2 The Bounded Difference Inequality Definition 1. Let A be some set and : An → R. We say o satisfies the bounded difference assumption if 3c₁,..., Cn0 s.t. Vi, 1 < i 0 and 212 Pr{p(X₁,..., Xn) - E[p(X₁,..., Xn)] > t} t} < 2e 14. - Remark. The bounded difference inequality recovers Hoeffding's inequality [2]. Assume X, E [a, bi] and take (X₁,..., Xn) = 1 X₁. Then c₁ = biai. Pluging everything into the BDI gives n n 212 Pr{| Σ X₁ - E[X]| ≥ t} < 2e_i=1(bi-ai)². i=1 How to get to this step?

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2 The Bounded Difference Inequality
Definition 1. Let A be some set and : An → R. We say & satisfies the bounded difference assumption if
₁,..., Cn0 s.t. Vi, 1 ≤ i ≤ n
|(21,..,i• ‚ xn) — (x₁, ..., x'i,
,xn)| ≤ Ci
That is, if we subsitute x; to x';, while keeping other x; fixed, changes by at most c₁.
Theorem 1. Let X₁,..., Xn be arbitrary independent random variables on set A and : An → R satisfy
the bounded difference assumption. Then Vt > 0
and
sup
x1,...,xnx EA
2+2
Pr{p(X₁,..., Xn) - E[o(X₁,..., Xn)] > t} <e_ [²=1²
е
2+2
Pr{p(X₁, ..., Xn) - E[o(X₁, ..., Xn)] < −t} < e¯ï=1&².
Remark. By combining the above two inequalities, we obtain:
2+2
Pr{|(X₁,..., Xn) - E[o(X₁,..., Xn)]| ≥ t} < 2e¯¯ï=1¢.
Remark. The bounded difference inequality recovers Hoeffding's inequality [2]. Assume X₁ € [ai, bi] and
take (X₁,..., Xn) = 1 X₁. Then = bị -
- ai. Pluging everything into the BDI gives
Pr{| Σ
n
2+2
X; – E[Σ X;]| ≥ t} < 2€¯¯¤ï=1(b;−0;)²
Xi
i=1
How to get to this step?
Transcribed Image Text:2 The Bounded Difference Inequality Definition 1. Let A be some set and : An → R. We say & satisfies the bounded difference assumption if ₁,..., Cn0 s.t. Vi, 1 ≤ i ≤ n |(21,..,i• ‚ xn) — (x₁, ..., x'i, ,xn)| ≤ Ci That is, if we subsitute x; to x';, while keeping other x; fixed, changes by at most c₁. Theorem 1. Let X₁,..., Xn be arbitrary independent random variables on set A and : An → R satisfy the bounded difference assumption. Then Vt > 0 and sup x1,...,xnx EA 2+2 Pr{p(X₁,..., Xn) - E[o(X₁,..., Xn)] > t} <e_ [²=1² е 2+2 Pr{p(X₁, ..., Xn) - E[o(X₁, ..., Xn)] < −t} < e¯ï=1&². Remark. By combining the above two inequalities, we obtain: 2+2 Pr{|(X₁,..., Xn) - E[o(X₁,..., Xn)]| ≥ t} < 2e¯¯ï=1¢. Remark. The bounded difference inequality recovers Hoeffding's inequality [2]. Assume X₁ € [ai, bi] and take (X₁,..., Xn) = 1 X₁. Then = bị - - ai. Pluging everything into the BDI gives Pr{| Σ n 2+2 X; – E[Σ X;]| ≥ t} < 2€¯¯¤ï=1(b;−0;)² Xi i=1 How to get to this step?
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