Success Boosting
23 Jan 2024
Motivation
Suppose we have a randomized algorithm A A A which outputs a value Z Z Z that is correct with probability 2 3 . \displaystyle{\frac{2}{3}}. 3 2 .
Suppose for our application, we want to obtain the correct result with probability 0.999 , 0.999, 0.999 , or 1 − 1 n 2 , 1 - \displaystyle{\frac{1}{n^2}}, 1 − n 2 1 , or ( 1 − δ ) . (1 - \delta). ( 1 − δ ) .
To illustrate the usefulness of Chernoff bounds and other concentration inequalities, we can take the aforementioned algorithm A A A . We run A A A a total of O ( log 1 δ ) \displaystyle{\mathcal{O}\left(\log \frac{1}{\delta}\right)} O ( log δ 1 ) times and take the median; with probability ( 1 − δ ) (1 - \delta) ( 1 − δ ) it arrives at the correct answer.
Framework
Suppose we design a randomized algorithm A A A to estimate a hidden statistic Θ \Theta Θ of a dataset and we know in advance 0 < Θ ≤ 1000. 0 < \Theta \leq 1000. 0 < Θ ≤ 1000.
Suppose each time we use A , A, A , it outputs a number X X X with E [ X ] = Θ \mathbb{E}[X] = \Theta E [ X ] = Θ and Var [ X ] = 100 Θ 2 . \text{Var}[X] = 100\,\Theta^2. Var [ X ] = 100 Θ 2 .
Suppose we want to estimate Θ \Theta Θ within tolerance ε \varepsilon ε and with probability 1 − δ . 1 - \delta. 1 − δ .
Accuracy boosting: Repeat A A A a total of 10 12 ε 2 \displaystyle{\frac{10^{12}}{\varepsilon^2}} ε 2 1 0 12 times and take the mean
Success boosting: Find the mean of a total of O ( log 1 δ ) \mathcal{O}\left(\log \frac{1}{\delta}\right) O ( log δ 1 ) trials and take the median to be correct with probability 1 − δ . 1 - \delta. 1 − δ .
Max Load
Suppose we have a fair n n n ‑sided die that is rolled n n n times. On average, what is the largest number of times any outcome is rolled?
Θ ( 1 ) \Theta(1) Θ ( 1 )
Θ ˜ ( log n ) \~{\Theta}(\log n) Θ ˜ ( log n )
Θ ˜ ( n ) \~{\Theta}(\sqrt{n}) Θ ˜ ( n )
Θ ˜ ( n ) \~{\Theta}(n) Θ ˜ ( n )
Let k ∈ [ n ] k \in [n] k ∈ [ n ] be a fixed value and
X i = { 1 if i -th roll is k 0 otherwise. X_i = \begin{cases}
1 &\text{if $i$-th roll is $k$}\\
0 &\text{otherwise.}
\end{cases}
X i = { 1 0 if i -th roll is k otherwise.
Then, E [ X i ] = 1 n . \displaystyle{\mathbb{E}[X_i] = \frac{1}{n}}. E [ X i ] = n 1 .
The total number of rolls with value k k k is X = ∑ i = 1 n X i X = \sum_{i=1}^{n}X_i X = ∑ i = 1 n X i and E [ X ] = 1. \mathbb{E}[X] = 1. E [ X ] = 1.
Using Chernoff bounds, we can let δ = 3 log n \delta = 3\log n δ = 3 log n and find that
Pr [ X ≥ 3 log n ] ≤ 1 n 2 \Pr[X \geq 3\log n] \leq \frac{1}{n^2}
Pr [ X ≥ 3 log n ] ≤ n 2 1
Using union bound, we have at least a ( 1 − 1 n ) \displaystyle{\left(1 - \frac{1}{n}\right)} ( 1 − n 1 ) probability that no outcome will be rolled more than 3 log n 3 \log n 3 log n times.
Coupon Collector
Suppose we have a fair n n n ‑sided die. On average, how many times should we roll the die before we see all possible outcomes among the rolls?
Θ ( n ) \Theta(n) Θ ( n )
Θ ( n log n ) \Theta(n \log n) Θ ( n log n )
Θ ( n n ) \Theta(n \sqrt{n}) Θ ( n n )
Θ ( n 2 ) \Theta\left(n^2\right) Θ ( n 2 )