Foundations of Data Science

g(x)
f(x)
G 1 = {1}; G 2 = {2, 3, 4}; G 3 = {5}.
γ 1 γ 2 γ 3 γ 4 γ 5
x
Figure 5.5: Bounding l 1 distance.
Proof: Let t = c ln(1/π min)
, for a suitable constant c. Let
Φ 2 ε 3
a = a t = 1 t (p 0 + p 1 + · · · + p t−1 )
be the running average distribution. We need to show that ||a − π|| 1
≤ ε. Let
v i = a i
π i
.
The proof has two parts - the first is technical reducing what we need to prove by a series
of manipulations to proving essentially that v i do not drop too fast as we increase i. [Note:
In the extreme, if all v i equal 1, ||a − π|| 1
= 0; so, intuitively, indeed, proving that v i do
not fall too much should give us what we want.] In the second part, we prove that v i do
not fall fast using the concept of “probability flows”.
Renumber states so that v 1 ≥ v 2 ≥ · · · . We call a state i for which v i > 1 “heavy”
since it has more probability according to a than its stationary probability. Let i 0 be the
maximum i such that v i > 1; it is the last heavy state. By Proposition (5.4):
||a − π|| 1
= 2
i 0
∑
i=1
(v i − 1)π i = 2 ∑
154
i≥i 0 +1
(1 − v i )π i . (5.3)