to obtain
! ∞
Z
∞
d
X
Y
X
1
cn P ||Sn || <
≥
g(xi )
cn ϕn (x)dx.
r
d
R
n=0
n=0
i=1
P∞ n n
1
d
As before, we know that given x ∈ R , n=0 c ϕ (x) = 1−cϕ(x)
. Thus, we have
!
Z
∞
d
X
Y
1
1
(11)
cn P ||Sn || <
≥
g(xi )
dx.
r
1
−
cϕ(x)
d
R
n=0
i=1
As noted above, the integral on the right-hand side of (11) must be real, so we can
take the real part of both sides to find
!
Z
∞
d
Y
X
1
1
n
c P ||Sn || <
≥
g(xi ) Re
dx.
r
1 − cϕ(x)
Rd
n=0
i=1
16
SIMON LAZARUS
By Remark 2.9, for all x ∈ R we have Re
d
1
1−cϕ(x)
≥ 0. So since g ≥ 0 on R and
d
since [−r, r] ⊆ R , we see
Z
∞
X
1
n
c P ||Sn || <
≥
r
[−r,r]d
n=0
d
Y
!
g(xi ) Re
i=1
1
1 − cϕ(x)
dx.
Now note that since for all t ∈ [−r, r] we have | rt | ≤ 1 < π/3, by Lemma 2.10,
2
for all t ∈ [−r, r] we have 1 − cos(t/r) ≥ (t/r)
4 . Thus, for all t ∈ [−r, r] \ {0},
r(1−cos(t/r))
1
1
1
we have g(t) =
≥ 4πr . So since g(0) = 2πr
> 4πr
, we see that for
πt2
1
d
all t ∈ [−r, r], g(t) ≥ 4πr . Thus,
for
all
(x
,
.
.
.
,
x
)
∈
[−r,
r]
,
we
certainly have
1
d
Qd
1
1
d
i=1 g(xi ) ≥ (4πr)d . Since Re 1−cϕ ≥ 0 on R , this means
Z
∞
X
1
1
1
cn P ||Sn || <
Re
≥
dx.
r
(4πr)d [−r,r]d
1 − cϕ(x)
n=0
The above inequality holds true for any c ∈ (0, 1). Therefore, taking the supremum
of both sides over all such c and employing Lemma 2.13 gives
Z
∞
X
1
1
1
P ||Sn || <
≥
sup
Re
dx = ∞,
r
(4πr)d c∈(0,1) [−r,r]d
1 − cϕ(x)
n=0
as desired.
3. Sufficient Conditions for Recurrence and Transience
Having established in the previous section two different necessary and sufficient
conditions for recurrence, we now employ these conditions to discover some simpler
conditions that imply recurrence or transience.
p
Theorem 3.1. Let {Sn } be a random walk on R. If Sn /n → 0, then {Sn } is
recurrent.
p
Proof. Let c ∈ N \ {1}. Since Sn /n → 0, by definition P Snn < 1c → 1, i.e.
n
P |Sn | <
→ 1.
c
Thus, we can choose N ∈ N\{1} such that for all n > N we have P |Sn | < nc > 12 .
By doing so, we find that
cN
cN
X
X
n
1
1
(12)
P |Sn | <
≥
= (c − 1)N · .
c
2
2
n=N +1
n=N +1
Now since N > 1, by Lemma 2.6,
∞
X
P (|Sn | < 1) ≥
n=0
∞
1 X
1
P (|Sn | < N ) ≥
2N n=0
2N
cN
X
P (|Sn | < N )
n=N +1
where the latter inequality follows since each of the terms of the sum is nonnegative.
Now note that for all n ∈ {N + 1, N + 2,. . . , cN } we have nc ≤ N . Thus, for all
n ∈ {N + 1, . . . , cN } we have P |Sn | < nc ≤ P (|Sn | < N ). Therefore,
∞
X
1
P (|Sn | < 1) ≥
2N
n=0
cN
X
n=N +1
n
P |Sn | <
,
c
ON THE RECURRENCE OF RANDOM WALKS
17
which by (12) above means
∞
X
1
1
c−1
· (c − 1)N · =
.
