Foundations of Data Science

Collect terms of the summation with t nonzero r i for t = 1, 2, . . . , r/2. There are ( )
n
t
subsets of {1, 2, . . . , n} of cardinality t. Once a subset is fixed as the set of t values of i
with nonzero r i , set each of the r i ≥ 2. That is, allocate two to each of the r i and then
( allocate the remaining r − 2t to the t r i arbitrarily. The number of such allocations is just
r−2t+t−1
) (
t−1 = r−t−1
)
t−1 . So,
r/2
∑
( )( )
n r − t − 1
E(x r ) ≤ r! f(t), where f(t) =
σ 2t .
t t − 1
t=1
Thus f(t) ≤ h(t), where h(t) = (nσ2 ) t
So, we get
t=1
t!
2 r−t−1 . Since t ≤ r/2 ≤ nσ 2 /4, we have
h(t)
h(t − 1) = nσ2
2t
≥ 2.
r/2
∑
E(x r ) = r! f(t) ≤ r!h(r/2)(1 + 1 2 + 1 4 + · · · ) ≤ r!
(r/2)! 2r/2 (nσ 2 ) r/2 .
Applying Markov inequality,
Pr(|x| > a) = Pr(|x| r > a r ) ≤ r!(nσ2 ) r/2 2 r/2
(r/2)!a r
( ) 2rnσ
2 r/2
= g(r) ≤
.
a 2
This holds for all r ≤ s, r even and applying it with r = s, we get the first inequality of
the theorem.
We now prove the second inequality. For even r, g(r)/g(r − 2) = 4(r−1)nσ2 and so
a 2
g(r) decreases as long as r − 1 ≤ a 2 /(4nσ 2 ). Taking r to be the largest even integer
less than or equal to a 2 /(6nσ 2 ), the tail probability is at most e −r/2 , which is at most
e · e −a2 /(12nσ 2) ≤ 3 · e −a2 /(12nσ 2) , proving the theorem.
12.6 Applications of the tail bound
Chernoff Bounds
Chernoff bounds deal with sums of Bernoulli random variables. Here we apply Theorem
12.5 to derive these.
Theorem 12.6 Suppose y 1 , y 2 , . . . , y n are independent 0-1 random variables with E(y i ) =
p for all i. Let y = y 1 + y 2 + · · · + y n . Then for any c ∈ [0, 1],
Pr (|y − E(y)| ≥ cnp) ≤ 3e −npc2 /8 .
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