Foundations of Data Science

Theorem 12.4 Let 0 < γ ≤ 1, then Pr ob ( s < (1 − γ)m ) ( ) m
e
<
−γ
(1+γ) < e
− γ2 m
(1+γ)
Proof: For any λ > 0
Prob ( s < (1 − γ)m ) = Prob ( − s > −(1 − γ)m ) = Prob ( e −λs > e −λ(1−γ)m) .
Applying Markov’s inequality
Now
Thus,
n∏
Prob ( s < (1 − γ)m ) E(e −λX i
)
< E(e−λx )
e < i=1
−λ(1−γ)m e .
−λ(1−γ)m
E(e −λx i
) = pe −λ + 1 − p = 1 + p(e −λ − 1) + 1.
Prob(s < (1 − γ)m) <
n∏
i=1
[1 + p(e −λ − 1)]
e −λ(1−γ)m .
Since 1 + x < e x Prob ( s < (1 − γ)m ) < enp(e−λ −1)
e −λ(1−γ)m .
Setting λ = ln 1
1−γ
enp(1−γ−1)
Prob ( s < (1 − γ)m ) <
(1 − γ)
(
(1−γ)m
)
e −γ m
<
.
(1 − γ) (1−γ)
But for 0 < γ ≤ 1, (1 − γ) (1−γ) γ2
−γ+
> e 2 . To see this note that
(1 − γ) ln (1 − γ) = (1 − γ)
(−γ ·)
− γ2
2 − γ3
3 − · ·
= −γ − γ2
2 − γ3
3 − · · · + γ2 + γ3
2 + γ4
3 + · · ·
) ( )
= −γ +
(γ 2 − γ2 γ
3
+
2 2 − γ3
+ · · ·
3
= −γ + γ2
2 + γ3
6 + · · ·
≥ −γ + γ2
2 .
It then follows that
Prob ( s < (1 − γ)m ) (
)
e −γ m
<
< e − mγ2
(1 − γ) (1−γ)
2 .
2 .
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