2N
2
4
n=0
P∞
So since c ∈ N\{1} was arbitrary, necessarily n=0 P (|Sn | < 1) = ∞. By Theorem
2.7, then, {Sn } is recurrent (for in R the norms | · | and || · || are equivalent).
P (|Sn | < 1) ≥
Corollary 3.2. Let {Sn } = {X1 + . . . + Xn } be a random walk on R. Suppose
E|X1 | < ∞. Then {Sn } is recurrent if and only if E[X1 ] = 0.
Proof. Note that E|X1 | < ∞ implies that E[X1 ] exists. By the Strong Law of Large
a.s.
Numbers, Sn /n → E[X1 ]. So since convergence almost surely implies convergence
in probability, we see that if E[X1 ] = 0, then by Theorem 3.1, {Sn } is recurrent. If
a.s.
instead E[X1 ] > 0, then since n1 → 0 from above and Sn /n → E[X1 ], necessarily
a.s.
Sn → ∞, which precludes {Sn } from having any recurrent points. Similarly, if
a.s.
E[X1 ] < 0, then Sn → −∞, meaning {Sn } has no recurrent points.
Theorem 3.3. Let {Sn } be a random walk on R2 . If there exists a random variable
√ d
Y on R2 such that Sn / n → Y and if Y has a density function f such that f (0) > 0
and f is continuous at 0, then {Sn } is recurrent.
P∞
Proof. We wish to prove that n=0 P (||Sn || < 1) = ∞; by Theorem 2.7, showing
this will establish the theorem. By Lemma 2.6, we know that for all m ∈ {2, 3, . . .},
∞
1 X
P (||Sn || < m).
P (||Sn || < 1) ≥
4m2 n=0
n=0
∞
X
Thus,
suffices to show that for all R > 0, there exists m ∈ {2, 3, . . .} such that
Pit
∞
1
n=0 P (||Sn || < m) > R. Therefore, it certainly suffices to show that
4m2
∞
1 X
P (||Sn || < m) = ∞,
m→∞ m2
n=0
P∞
as in this case we can always choose
an m such that m12 n=0 P (||Sn || < m) > 4R.
R
Now for all y > 0, let g(y) = (−y,y)2 f (x)dx = P (||Y || < y). Note that g ≥ 0
on (0, ∞). Now since f is continuous at 0 and f (0) > 0, by definition there exists
r > 0 such that for all x ∈ [−r, r]2 we have |f (x) − f (0)| < f (0)
2 . Thus, certainly
for all x ∈ [−r, r]2 we have f (x) > f (0)
.
Therefore,
for
all
y
∈
(0,
r],
2
Z
Z
f (0)
f (0)
dx =
· (2y)2 = 2f (0)y 2 .
g(y) =
f (x)dx ≥
2
2
(−y,y)2
(−y,y)2
(S1)
lim inf
Now note that since g ≥ 0 on (0, ∞),
Z ∞
Z ∞
Z
g(t−1/2 )dt ≥
g(t−1/2 )dt ≥
0
1/r 2
∞
2f (0)(t−1/2 )2 dt = 2f (0)
1/r 2
where the second inequality follows since t ≥
suffices to prove
(S2)
Z
∞
1/r 2
1
r2
1
dt = ∞,
t
implies t−1/2 ∈ (0, r]. Thus, it
Z ∞
∞
1 X
lim inf 2
P (||Sn || < m) ≥
g(t−1/2 )dt
m→∞ m
0
n=0
18
SIMON LAZARUS
since the integral on the right-hand side of the above is ∞, meaning (S2) implies
(S1). Now note that for all m ∈ N we have
Z ∞
∞
X
P (||Sn || < m) =
P (||Sbyc || < m)dy,
0
n=0
where b·c is the floor
y = tm2 in
R ∞ m ∈ N, we can substitute
P∞function. Thus, given
2
the above to find n=0 P (||Sn || < m) = 0 P (||Sbtm2 c || < m) · m dt; that is,
Z ∞
∞
1 X
(13)
P
(||S
||
<
m)
=
P (||Sbtm2 c || < m)dt.
n
m2 n=0
0
Now suppose for a moment that for all t > 0 we have
lim P (||Sbtm2 c || < m) = g(t−1/2 ).
