Random walks
EPFL - Spring Semester 2014-2015
Solutions 7
1. a) This is a direct computation:
(k)
(P φ
)j =
N
−1
X
pjl exp(2πilk/N ) = . . . =
N
−1
X
!
cm exp(2πimk/N )
(k)
exp(2πijk/N ) = λk φj
m=0
l=0
b1) The eigenvalues are λk =
1
2
(exp(2πik/N ) + exp(−2πik/N )) = cos(2πk/N )
b2) Recalling that
N
−1
X
j=0
(
N
exp(2πijk/N ) =
0
if k = 0
otherwise
we obtain that the eigenvalues are

 (
N −1
N −1
1
1 X
1 X
λk =
exp(2πikj/N ) =
exp(2πikj/N ) − 1 =
−1
N −1
N −1
N −1
j=1
j=0
if k = 0
otherwise
c) Observe that if the spectral gap is determined by the second largest eigenvalue (i.e., the one
closest to +1), then adding self-loops of weight α to every state can only reduce the spectral gap,
as this addition increases all eigenvalues. Adding self-loops might only increase the spectral gap
(and therefore speed up the convergence to equilibrium) when the gap is determined by the least
eigenvalue (i.e., the one closest to −1).
c1) In this case, the second largest eigenvalue is given by cos(2π/N ), while the least eigenvalue is
given by cos(2π(N −1)/2N ) = cos(π−π/N ) = − cos(π/N ). So the spectral gap is γ = 1−cos(π/N ).
Adding self-loops of weight α changes these eigenvalues to
α + (1 − α) cos(2π/N )
and α − (1 − α) cos(π/N )
respectively. The weight α realizing the largest spectral gap is therefore solution of the equation
1 − α − (1 − α) cos(2π/N ) = 1 − |α − (1 − α) cos(π/N )| = 1 + α − (1 − α) cos(π/N )
so α (2 + cos(π/N ) − cos(2π/N )) = cos(π/N ) − cos(2π/N ). For N large, the value of α (which
represents approximately the increase of the spectral gap in this case) is therefore roughly given by
1
π2
2π 2
3π 2
α'
1−
−
1
+
=
2
2N 2
N2
4N 2
(and recall that the spectral gap of the original chain is '
π2
)
2N 2
c2) For the complete graph, the computation is (much) simpler. The only eigenvalue not equal
to 1 is − N 1−1 . If one therefore sets α = N1 (which one can check leads to the transition matrix
pij = 1/N for all i, j), this increases the value of the spectral gap from 1 − N 1−1 to 1. In this case,
the convergence to the uniform distribution is immediate: after one step only, the distribution of
the chain is indeed uniform.
1
2. a) The transition matrix is doubly stochastic, so the stationary distribution is uniform (i.e.,
1
π(ij) = KM
= N1 for every (ij) ∈ S). The chain is clearly irreducible (and finite), so also positiverecurrent. It is aperiodic (and therefore ergodic) if and only if either K or M is odd, in which case
π is also a limiting distribution.
b) First observe that when both K and M are even, then one eigenvalue is equal to −1 (take
k = K/2 and m = M/2), so the spectral gap is 0. In the case were K is odd and M is even, the
eigenvalue which is the closest to either +1 or −1 is
π
1
π(K − 1)
1
π2
λ( K−1 , M ) =
cos
− 1 = − cos
+ 1 ' −1 +
2
2
2
K
2
K
4K 2
for K large. We have a symmetric situation for K even and M odd. Finally, when both K and M
are odd, the eigenvalue which are candidates for being the closest to either +1 or −1 are
π
π π(K − 1)
π(M − 1)
1
1
λ( K−1 , M −1 ) =
cos
+ cos
= − cos
+ cos
2
2
2
K
M
2
K
M
2
2
π
π
' −1 +
+
2
2
4K
4M 1
2π
2π 2
λ(1,0) =
cos
+1 '1− 2
2
K
K
1
2π
2π 2
λ(0,1) =
1 + cos
'1− 2
2
M
M
for K, M large. When K is close to M , the first of these three eigenvalues “wins” and determines
the spectral gap.
c) In the case where K = M is an odd (and large) number, N = K 2 = M 2 and the spectral gap is
given approximately by
π2
γ'
2N
so by the theorem seen in class,
√
N
1
log N
nπ 2
n
max kP(ij) − πkTV ≤
exp(−γn) ' exp
−
2
2
2
2N
(ij)∈S
2N log N
which falls below ε for n = 2
2 + c and c = log(1/2ε).
π
2