Chernoff-Hoeffding Bounds for Markov Chains:
Generalized and Simplified
Kai-Min Chung1 , Henry Lam2 , Zhenming Liu3 , andMichael Mitzenmacher4
1

2
3

4

Computer Science Department, Cornell University. Supported by a Simons
Foundation Fellowship.
chung@cs.cornell.edu
Mathematics and Statistics Department, Boston University.
khlam@bu.edu
Harvard School of Engineering and Applied Sciences. Supported by NSF grantCCF-0915922.
zliu@fas.harvard.edu
Harvard School of Engineering and Applied Sciences. Supported in part by
NSF grants CCF-0915922 and IIS-0964473.
michaelm@eecs.harvard.edu

Abstract
We prove the first Chernoff-Hoeffding bounds for general (nonreversible) finite-state Markov
chains based on the standard L1 (variation distance) mixing-time of the chain. Specifically,
consider an ergodic Markov chain M and a weight functionf : [n] → [0, 1] on the state space[n] of M with mean µ , Ev←π [f (v)], where π is the stationary distribution of M . A t-steprandom walk (v1 , . . . , vt ) on M starting from the stationary distribution π has expected totalPtweight E[X] = µt, where X , i=1 f (vi ). Let T be the L1 mixing-time of M . We show that theprobability of X deviating fromits mean by a multiplicative factor of δ, i.e., Pr [|X − µt| ≥ δµt],
is at most exp(−Ω δ 2 µt/T ) for 0 ≤ δ ≤ 1, and exp(−Ω (δµt/T )) for δ > 1. In fact, the bounds
hold even if the weight functions fi ’s for i ∈ [t] are distinct, provided that all of them have thesame mean µ.
We also obtain a simplified proof for the Chernoff-Hoeffding bounds based on the spectral expansion λ of M , which is the square root of the second largest eigenvalue (in absolutevalue) of M M̃ , where M̃ is the time-reversal Markov
 chain of M . We show that the probability
2
Pr [|X − µt| ≥ δµt] is at most exp(−Ω δ (1 − λ)µt ) for 0 ≤ δ ≤ 1, and exp(−Ω (δ(1 − λ)µt))
for δ > 1.
Both of our results extend to continuous-time Markov chains, and to the case where thewalk starts from an arbitrary distribution x, at a price of a multiplicative factor depending on
the distribution x in the concentration bounds.
1998 ACM Subject Classification G.3 Probability and Statistics
Keywords and phrases probabilistic analysis, tail bounds, Markov chains
Digital Object Identifier 10.4230/LIPIcs.STACS.2012.124

1Introduction

In this work, we establish large deviation bounds for random walks on general (irreversible)
finite state Markov chains based on mixing properties of the chain in both discrete andcontinuous time settings. To introduce our results we focus on the discrete time setting,which we now describe.
© K-M. Chung, H. Lam, Z. Liu, and M. Mitzenmacher;licensed under Creative Commons License NC-ND
29th Symposium on Theoretical Aspects of Computer Science (STACS’12).
Editors: Christoph Dürr, Thomas Wilke; pp. 124–135
Leibniz International Proceedings in Informatics
Schloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany

K-M. Chung, H. Lam, Z. Liu, and M. Mitzenmacher

125

Let M be an ergodic Markov chain with finite state space V = [n] and stationary
distribution π. Let (v1 , . . . , vt ) denote a t-step random walk on M starting from a distribution
ϕ on V . For every i ∈ [t], let fi : V → [0, 1] be a weight function at step i so that Ev←π [fi (v)] =Ptµ > 0 for all i. Define the total weight of the walk (v1 , . . . , vt ) by X , i=1 fi (vi ). The
expected total weight of the random walk (v1 , . . . , vt ) is E[ 1t X] ≈ µ as t → ∞.
When the vi ’s are drawn independently according to the stationary distribution π, astandard Chernoff-Hoeffding bound says that(2
Pr [|X − µt| ≥ δµt] ≤e−Ω(δ µt)

for 0 ≤ δ ≤ 1,e−Ω(δµt)for δ > 1.

However, when (v1 , . . . , vt ) is a random walk on a Markov chain M , it is known that theconcentration bounds depend inherently on the mixing properties of M , that is the speed atwhich a random walk converges toward its stationary distribution.
Variants of Chernoff-Hoeffding bounds for random walk on Markov chains have beenstudied in several fields with various motivations [5, 10, 11, 12, 17, 16, 7]. For instance,these bounds are linked to the performance of Markov chain Monte Carlo integration
techniques [11, 9]. They have also been applied to various online learning problem [15],testing properties of a given graph [6], leader election problems [10], analyzing the structureof the social networks [2, 13], understanding the performance of data structures [4], andcomputational complexity [7]. Improving such bounds is therefore of general interest.
We improve on previous work in two ways. First, all the existing deviation bounds, as
far as we know, are based on the spectral expansion λ(M ) of the chain M . This spectral
expansion λ(M ) characterizes how much M can stretch vectors in Rn under a normed space
defined by the stationary distribution π, which coincides with the second largest absolute
eigenvalue of M when M is reversible. (A formal definition is deferred to Section 2.) The
most general result for Markov chains in this form (see, e.g. [12, 16]) is(2
Pr [|X − µt| ≥ δµt] ≤kϕkπ e−Ω((1−λ)δ µt)−Ω((1−λ)δµt)

kϕkπ efor 0 ≤ δ ≤ 1,for δ > 1.