(S3)
m→∞
Then by taking the limit inferior of both sides of (13) and applying Fatou’s Lemma
(which we can do since probabilities are nonnegative), we find
Z ∞
Z ∞
∞
1 X
lim inf 2
P (||Sn || < m) ≥
lim inf P (||Sbtm2 c || < m)dt =
g(t−1/2 )dt,
m→∞ m
m→∞
0
0
n=0
which is precisely (S2). Therefore, it suffices to show that (S3) holds. So let t > 0.
p
√ d
d
Note that since Sn / n → Y as n → ∞, certainly Sbtm2 c / btm2 c → Y as m → ∞
since the sequence {btm2 c} tends to ∞ as m → ∞ and is a sequence of natural
numbers. So since √ m 2 → t−1/2 as m → ∞, by the definition of convergence in
btm c
distribution we see that
P (||Sbtm2 c || < m) = P
!
S 2 m
btm c p
< p
btm2 c btm2 c
m→∞
−→ P (||Y || < t−1/2 ) = g(t−1/2 ),
thus establishing (S3).
Remark 3.4. If {Sn } = {X1 + . . . + Xn } is a random walk on R2 and if E[X1 ] = 0
and the coordinates √
of X1 have finite covariances, then the Central Limit Theorem
guarantees that Sn / n converges in distribution to a multivariate normal random
variable on R2 with mean 0. If this normal random variable on R2 is non-degenerate,
then its density function is positive and continuous at 0, so by Theorem 3.3 {Sn }
is recurrent. If instead the normal random variable on R2 is degenerate, then
necessarily {Sn } is not truly 2-dimensional (in the sense of Definition 3.6 below)
and hence we have reduced {Sn } to a random walk on R. In this case as well,
since E[X1 ] = 0 and finite covariances imply E|X1 | < ∞, by Corollary 3.2 {Sn } is
recurrent.
Example 3.5. Let {Sn } = {X1 + . . . + Xn } be a simple random walk on Z and
{Tn } = {Y1 + . . . + Yn } be a simple random walk on Z2 . Then both {Sn } and {Tn }
are recurrent, as follows. First, {Sn } is recurrent by Corollary 3.2 since E[X1 ] = 0
and E|X1 | = 1 < ∞. Second, {Tn } is recurrent by Remark 3.4 since E[Y1 ] = 0 and,
writing Y1 = (Y1,1 , Y1,2 ), we have Cov(Y1,1 , Y1,2 ) = 0 < ∞ (since Y1,1 = 0 when
Y1,2 6= 0 and vice-versa), V ar(Y1,1 ) = 12 < ∞, and V ar(Y1,2 ) = 21 < ∞.
ON THE RECURRENCE OF RANDOM WALKS
19
We have now shown that under the appropriate conditions, random walks on R
and on R2 are recurrent. One may now ask: under what conditions are random
walks on R3 recurrent? What about random walks on Rd in general? As we shall
prove below, if d ≥ 3, then no random walks on Rd are recurrent. However, since,
for example, one could easily embed a simple random walk on Z (which is recurrent)
into R3 and call it a random walk on R3 , we see that we must be at least somewhat
careful with our definition of a random walk on Rd . Hence, we shall introduce the
notion of truly d-dimensional random walks, as defined below.
Definition 3.6. Let {Sn } = {X1 + . . . + Xn } be a random walk on Rd . If there
exists θ ∈ Rd such that θ 6= 0 and P (X1 · θ = 0) = 1, then we say {Sn } is not truly
d-dimensional.
Essentially, if with probability 1 the steps of a random walk reside in some
space with dimension strictly less than d, then that random walk is not truly ddimensional. For example, any simple random walk on Zd is truly d-dimensional;
but embedding a simple random walk {Sn } = {X1 + . . . + Xn } on Zk into Rm
when k < m results in a random walk on Rm which is not truly m-dimensional, as
necessarily (and hence with probability 1) X1 · ek+1 = 0.
