|
|
Line 11: |
Line 11: |
| == Chernoff Bound == | | == Chernoff Bound == |
|
| |
|
| | === Moment generating functions === |
| | |
| | === Binomial distribution === |
| | |
| | |
| | === The Chernoff bound === |
| {|border="1" | | {|border="1" |
| |'''Chernoff bound (the upper tail):''' | | |'''Chernoff bound (the upper tail):''' |
Revision as of 05:24, 26 January 2010
Select the Median
The selection problem is the problem of finding the [math]\displaystyle{ k }[/math]th smallest element in a set [math]\displaystyle{ S }[/math]. A typical case of selection problem is finding the median, the [math]\displaystyle{ (\lceil n/2\rceil) }[/math]th element in the sorted order of [math]\displaystyle{ S }[/math].
The median can be found in [math]\displaystyle{ O(n\log n) }[/math] time by sorting. There is a linear-time deterministic algorithm, "median of medians" algorithm, which is very sophisticated. Here we introduce a much simpler randomized algorithm which also runs in linear time. The idea of this randomized algorithm is by sampling.
Randomized median algorithm
Analysis
Chernoff Bound
Moment generating functions
Binomial distribution
The Chernoff bound
Chernoff bound (the upper tail):
- Let [math]\displaystyle{ X=\sum_{i=1}^n X_i }[/math], where [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are independent Poisson trials. Let [math]\displaystyle{ \mu=\mathbf{E}[X] }[/math].
- Then for any [math]\displaystyle{ \delta\gt 0 }[/math],
- [math]\displaystyle{ \Pr[X\ge (1+\delta)\mu]\lt \left(\frac{e^{\delta}}{(1+\delta)^{(1+\delta)}}\right)^{\mu}. }[/math]
|
Chernoff bound (the lower tail):
- Let [math]\displaystyle{ X=\sum_{i=1}^n X_i }[/math], where [math]\displaystyle{ X_1, X_2, \ldots, X_n }[/math] are independent Poisson trials. Let [math]\displaystyle{ \mu=\mathbf{E}[X] }[/math].
- Then for any [math]\displaystyle{ 0\lt \delta\lt 1 }[/math],
- [math]\displaystyle{ \Pr[X\le (1-\delta)\mu]\lt \left(\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}\right)^{\mu}. }[/math]
|
Chernoff-Hoeffding bound:
- Let [math]\displaystyle{ X=\sum_{i=1}^n X_i }[/math], where for each [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ X_i }[/math] is independently distributed over the range [math]\displaystyle{ [0,1] }[/math]. Let [math]\displaystyle{ \mu=\mathbf{E}[X] }[/math].
- Then for any [math]\displaystyle{ \delta\gt 0 }[/math],
- [math]\displaystyle{ \Pr[X\ge (1+\delta)\mu]\lt \left(\frac{e^{\delta}}{(1+\delta)^{(1+\delta)}}\right)^{\mu}; }[/math]
- and for any [math]\displaystyle{ 0\lt \delta\lt 1 }[/math],
- [math]\displaystyle{ \Pr[X\ge (1-\delta)\mu]\lt \left(\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}}\right)^{\mu}. }[/math]
|
Useful forms of the Chernoff bound
- Let [math]\displaystyle{ X=\sum_{i=1}^n X_i }[/math], where for each [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ X_i }[/math] is independently distributed over the range [math]\displaystyle{ [0,1] }[/math]. Let [math]\displaystyle{ \mu=\mathbf{E}[X] }[/math]. Then
- 1. for [math]\displaystyle{ 0\lt \delta\le 1 }[/math],
- [math]\displaystyle{ \Pr[X\ge (1+\delta)\mu]\lt \exp\left(-\frac{\mu\delta^2}{3}\right); }[/math]
- [math]\displaystyle{ \Pr[X\le (1-\delta)\mu]\lt \exp\left(-\frac{\mu\delta^2}{2}\right); }[/math]
- 2. for [math]\displaystyle{ t\gt 0 }[/math],
- [math]\displaystyle{ \Pr[X\ge\mu+t]\le \exp\left(-\frac{2t^2}{n}\right); }[/math]
- [math]\displaystyle{ \Pr[X\le\mu-t]\le \exp\left(-\frac{2t^2}{n}\right); }[/math]
- 3. for [math]\displaystyle{ t\ge 2e\mu }[/math],
- [math]\displaystyle{ \Pr[X\ge t]\le 2^{-t}. }[/math]
|
Permutation Routing