(1)

where ϕ is an arbitrary initial distribution and k · kπ is the π-norm (which we define formallylater).
However, for general irreversible Markov chains, the spectral expansion λ does notdirectly characterize the mixing time of a chain and thus may not be a suitable parameter forsuch bounds. A Markov chain M could mix rapidly, but have a spectral expansion λ close to
1, in which case Eq. (1) does not yield meaningful bound. In fact there is a way to modify
any given Markov chain M so that the modified Markov chain M 0 has (asymptotically) thesame mixing-time as M , but the spectral expansion of M 0 equals 1 (Appendix A gives adetailed construction). It is therefore natural to seek a Chernoff-type bound for Markov
chains directly parameterized by the chain’s mixing time T .
Second, most previous analyses for deviation bounds such as Eq. (1) are based on nonelementary methods such as perturbation theory [5, 12, 11, 17]. Kahale [10] and Healy [7]provided two elementary proofs for reversible chains, but their results yield weaker bounds
than those in Eq. (1). Recently, Wagner [16] provided another elementary proof for reversible
chains matching the form in Eq. (1). Together with the technique of “reversiblization” [3, 12],Wagner’s analysis can be generalized to irreversible chains. However, his use of decoupling on
the linear projections outright arguably leads to a loss of insight; here we provide an approachbased on directly tracing the corresponding sequence of linear projections, in the spirit of [7].
This more elementary approach allows us to tackle both reversible and irreversible chains ina unified manner that avoids the use of “reversiblization".

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Chernoff-Hoeffding Bounds for Markov Chains: Generalized and Simplified

As we describe below, we prove a Chernoff-type bound for general irreversible Markov
chains with general weight functions fi based on the standard L1 (variation distance) mixing
time of the chain, using elementary techniques based on extending ideas from [7]. The
exponents of our bounds are tight up to a constant factor. As far as we know, this is thefirst result that shows that the mixing time is sufficient to yield these types of concentration
bounds for random walks on Markov chains. Along the way we provide a unified proof for(1) for both reversible and irreversible chains based only on elementary analysis. This proof
may be of interest in its own right.

2

Preliminaries

Throughout this paper we shall refer M as the discrete time Markov chain under consideration.
Depending on the context, M shall be interpreted as either the chain itself or the correspondingtransition matrix (i.e. it is an n by n matrix such that Mi,j represents the probability a walkat state i will move to state j in the next step). For the continuous time counterpart, we
write Λ as the generator of the chain and let M (t) = etΛ , which represents the transition
probability matrix from t0 to t0 + t for an arbitrary t0 .
Let u and w be two distributions over the state space V. The total variation distancePP
1
between u and w is ku − wkT V = maxA⊆Vi∈A ui −i∈A wi = 2 ||u − w||1 .
Let  > 0. The mixing time of a discrete time Markov chain M is
T () = min {t : maxx kxM t − πkT V ≤ }, where x is an arbitrary initial distribution. The
mixing time of a continuous time Markov chain specified by the generator Λ is T () =min {t : maxx kxM (t) − πkT V ≤ }, where M (t) = eΛt .
We next define an inner product space specified by the stationary distribution π:
I Definition 1 (Inner product under π-kernel). Let M be an ergodic Markov chain with state
space [n] and π be its stationary distribution. Let u and v be two vectors in Rn . The innerP
i vi
product under the π-kernel is hu, viπ = x∈[n] uπ(i).
We may verify that h·, ·iπ indeed forms an inner product space by checking it is symmetric,
linear inpthe first argument, and positive definite. The π-norm of a vector u in Rn iskukπ = hu, uiπ . Note that kπkπ = 1. For a vector x ∈ Rn , we write xk = hx, πiπ π for its
component along the direction of π and x⊥ = x − xk for its component perpendicular to π.
We next define the spectral norm of a transition matrix.
I Definition 2 (Spectral norm). Let M the transition matrix of an ergodic Markov chain.
kπDefine the spectral norm of M as λ(M ) = maxhx,πiπ =0 kxM
kxkπ .
When M is clear from the context, we shall simply write λ for λ(M ). We shall also
refer 1 − λ(M ) as the spectral gap of the chain M . In the case when M is reversible, λ(M )coincides with the second largest eigenvalue of M (the largest eigenvalue of M is always1). However, when M is irreversible, such relation does not hold (one hint to realize thatthe eigenvalues of M for an irreversible chain can be complex, and the notion of being
the second largest may not even be well defined). Nevertheless, we can still connect λ(M )with an eigenvalue of a matrix related to M . Specifically, let M̃ be the time reversal of M :(y,x)
M̃ (x, y) = π(y)M. The multiplicative reversiblization R(M ) of M is R(M ) ≡ M M̃ . Theπ(x)value of λ(M p) then coincides with the square root of the second largest eigenvalue of R(M ),i.e. λ(M ) = λ(R(M )). Finally, notice that the stationary distribution of M , M̃ , and R
are all the same. These facts can be found in [3].

