Foundations of Data Science

Let
γ i = π 1 + π 2 + · · · + π i .
Define a function f : [0, γ i0 ] →Reals by f(x) = v i − 1 for x ∈ [γ i−1 , γ i ). See figure (5.5).
Now,
i 0
∑
i=1
(v i − 1)π i =
∫ γi0
0
f(x) dx. (5.4)
We make one more technical modification. We divide {1, 2, . . . , i 0 } into groups G 1 , G 2 , G 3 , . . . , G r ,
of contiguous subsets. [We specify the groups later.] Let u t = Max i∈Gt v i . Then we define
a new function g(x) by g(x) = u t by (see figure (5.5)): for x ∈ ∪ i∈Gt [γ i−1 , γ i ). Clearly,
We now assert (with u r+1 = 0):
∫ γi0
0
∫ γi0
0
g(x) dx =
f(x) dx ≤
∫ γi0
0
g(x) dx. (5.5)
r∑
π(G 1 ∪ G 2 ∪ . . . ∪ G t )(u t − u t+1 ). (5.6)
t=1
This is just the statement that the area under g(x) in the figure is exactly covered by the
rectangles whose bottom sides are the dotted lines. We leave the formal proof of this to
the reader. We now focus on proving that
r∑
π(G 1 ∪ G 2 ∪ . . . ∪ G t )(u t − u t+1 ) ≤ ε/2, (5.7)
t=1
(for a sub-division into groups we specify) which suffices by (5.3, 5.4, 5.5 and 5.6). While
we start the proof of (5.7) with a technical observation (5.8), its proof will involve two nice
ideas: the notion of probability flow and reckoning probability flow in two different ways.
First, the technical observation: if 2 ∑ i≥i 0 +1 (1−v i)π i ≤ ε and we would be done by (5.3).
So assume now that ∑ i≥i 0 +1 (1 − v i)π i > ε/2 from which it follows that ∑ i≥i 0 +1 π i ≥ ε/2
and so, for any subset A of heavy nodes,
Min(π(A), π(Ā)) ≥ ε π(A). (5.8)
2
We now define the subsets. G 1 will be just {1}. In general, suppose G 1 , G 2 , . . . , G t−1 have
already been defined. We start G t at i t = 1+ (end of G t−1 ). Let i t = k. We will define l,
the last element of G t to be the largest integer greater than or equal to k and at most i 0
so that
l∑
π j ≤ εΦγ k /4.
j=k+1
Lemma 5.6 Suppose groups G 1 , G 2 , . . . , G r , u 1 .u 2 , . . . , u r , u r+1 are as above. Then,
π(G 1 ∪ G 2 ∪ . . . G r )(u t − u t+1 ) ≤ 8
tΦε .
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