We shall now prove that restricting ourselves to truly d-dimensional random
walks on Rd is a sufficient condition for transience when d ≥ 3. This will resolve
the remainder of the question posed above, for if a random walk on Rd is not truly
d-dimensional, then it reduces to the case of a random walk on Rd−1 . We begin by
proving two lemmas, the latter of which will be necessary later.
Lemma 3.7. Let a, b ∈ R such that a ≤ 1. Then (taking
1
1
0 ≤ Re
.
≤
1 − (a + bi)
1−a
1
0
= ∞)
Proof. Let z = a + bi. Then
1
1−z
1−z
1
1 − Re(z)
= Re
·
= Re
;
Re
=
2
1−z
1−z 1−z
|1 − z|
|1 − z|2
that is,
1
1−a
Re
=
.
1 − (a + bi)
(1 − a)2 + (−b)2
Since a ≤ 1, the right-hand side of the above equation has both a nonnegative
numerator and a nonnegative denominator and hence is nonnegative even if a = 1.
Additionally, if a < 1, since (−b)2 ≥ 0 we certainly have
1
1−a
1
0 ≤ Re
≤
=
2
1 − (a + bi)
(1 − a)
1−a
1
As stated above, when a = 1 we still have 0 ≤ Re 1−(a+bi)
; and also when a = 1
1
1
we trivially have Re 1−(a+bi)
< 1−a
= ∞.
Lemma 3.8. Let a, b ∈ R such that 0 ≤ a ≤ 1 and let c ∈ (0, 1). Then
1
1
0 ≤ Re
≤
.
1 − c(a + bi)
1−a
20
SIMON LAZARUS
Proof. Since a ≥ 0 and c < 1, certainly ac ≤ a, meaning 1 − ac ≥ 1 − a. Thus,
1
1
1−ac ≤ 1−a . Also, since a ≤ 1 and c < 1, Re(c(a + bi)) = ca < 1. So by Lemma
3.7,
1
1
1
0 ≤ Re
≤
≤
.
1 − c(a + bi)
1 − ca
1−a
Theorem 3.9. Let d ∈ N such that d ≥ 3. Let {Sn } = {X1 + . . . + Xn } be a truly
d-dimensional random walk on Rd . Then {Sn } is transient.
Proof. Let ϕ be the characteristic function of X1 , let µ be the probability measure
associated with X1 , and let A = {x ∈ Rd : |x| = 1}. We seek to bound the quantity
1 − Re(ϕ) from below.
We first prove that there exist p > 0 and s0 > 0 such that for all s ∈ (0, s0 ),
Z
(θ · y)2 dµ(y) ≥ p.
(14)
inf
θ∈A
R
|θ·y|<π/(3s)
R
(Here, |θ·y|<π/(3s) means {y∈Rd :|θ·y|<π/(3s)} .) The above integral is clearly nonnegative for all s > 0 and θ ∈ A since for all θ, y ∈ Rd we have (θ · y)2 ≥ 0. So
to prove (14), assume for a contradiction that (*) for all p > 0 and s0 > 0, there
exists s ∈ (0, s0 ) such that the infimum in (14) is < p.
Let n ∈ N. We claim that
Z
(θ · y)2 dµ(y) = 0.
(15)
inf 1 inf
s∈(0, n ) θ∈A
|θ·y|<π/(3s)
For if this were not the case, there would exist c > 0 such that
Z
inf 1 inf
(θ · y)2 dµ(y) = c,
s∈(0, n ) θ∈A
|θ·y|<π/(3s)
which would in turn imply that for all s ∈ (0, n1 ), the infimum in (14) is ≥ c, thus
contradicting our assumption (*) above when we take s0 = n1 . Thus, (15) holds.