K-M. Chung, H. Lam, Z. Liu, and M. Mitzenmacher

3127Chernoff-Hoeffding Bounds for Discrete Time Markov Chains

We now present our main result formally.
I Theorem 3. Let M be an ergodic Markov chain with state space [n] and stationary
distribution π. Let T = T () be its ε-mixing time for ε ≤ 1/8. Let (V1 , . . . , Vt ) denote at-step random walk on M starting from an initial distribution ϕ on [n], i.e., V1 ← ϕ. For
every i ∈ [t], let fi : [n] → [0, 1] be a weight function at step i such that the expected weightPt
Ev←π [fi (v)] = µ for all i. Define the total weight of the walk (V1 , . . . , Vt ) by X , i=1 fi (Vi ).
There exists some constant c (which is independent of µ, δ and ) such that(2
1. Pr[X ≥ (1 + δ)µt]
2. Pr[X ≤ (1 − δ)µt]≤≤

ckϕkπ exp −δ µt/(72T )for 0 ≤ δ ≤ 1

ckϕkπ exp (−δµt/(72T ))for δ > 1ckϕkπ exp −δ 2 µt/(72T )

for 0 ≤ δ ≤ 1

Before we continue our analysis, we remark on some aspects of the result.
Optimality of the bound The bound given in Theorem 3 is optimal among all bounds
based on the mixing time of the Markov chain, in the sense that for any given T and constantε, one can find a δ, a family of functions {fi : V → [0, 1]}, and a Markov chain with mixing
time T (ε) = T that has deviation probabilities matching the exponents displayed in Theorem
3, up to a constant factor. In this regard, the form of our dependency on T is tight for
constant ε. For example, consider the following Markov chain:
The chain consists of 2 states s1 and s2 .
At any time step, with probability p the random walk jumps to the other state and withprobability 1 − p it stays in its current state, where p is determined below.
for all fi , we have fi (s1 ) = 1 and fi (s2 ) = 0.
Notice that the stationary distribution is uniform and T () = Θ(1/p) when  is a constant.
Thus, we shall set p = Θ(1/T ) so that the mixing-time T (ε) = T . Let us consider a walkstarting from s1 for sufficiently large length t. The probability that the walk stays entirely
in s1 up to time t is (1 − p)t ≈ e−tp = exp(−Θ(t/T )). In other words, for δ = 1 we have
Pr[X ≥ (1 + δ)µt] = Pr[X ≥ t] = Pr[the walk stays entirely in s1 ] = exp(−Θ(t/T ())). This
matches the first bound in Theorem 3 asymptotically, up to a constant factor in the exponent.
The second bound can be matched similarly by switching the values of fi (·) on s1 and s2 .
Finally, we remark that this example only works for  = Ω(1), which is how mixing times
appear in the usual contexts. It remains open, though, whether our bounds are still optimal
when  = o(1).
Dependency on the threshold  of the mixing time Note that the dependence of 
only lies on T (). Since T () is non-decreasing in , it is obvious that  = 1/8 gives the best
bound in the setting of Theorem 3. In fact, a more general form of our bound,√ as will be seenalong our derivation later, replaces 1/72 in the √exponent by a factor (1 − 2)/36. Hencethe optimal choice of  is the maximizer of (1 − 2)/T () (with  < 1/2), which differs fordifferent Markov chains. Such formulation seems to offer incremental improvement and so
we choose to focus on the form in Theorem 3.
Comparison with spectral expansion based Chernoff bound The bound given in
Theorem 3 is not always stronger than spectral expansion based Chernoff bounds (1) that ispresented in, for example, Lezaud [12] and Wagner [16]. Consider, for instance, a random
constant degree regular graph G. One can see that the spectral gap of the Markov chain
induced by a random walk over G is a constant with high probability. On the other

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hand, the mixing time of the chain is at least Ω(log n) because the diameter of a constantdegree graph is at least Ω(log n). Lezaud [12] or Wagner [16] gives us a concentration
bound Pr[X ≥ (1 + )µt] ≤ ckϕkπ exp −Θ(δ 2 µt) when δ < 1 while Theorem 3 gives us
Pr[X ≥ (1 + )µt] ≤ ckϕkπ exp −Θ(δ 2 µt/(log n)) .
Comparison with a union bound Assuming the spectral expansion based Chernoff
bound in Lezaud [12] and Wagner [16], there is a simpler analysis to yield a mixing time
based bound in a similar but weaker form than Theorem 3: we first divide the random walk(V1 , ..., Vt ) into T () groups for a sufficiently small  such that the ith group consists of the
sub-walk Vi , Vi+T () , Vi+2T () , .... The walk in each group is then governed by the Markov
chain M T () . This Markov chain has unit mixing time and as a result, its spectral expansion
can be bounded by a constant (by using our Claim 3.1 below). Together with a union boundacross different groups, we obtain(2
1. Pr[X ≥ (1 + δ)µt]
2. Pr[X ≤ (1 − δ)µt]≤≤

cT kϕkπ exp −δ µt/(72T )for 0 ≤ δ ≤ 1

cT kϕkπ exp (−δµt/(72T ))for δ > 1

cT kϕkπ exp −δ 2 µt/(72T )

for 0 ≤ δ ≤ 1(2)

Theorem 3 shaves off the extra leading factors of T in these inequalities, which hassignificant implications. For example, Eq. (2) requires the walk to be at least Ω(T log T ),while our bounds address walk lengths between T and T log T . Our tighter bound further
can become important when we need a tighter polynomial tail bound.
As a specific example, saving the factor of T becomes significant when we generalize thesebounds to continuous-time chains using the discretization strategy in Fill [3] and Lezaud [12].
The strategy is to apply known discrete time bound on the discretized continuous time chain,say in a scale of b units of time, followed by taking limit as b → 0 to yield the corresponding
continuous time bound. Using this to obtain a continuous analog of Eq. (2) does not work,
since under the b-scaled discretization the mixing time becomes T /b, which implies that theleading factor in Eq. (2) goes to infinity in the limit as b → 0.
We now proceed to prove Theorem 3.
Proof. (of Theorem 3) We partition the walk V1 , ..., Vt into T = T () subgroups so thatthe i-th sub-walk consists of the steps (Vi , Vi+T , ...). These sub-walks can be viewed asP
generated from Markov chain N , M T . Also, denote X (i) , 0≤j≤t/T fi+jT (Vi+jT ) as thePTtotal weight for each sub-walk and X̄ = i=1 X (i) /T as the average total weight.
Next, we follow Hoeffding’s approach [8] to cope with the correlation among the X (i) .
To start,h

Pr[X ≥ (1 + δ)µt] = Pr X̄ ≥ (1 + δ)

E[erX̄ ]µt≤ r(1+δ)µt/T .
T
e

i

(3)

Now noting that exp(·) is a convex function, we use Jensen’s inequality to obtain
X1
r X̄
rX (i)
E[e

]≤

i≤T

T

E[e

].

(4)

(i)

We shall focus on giving an upper bound on E[erX ]. This requires two steps:
First, we show the chain N has a constant spectral gap based on the fact that it takes
one step to mix.
Second, we appy a bound on the moment generating function of X (k) using its spectral
expansion.

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129

Specifically, we shall prove the following claims, whose proofs will be deferred to the next
two subsections.
I Claim 3.1.√Let M be a general ergodic Markov chain with -mixing time T (). We haveλ(M T () ) ≤ 2.
I Claim 3.2. Let M be an ergodic Markov chain with state space [n], stationary distribution
π, and spectral expansion λ = λ(M ). Let (V1 , . . . , Vt ) denote a t-step random walk on M
starting from an initial distribution ϕ on [n], i.e., V1 ← ϕ. For every i ∈ [t], let fi : [n] → [0, 1]
be a weight function at step i such that the expected weight Ev←π [fi (v)] = µ for all i. DefinePt
the total weight of the walk (V1 , . . . , Vt ) by X , i=1 fi (Vi ). There exists some constant c
and a parameter r > 0 that depends only on λ and δ such that
(

2
rX
1.

E[e ]
er(1+δ)µt≤

ckϕkπ exp −δ (1 − λ)µt/36

for 0 ≤ δ ≤ 1

ckϕkπ exp (−δ(1 − λ)µt/36)

for δ > 1.

−rX

2.

E[e
]
e−r(1−δ)µt≤

ckϕkπ exp −δ 2 (1 − λ)µt/36



for 0 ≤ δ ≤ 1.

Claim 3.1 gives a bound on the spectral expansion of each sub-walk X (i) , utilizing thefact that they have unit mixing times. Claim 3.2 is a spectral version of Chernoff bounds
for Markov chains. As stated previously, while similar results exist, we provide our own
elementary proof of claim 3.2, both for completeness and because it may be of independentinterest.
We now continue the proof assuming these two claims. Using Claim 3.1, we knowλ(N ) ≤ 12 . Next, by Claim 3.2, for the i-th sub-walk, we have(i2
rX (i)E[e]er(1+δ)µt/T≤

ckϕM kπ exp −δ µt/(72T )i

ckϕM kπ exp (−δµt/(72T ))for 0 ≤ δ ≤ 1(5)for δ > 1

for an appropriately chosen r (which depends only on λ and δ and hence the same for alli). Note that M i arises because X (i) starts from the distribution ϕM i . On the other hand,notice that kϕM i k2π = kϕk M i k2π + kϕ⊥ M i k2π ≤ kϕk k2π + λ2 (M i )kϕ⊥ k2π ≤ kϕk2π (by using
Lemma 5), or in other words kϕM i kπ ≤ kϕkπ . Together with (3) and (4), we obtain(
2
Pr[X ≥ (1 + δ)µt] ≤ckϕkπ exp −δ µt/(72T )for 0 ≤ δ ≤ 1

ckϕkπ exp (−δµt/(72T ))for δ > 1

This proves the first half of the theorem. The second case can be proved in a similar manner,namely that
Pr[X ≤ (1 − δ)µt] ≤