π
1
Now note that if s ∈ (0, n1 ) then 3s
> nπ
3 . Thus, given any θ ∈ A and s ∈ (0, n )
we certainly have
Z
Z
(θ · y)2 dµ(y) ≤
(θ · y)2 dµ(y)
|θ·y|<nπ/3
|θ·y|<π/(3s)
since the integrand is nonnegative and the integral on the left is over a smaller
set. Taking the infimum of both sides over all such θ and s and recalling (15), this
means that
Z
Z
2
inf 1 inf
(θ · y) dµ(y) ≤ inf 1 inf
(θ · y)2 dµ(y) = 0.
s∈(0, n ) θ∈A
|θ·y|<nπ/3
s∈(0, n ) θ∈A
|θ·y|<π/(3s)
The integral on the left-hand side of the above is independent of s, so we can
drop the infimum over s ∈ (0, n1 ) on the left-hand side. Furthermore, the above
inequality holds for every
R n ∈ N since our fixed n was arbitrary. That is, for all
n ∈ N we have inf θ∈A |θ·y|<nπ/3 (θ · y)2 dµ(y) = 0. By definition, this implies that
for all n ∈ N, there exists θn ∈ A such that
Z
1
(16)
(θn · y)2 dµ(y) < .
n
|θn ·y|<nπ/3
ON THE RECURRENCE OF RANDOM WALKS
21
Since A is compact, this means there exists θ ∈ A and a subsequence {θnk } of {θn }
such that θnk → θ. Now for all k ∈ N and for all y ∈ Rd , let
(
(θnk · y)2 if |θnk · y| < nk3π
fk (y) =
0
else.
Since nk3π → ∞ and θnk → θ as k → ∞, clearly for all y ∈ Rd we have that
fk (y) → (θ · y)2 as k → ∞. Since each fk is nonnegative on Rd , by Fatou’s Lemma
this means
(17)
Z
Z
Z
(θ · y)2 dµ(y) ≤ lim inf
k→∞
Rd
k→∞
By (16), we have that
Z
lim inf
k→∞
(θnk · y)2 dµ(y).
fk (y)dµ(y) = lim inf
Rd
|θnk ·y|<nk π/3
(θnk · y)2 dµ(y) ≤ lim inf
k→∞
|θnk ·y|<nk π/3
1
= 0,
nk
so (17) reads
Z
(θ · y)2 dµ(y) = 0.
(18)
Rd
But this contradicts the fact that {Sn } is truly d-dimensional, as follows. Since
θ 6= 0, so by definition we have P (θ · X1 6= 0) > 0, i.e. P ((θ · X1 )2 > 0) > 0. That
is,
0 < P (X1 ∈ {y ∈ Rd : (θ · y)2 > 0}) = µ({y ∈ Rd : (θ · y)2 > 0})
meaning
Z
(θ · y)2 dµ(y)
0<
{y∈Rd :(θ·y)2 >0}
since (θ · y)2 is strictly positive on {y ∈ Rd : (θ · y)2 > 0} and that set has positive
measure. So since for all y ∈ Rd , (θ · y)2 ≥ 0 and since {y ∈ Rd : (θ · y)2 > 0} ⊆ Rd ,
we certainly have
Z
Z
0<
(θ · y)2 dµ(y) ≤
(θ · y)2 dµ(y),
{y∈Rd :(θ·y)2 >0}
Rd
a contradiction to (18). Thus, (14) holds.
√
Now for all r > 0, let Br = {x ∈ Rd : |x| ≤ r d}. Since (by Theorem 1.7)
s0
ϕ(0) = 1 and ϕ is uniformly continuous on Rd , we can choose r ∈ (0, √
) such that
d
Re(ϕ(x)) > 0 whenever x ∈ Br . Then certainly Re(ϕ(x)) > 0 for all x ∈ [−r, r]d ,
as we know [−r, r]d ⊆ Br .
√
√
Now let θ1 ∈ A and s ∈ (0, r d]. Then by ourR choice of r we know s ≤ r d < s0 .
Since µ is a probability measure, we also have Rd 1dµ = 1. So since for all z ∈ R,
1 − cos(z) ≥ 0, we have
Z
Z
1 − Re(ϕ(sθ1 )) =
(1 − cos(sθ1 · y))dµ(y) ≥
(1 − cos(sθ1 · y))dµ(y).