(k)
T
X

1 E[e−rX ]≤≤ ckϕkπ exp −δ 2 µt/(72T )−r(1−δ)µt/T−r(1−δ)µt/T
Te
e
k=1

E[e−rX̄ ]again by Jensen’s inequality applied to exp(·).
J

3.1

Mixing Time v.s. Spectral Expansion

In this subsection we prove Claim 3.1. We remark that Sinclair [14] presents a similar result
for reversible Markov chains: for every parameter ε ∈ (0, 1),
1 λ(M )1log
≤ T (ε),2 1 − λ(M )2ε(6)where T (ε) is the ε-mixing-time of M . However, in general it is impossible to get a boundon λ(M ) based on mixing time information for general irreversible chains because a chain M

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Chernoff-Hoeffding Bounds for Markov Chains: Generalized and Simplified

can have λ(M ) = 1 but the ε-mixing-time of M is, say, T (ε) = 2 for some constant  (andλ(M 2 )  1).
In light of this issue, our proof of Claim 3.1 depends crucially on the fact that M T (ε) has
mixing time 1, which, as we shall see, translates to a bound on its spectral expansion that
holds regardless of reversibility. We need the following result on reversible Makrov chains,which is stronger result than Eq. (6) from [14].
I Lemma 4. Let 0 < ε ≤ 1/2 be a parameter. Let M be an ergodic reversible Markov chain
with ε-mixing time T (ε) and spectral expansion λ(M ). It holds that λ(M ) ≤ (2ε)1/T (ε) .
We remark that it appears possible to prove Lemma 4 by adopting an analysis similar
to Aldous’ [1], who addressed the continuous time case. We present an alternative proof thatis arguably simpler; in particular, our proof does not use the spectral representation theorem
as used in [1] and does not involve arguments that take the number of steps to infinity.
Proof. (of Lemma 4) Recall that for an ergodic reversible Markov chain M , it holds thatλ(M t ) = λt (M ) for every t ∈ N. Hence, it suffices to show that λ(M T () ) ≤ 2. Also, recall
that λ(M T () ) is simply the second largest eigenvalue (in absolute value) of M T () . Let v be
the corresponding eigenvector, i.e. v satisfies vM T () = λ(M T () )v. Since M is reversible, theentries of v are real-valued. Also, notice that v is a left eigenvector of M while (1, 1, ..., 1)T is
a right eigenvector of M (using the fact that each row of M sums to one). Furthermore, v andP(1, ..., 1)T do not share the same eigenvalue. So we have hv, (1, ..., 1)T i = 0 , i.e. i vi = 0.
Therefore, by scaling v, we can assume w.l.o.g. that x , v + π is a distribution. We have thefollowing claim.
I Claim 3.3. Let x be an arbitrary initial distribution. Let M be an ergodic Markov chain withstationary distribution π and mixing time T (). We have kxM T () − πkT V ≤ 2kx − πkT V .
The key idea for proving this claim is to split the difference x−π into positive and negative
components, and analyze kxM T () − πkT V = k(x − π)M T () kT V using these components
together with a rescaling to transform the components into probability distributions andthen invoke the definition of mixing time. Details are available in the full version of thispaper.
By Claim 3.3, kxM T (ε) − πkT V ≤ 2εkx − πkT V , i.e. kxM T (ε) − πk1 ≤ 2εkx − πk1 .
Observing that (xM T (ε) − π) and (x − π) are simply λ(M T (ε) )v and v, the above inequality
means λ(M T (ε) )kvk1 ≤ 2εkvk1 , which implies λ(M T (ε) ) ≤ 2ε, as desired.
J
Proof. (of Claim 3.1) The idea is to reduce to the reversible case by considering the reversiblization of M T (ε) . Let M̃ T (ε) be the time reversal ofpM T (ε) , and R , M T (ε) M̃ T (ε) be thereversiblization of M T (ε) . By Claim 3.1, λ(M T (ε) ) = λ(R). Let us recall (from Section 2)that M , M T (ε) , and M̃ T (ε) all share the same stationary distribution π. Next, we claim thatthe ε-mixing-time of R is 1. This is because kϕM T (ε) M̃ T (ε) − πkT V ≤ kϕM T (ε) − πkT V ≤ ε,where the second inequality uses the definition of T (ε) and the first inequality holds since
any Markov transition is a contraction mapping: for any Markov transition, say S = (s(i, j)),
P P
P P
P
and any vector x, kxSk1 = j | i xi s(i, j)| ≤ j i |xi |s(i, j) = i |xi | = kxk1 ; puttingx = ϕM T (ε) − π and Sp= M̃ T (ε) √gives the first inequality. Now, by Lemma 4, λ(R) ≤ 2ε,T (ε)
and hence λ(M) = λ(R) ≤ 2ε, as desired.
J

3.2

Bounding the Moment Generating Function

We now prove Claim 3.2. We focus on the first inequality in the claim; the derivation of thesecond inequality is similar.

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131

Claim 3.2 leads directly to a spectral version of the Chernoff bound for Markov chains.
Lezaud [12] and Wagner [16] give similar results for the case where fi are the same for alli. The analysis of [16] in particular can be extended to the case where the functions fi aredifferent. Here we present an alternative analysis and along the way will discuss the merit of
our approach compared to the previous proofs.
Pt
Recall that we define X = i=1 fi (Vi ). We start with the following observation, which
has been used previously [7, 12, 16]:E[erX ] = kϕP1 M P2 ...M Pt k1 ,(7)

where the Pi are diagonal matrices with diagonal entries (Pi )j,j , erfi (j) for j ∈ [n]. One
can verify this fact by observing that each walk V1 , . . . , Vt is assigned
the correspondingP
probability in the product of M ’s with the appropriate weight er i fi (Vi ) .
For ease of exposition, let us assume Pi are all the same at this moment. Let P =P1 = ... = Pt , then (7) becomes kϕ(P M )t−1 P k1 = hϕ(P M )t−1 P, πiπ = hϕ(P M )t , πiπ =kϕ(P M )t k1 (see Lemma 5 below). Up to this point, our analysis is similar to previous
work [5, 12, 7, 16]. Now there are two natural possible ways of bounding kϕ(P M )t k1 =hϕ(P M )t , πiπ .
Approach 1. Bounding the spectral norm of the matrix P M . In this approach,
we observe that hϕ(P M )t , πiπ ≤ kϕkπ kP M ktπ where kP M kπ is the operator norm of thematrix P M induced by k · kπ (see, for example, the proof of Theorem 1 in [16]). Thismethod decouples the effect of each P M as well as the initial distribution. When M isreversible, kP M kπ can be bounded through Kato’s spectral perturbation theory [5, 12, 11].
Alternatively, Wagner [16] tackles the variational description of kP M kπ directly, using
only elementary techniques, whose analysis can be generalized to irreversible chains.
Approach 2. Inductively giving a bound for x(P M )i for all i ≤ t. In thisapproach, we do not decouple the product ϕ(P M )t . Instead, we trace the change of thevector ϕ(P M )i for each i ≤ t. As far as we know, only Healy [7] adopts this approach andhis analysis is restricted to regular graphs, where the stationary distribution is uniform.
His analysis also does not require perturbation theory.
Our proof here generalizes the second approach to any ergodic chains by only using
elementary methods. We believe this analysis is more straightforward for the following
reasons. First, directly tracing the change of the vector ϕ(P M )i for each step keeps thegeometric insight that would otherwise be lost in the decoupling analysis as in [12, 16].
Second, our analysis studies both the reversible and irreversible chains in a unified manner.
We do not use the reversiblization technique to address the case for irreversible chains. While
the reversiblization technique is a powerful tool to translate an irreversible Markov chain
problem into a reversible chain problem, this technique operates in a blackbox manner; proofs
based on this technique do not enable us to directly measure the effect of the operator P M .
We now continue our analysis by using a framework similar to the one presented by
Healy [7]. We remind the reader that we no longer assume Pi ’s are the same. Also, recall
that E[erX ] = kϕP1 M P2 ...M Pt k1 = hϕP1 M P2 ...M Pt , πiπ = k(ϕP1 M P2 ...M Pt )k kπ . Let us
briefly review the strategy from [7].
First, we observe that an arbitrary vector x in Rn can be decomposed into its parallel
component (with respect to π) xk = hx, πiπ and the perpendicular component x⊥ = x−xk
in the Lπ space. This decomposition helps tracing the difference (in terms of the norm)
between each pair of ϕP1 M...Pi M and ϕP1 M...Pi+1 M for i ≤ t, i.e. two consecutive

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Chernoff-Hoeffding Bounds for Markov Chains: Generalized and Simplified