Rd
By Lemma 2.10, this means
Z
1 − Re(ϕ(sθ1 )) ≥
|sθ1 ·y|<π/3
|sθ1 ·y|<π/3
(sθ1 · y)2
s2
dµ(y) =
4
4
Z
|θ1 ·y|<π/(3s)
(θ1 · y)2 dµ(y).
22
SIMON LAZARUS
Since s ∈ (0, s0 ), applying (14) to the right-hand side of the above thus gives
s2
1 − Re(ϕ(sθ1 )) ≥
· p > 0.
4
√
Since θ1√∈ A and s ∈ (0, r d] were arbitrary, we see that for all θ ∈ A and for all
s ∈ [0, r d] we have
1
4
(19)
0<
≤ 2
1 − Re(ϕ(sθ))
ps
(for if s = 0 then this follows trivially from ∞ ≤ ∞).
Now let c ∈ (0, 1) and recall that Re(ϕ(x)) > 0 for all x ∈ [−r, r]d . So since
|ϕ| ≤ 1 on Rd , we see that for all x ∈ [−r, r]d , the hypotheses of Lemma 3.8 are
satisfied with a + bi = ϕ(x). Thus, for all x ∈ [−r, r]d we have
1
1
≤
.
Re
1 − cϕ(x)
1 − Re(ϕ(x))
This means that
Z
Re
[−r,r]d
1
1 − cϕ(x)
Z
1
dx.
1 − Re(ϕ(x))
dx ≤
[−r,r]d
1
Now since |ϕ| ≤ 1 on Rd , certainly for all x ∈ Br we have 1−Re(ϕ(x))
≥ 0. So since
d
[−r, r] ⊆ Br , the above inequality implies
Z
Z
1
1
Re
dx ≤
dx.
1 − cϕ(x)
[−r,r]d
Br 1 − Re(ϕ(x))
By changing to spherical coordinates (and noting that 0 ≤ sin(t) ≤ 1 if t ∈ [0, π],
meaning we can ignore all of the volume element of this change except for the sd−1
below) and recalling that A = {x ∈ Rd : |x| = 1}, we therefore have
Z
Z r √d
Z
1
1
Re
dx ≤
dθds,
sd−1
1
−
cϕ(x)
1
−
Re(ϕ(sθ))
d
[−r,r]
0
A
which by (19) means
Z
Z
√
r d
Z
4
dθds.
2
ps
0
A
R
R
Now let C = p4 A dθ. Then C ∈ (0, ∞) since p > 0 and since A dθ is simply the
surface area of the (d − 1)-dimensional unit sphere. Thus,
Z
Z r √d
1
Re
dx ≤ C
sd−3 ds.
1 − cϕ(x)
[−r,r]d
0
Re
[−r,r]d
1
1 − cϕ(x)
dx ≤
d−1
s
Since d ≥ 3, the integral on the right-hand side of the above
equalssome D ∈ (0, ∞).
R
1
Thus, we have that for any given c ∈ (0, 1), [−r,r]d Re 1−cϕ(x)
dx ≤ C · D. So
certainly
Z
1
sup
Re
dx ≤ C · D < ∞.
1 − cϕ(x)
c∈(0,1) [−r,r]d
By Theorem 2.8, this means {Sn } is transient.
ON THE RECURRENCE OF RANDOM WALKS
23
Acknowledgments. I would like to thank Gregory Lawler for giving the lectures
that sparked my interest in random walks. I would like to thank Bowei Zheng for
recommending materials on probability theory and random walks for me to study,
for patiently answering my questions regarding said material and other topics, and
for providing valuable input on drafts of this paper. Finally, I would like to thank
all of the instructors and organizers of the 2013 University of Chicago Mathematics
REU for making the program as valuable and engaging as it was.
References
[1] K. L. Chung and W. H. J. Fuchs. “On the distribution of values of sums of random variables.”
Mem. Amer. Math. Soc. 1951 : 6 (1951).
[2] Rick Durrett. Probability: Theory and Examples, 4th ed. Cambridge University Press, 2010.