steps of the random walk. For this purpose, we need to understand the effects of thelinear operators M and Pi when they are applied to an arbitrary vector.
Second, after we compute the difference between each pair xP1 M...Pi M and xP1 M...Pi+1 M ,we set up a recursive relation, the solution of which yields the Chernoff bound.
We now follow this step step framework to prove Claim 3.2
The effects of the M and Pi operators Our way of tracing the vector ϕP1 M P2 ...M Pt
relies on the following two lemmas.
I Lemma 5. (The effect of the M operator) Let M be an ergodic Markov chain withstate space [n], stationary distribution π, and spectral expansion λ = λ(M ). Then
1. πM = π.
2. For every vector y with y⊥π, we have yM ⊥π and kyM kπ ≤ λkykπ .
Note that Lemma 5 is immediate from the definitions of π and λ.
I Lemma 6. (The effect of the P operator) Let M be an ergodic Markov chain withstate space [n] and stationary distribution π. Let f : [n] → [0, 1] be a weight function withEv←π [f (v)] = µ. Let P be a diagonal matrix with diagonal entries Pj,j , erf (j) for j ∈ [n],where r is a parameter satisfying 0 ≤ r ≤ 1/2. Then
1. k(πP )k kπ ≤ 1 + (er − 1)µ.
√2. k(πP )⊥ kπ ≤ 2r µ.
√
3. For every vector y⊥π, k(yP )k kπ ≤ 2r µkykπ .
4. For every vector y⊥π, k(yP )⊥ kπ ≤ er kykπ
Items 1 and 4 of Lemma 6 state that P can stretch both the perpendicular and parallel
components along their original directions moderately. Specifically, a parallel vector is
stretched by at most a factor of (1 + (er − 1)µ) ≈ 1 + O(rµ) and a perpendicular vector
is stretched by a factor of at most er ≈ 1 + O(r). (Recall r will be small.) On the other
hand, items 2 and 3 of the lemma state that P can create a new perpendicular component
from a parallel component and vice versa, but the new component is of a much smaller size√
compared to the original component (i.e. only of length at most 2r µ times the originalcomponent).
We note that the key improvement of our analysis (which can be found in the full versionof this work) over that of Healy [7] stems from items 2 and 3 of Lemma 6. Healy [7] proved abound with a factor of (er − 1)/2 = O(r) for both items for the special case of undirected and√regular graphs. Our quantitative improvement to O(r µ) (which is tight) is the key for us
to prove a multiplicative Chernoff bound without any restriction on the spectral expansionof M .
Recursive analysis We now provide a recursive analysis for the terms xP1 M...M Pi fori ≤ t based on our understanding of the effects from the linear operators M and Pi . This
completes the proof for Claim 3.2.
Sketch of proof of Claim 3.2. First, recall that
rX

E[e

k

k

] = k(ϕP1 M P2 ...M Pt ) kπ = k(ϕP1 M P2 ...M Pt M ) kπ =

ϕ

t
Y

!k

(Pi M )

i=1

πwhere the second equality comes from Lemma 5. Our choice of r is r = min{1/2, log(1/λ)/2, 1−√
λ, (1 − λ)δ/18}. We shall explain how we make such a choice as we walk through ouranalysis.

K-M. Chung, H. Lam, Z. Liu, and M. Mitzenmacher

133

We now trace the π-norm of both parallel and perpendicular components of the random
walk for each application of Pi M . Let z0 , ϕ and zi = zi−1 Pi M for i ∈ [t]. By triangle
inequality and Lemma 5 and 6, for every i ∈ [t],kk⊥kzi kπ = k(zi−1 Pi M )k kπ = k((zi−1 + zi−1
)Pi M )k kπ

≤
≤

k

⊥
k(zi−1 Pi M )k kπ + k(zi−1
Pi M )k kπ
√
k
⊥
kπ ,
(1 + (er − 1)µ) kzi−1 kπ + (2r µ) kzi−1

and similarly,√k⊥≤ (2rλ µ) kzi−1 kπ + (er λ) kzi−1
kπ
√
√
k
⊥
≤ (2rλ µ) kzi−1 kπ + λkzi−1 kπ ,√where the last inequality holds when r ≤ (1/2) log(1/λ) i.e. er ≤ 1/ λ. The reason torequire r ≤ (1/2)√ log(1/λ) is that we can guarantee the perpendicular component is shrinking(by a factor of λ < 1) after each step.
k
Now let α0 = kz0 kπ = 1 and β0 = kz0⊥ kπ , and define for i ∈ [t],√√√αi = (1 + (er − 1)µ) αi−1 + (2r µ) βi−1 and βi = (2rλ µ) αi−1 + λβi−1 .
k⊥kzi⊥ kπ ≤ k(zi−1 Pi M )⊥ kπ + k(zi−1
Pi M )⊥ kπk

One can prove by induction easily that kzi kπ ≤ αi and kzi⊥ kπ ≤ βi for every i ∈ [t],and αi ’s are strictly increasing. Therefore, bounding the moment generating functionkE[erX ] = kzt kπ ≤ αt boils down to bounding the recurrence relation for αi and βi .
rObserve that in the recurrence relation, only
√ the coefficient (1 + (e − 1)µ) > 1 while the√√remaining coefficients (2r µ), (2rλ µ), and λ are all less than 1 if r is chosen sufficiently
small. This suggests, intuitively, αi ’s terms will eventually dominate. This provides us a
guide toreduce the recurrence
relation to a single variable. In particular, one can show that√
Pi−1 p j+2 
βi ≤ 2r
λ µ αi−1 + λi β0 for every i ∈ [t], by expanding the recurrence relation
j=0
r
and using the fact that αi’s are increasing. Also, by
substituting βi−11 ≤ (1 + (e −P , we get αp
p
√
√
i−2
1)µ) + 2r µβ0 and αi ≤ 1 + (er − 1)µ + 4r2 µλj+2 µ αi−1 + 2r λi−1 µβ0 for
j=0
every 2 ≤ i ≤ t. One can then show that





√
i−2 p
t
X
Y
8r µβ0√1 + (er − 1)µ + 4r2 µ 
αt ≤ 1 +λj+2 µ(1 + (er − 1))
1−λ
j=0
i=2

which can be further shown to be bounded by

 

√ 
8r µ
8r2
kϕkπ exp(er − 1) +µt .
2 max 1,1−λ(1 − λ)through elementary analysis. Recall that √our goal is to choose an r to bound E[erX ]/er(1+δ)µt .
Choosing r = min{1/2, log(1/λ)/2, 1 − λ, (1 − λ)δ/18} = (1 − λ)δ/18, we complete theproof of Claim 3.2.
J
Before completing this subsection, we make a final remark. Our proof also works evenfor the case Eπ [fi (v)] are different for different values of i, which results in a more general
Chernoff type bound based on spectral expansions. This more general result, as far as weknow, has not been noted in existing literatures with the exception of Healy [7], who gavea Chernoff bound of this kind with stronger assumptions for regular graphs, although theanalysis given by Lezaud [12] or Wagner [16] also appears to be generalizable as well. On theother hand, this strengthened result of Claim 3.2 does not seem to be sufficient to remove
the requirement that Eπ [fi (v)] are the same for Theorem 3.

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Chernoff-Hoeffding Bounds for Markov Chains: Generalized and Simplified

3.3

Continuous Time Case

We now generalize our main result to cover the continuous time chains. The analysis issimilar to the one presented by Lezaud [12] and is presented in the full version of this paper.
I Theorem 7. Let Λ be the generator of an ergodic continuous time Markov chain withstate space [n] and mixing time T = T (). Let {vt : t ∈ R+ } be a random walk on the chain
starting from an initial distribution ϕ such that vt represents the state where the walk stay attime t. Let {ft : [n] → [0, 1] | t ∈ R+ } be a family of functions such that µ = Ev←π [ft (v)]Rt
for all t. Define the weight over the walk {vs : s ∈ R+ } up to time t by Xt , 0 fs (vs )ds.
There exists a constant c such that(2
1. Pr[X ≥ (1 + δ)µt]
2. Pr[X ≤ (1 − δ)µt]≤≤

ckϕkπ exp −δ µt/(72T )for 0 ≤ δ ≤ 1

ckϕkπ exp (−δµt/(72T ))for δ > 1ckϕkπ exp −δ 2 µt/(72T )

for 0 ≤ δ ≤ 1

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A

Construction of Mixing Markov Chain with No Spectral Expansion

135

In this section, we show that any ergodic Markov chain M with mixing time T = T (1/4)
can be modified to a chain M 0 such that M 0 has mixing time O(T ) but spectral expansionλ(M 0 ) = 1.
Our modification is based on the following simple observation. Let M 0 be an ergodic
Markov chain with stationary distribution π 0 . If there exist two states v and v 0 such that (i)
0
0
0
Mv,v
6 v,
0 = 1, i.e., state v leaves to state v with probability 1, and (ii) Mu,v 0 = 0 for all u =i.e., the only state transits to v 0 is v, then λ(M 0 ) = 1: Note that in this case, π 0 (v) = π 0 (v 0 )since all probability mass from v leaves to v 0 , which receives probability mass only fromv. Consider a distribution x whose probability mass all concentrates at v, i.e., xv = 1 andxu = 0 for all u 6= v. One step walk from x results in the distribution xM 0 whose probability
mass all concentrates at v 0 . By definition, kxkπ0 = kxM 0 kπ0 and thus λ(M 0 ) = 1.
Now, let M be an ergodic Markov chain with mixing time T = T (1/4) and stationary
distribution π. We shall modify M to a Markov chain M 0 that preserves the mixing-timeand satisfies the above property. We mention that it is not hard to modify M to satisfy theabove property. The challenge is to do so while preserving the mixing-time. Our construction
is as follows.
For every state v in M , we “split” it into three states (v, in), (v, mid), (v, out) in M 0 .
00For every state (v, in) in M 0 , we set M(v,in),(v,in)= M(v,in),(v,mid)= 1/2, i.e., (v, in)stays in the same state with probability 1/2 and transits to (v, mid) with probability 1/2.
0
For every state (v, mid) in M 0 , we set M(v,mid),(v,out)= 1, i.e., (v, mid) always leaves to(v, out).
0
For every pairs of states u, v in M , we set the transition probability M(u,out),(v,in)from(u, out) to (v, in) to be Mu,v .
It is not hard to verify that the modified chain M 0 is well-defined, ergodic, and satisfies
the aforementioned property (namely, (v, mid) leaves to (v, out) with probability 1 and is
the only state that transits to (v, out)). It remains to show that M 0 has mixing-time O(T ).
Toward this goal, let us define yet another Markov chain C that consists of three states{in, mid, out} with transition probability Cin,in = Cin,mid = 1/2, and Cmid,out = Cout,in = 1.
Clearly, C is ergodic and has constant mixing-time. Now, the key observation is that arandom walk on M 0 can be decomposed into walks on M and C in the following sense: every
step on M 0 corresponding to a step on C in a natural way, and one step on M 0 from (u, out)to (v, in) can be identified as a step from u to v in M . Note that the walks on M and C are
independent, and in expectation, every 4 steps of walk on M 0 induce one step of walk on M .
It is not hard to see from these observation that the mixing time of M 0 is at most 8T .

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