随机算法 (Fall 2015)/Problem Set 3: Difference between revisions
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Let <math>X_1,X_2,\dots,X_n</math> be independent geometrically distributed random variables each having expectation 2 (each of the <math>X_i</math> is an independent experiment counting the number of tosses of an unbiased coin up to and including the first HEADS). Let <math>X=\sum_{i=1}^nX_i</math> and <math>\delta</math> be a positive real constant. Derive the best upper bound you can on <math>\Pr[X>(1+\delta)(2n)]</math>. | Let <math>X_1,X_2,\dots,X_n</math> be independent geometrically distributed random variables each having expectation 2 (each of the <math>X_i</math> is an independent experiment counting the number of tosses of an unbiased coin up to and including the first HEADS). Let <math>X=\sum_{i=1}^nX_i</math> and <math>\delta</math> be a positive real constant. Derive the best upper bound you can on <math>\Pr[X>(1+\delta)(2n)]</math>. | ||
== Problem 4 == | |||
(Due to J. Naor.) | |||
The <i>Chernoff bound</i> is an exponentially decreasing bound on tail distributions. Let <math>X_1,\dots,X_n</math> be independent random variables and <math>\mathbf{E}[X_i]=0</math> for all <math>1\le i\le n</math>. Define <math>X=X_1+X_2+\dots+X_n</math>. We can use the following two kinds of tail inequalities for <math>X</math>. | |||
* '''Chernoff Bounds:''' | |||
:<math>\Pr[|X|\ge\delta]\le\min_{t\ge 0}\frac{\mathbf{E}[e^{t|X|}]}{e^{t\delta}}</math>. | |||
* '''<math>k</math>th-Moment Bound: | |||
:<math>\Pr[|X|\ge\delta]\le\frac{\mathbf{E}[|X|^k]}{\delta^k}</math>. | |||
# Show that for each <math>\delta</math>, there exists a choice of <math>k</math> such that the <math>k</math>th-moment bound is stronger than the Chernoff bound. (Hint: You may use the probabilistic method.) | |||
# Why would we still prefer the Chernoff bound to the seemingly stronger <math>k</math>th-moment bound? |
Revision as of 05:07, 4 December 2015
Problem 1
Use the Chernoff bounds instead of Chebyshev's inequality in the analysis of the LazySelect Algorithm and try to use as few random samples as possible.
Problem 2
A boolean code is a mapping [math]\displaystyle{ C:\{0,1\}^k\rightarrow\{0,1\}^n }[/math]. Each [math]\displaystyle{ x\in\{0,1\}^k }[/math] is called a message and [math]\displaystyle{ y=C(x) }[/math] is called a codeword. The code rate [math]\displaystyle{ r }[/math] of a code [math]\displaystyle{ C }[/math] is [math]\displaystyle{ r=\frac{k}{n} }[/math]. A boolean code [math]\displaystyle{ C:\{0,1\}^k\rightarrow\{0,1\}^n }[/math] is a linear code if it is a linear transformation, i.e. there is a matrix [math]\displaystyle{ A\in\{0,1\}^{n\times k} }[/math] such that [math]\displaystyle{ C(x)=Ax }[/math] for any [math]\displaystyle{ x\in\{0,1\}^k }[/math], where the additions and multiplications are defined over the finite field of order two, [math]\displaystyle{ (\{0,1\},+_{\bmod 2},\times_{\bmod 2}) }[/math].
The distance between two codeword [math]\displaystyle{ y_1 }[/math] and [math]\displaystyle{ y_2 }[/math], denoted by [math]\displaystyle{ d(y_1,y_2) }[/math], is defined as the Hamming distance between them. Formally, [math]\displaystyle{ d(y_1,y_2)=\|y_1-y_2\|_1=\sum_{i=1}^n|y_1(i)-y_2(i)| }[/math]. The distance of a code [math]\displaystyle{ C }[/math] is the minimum distance between any two codewords. Formally, [math]\displaystyle{ d=\min_{x_1,x_2\in \{0,1\}^k\atop x_1\neq x_2}d(C(x_1),C(x_2)) }[/math].
Usually we want to make both the code rate [math]\displaystyle{ r }[/math] and the code distance [math]\displaystyle{ d }[/math] as large as possible, because a larger rate means that the amount of actual message per transmitted bit is high, and a larger distance allows for more error correction and detection.
- Use the probabilistic method to prove that there exists a boolean code [math]\displaystyle{ C:\{0,1\}^k\rightarrow\{0,1\}^n }[/math] of code rate [math]\displaystyle{ r }[/math] and distance [math]\displaystyle{ \left(\frac{1}{2}-\Theta\left(\sqrt{r}\right)\right)n }[/math]. Try to optimize the constant in [math]\displaystyle{ \Theta(\cdot) }[/math].
- Prove a similar result for linear boolean codes.
Problem 3
Let [math]\displaystyle{ X_1,X_2,\dots,X_n }[/math] be independent geometrically distributed random variables each having expectation 2 (each of the [math]\displaystyle{ X_i }[/math] is an independent experiment counting the number of tosses of an unbiased coin up to and including the first HEADS). Let [math]\displaystyle{ X=\sum_{i=1}^nX_i }[/math] and [math]\displaystyle{ \delta }[/math] be a positive real constant. Derive the best upper bound you can on [math]\displaystyle{ \Pr[X\gt (1+\delta)(2n)] }[/math].
Problem 4
(Due to J. Naor.)
The Chernoff bound is an exponentially decreasing bound on tail distributions. Let [math]\displaystyle{ X_1,\dots,X_n }[/math] be independent random variables and [math]\displaystyle{ \mathbf{E}[X_i]=0 }[/math] for all [math]\displaystyle{ 1\le i\le n }[/math]. Define [math]\displaystyle{ X=X_1+X_2+\dots+X_n }[/math]. We can use the following two kinds of tail inequalities for [math]\displaystyle{ X }[/math].
- Chernoff Bounds:
- [math]\displaystyle{ \Pr[|X|\ge\delta]\le\min_{t\ge 0}\frac{\mathbf{E}[e^{t|X|}]}{e^{t\delta}} }[/math].
- [math]\displaystyle{ k }[/math]th-Moment Bound:
- [math]\displaystyle{ \Pr[|X|\ge\delta]\le\frac{\mathbf{E}[|X|^k]}{\delta^k} }[/math].
- Show that for each [math]\displaystyle{ \delta }[/math], there exists a choice of [math]\displaystyle{ k }[/math] such that the [math]\displaystyle{ k }[/math]th-moment bound is stronger than the Chernoff bound. (Hint: You may use the probabilistic method.)
- Why would we still prefer the Chernoff bound to the seemingly stronger [math]\displaystyle{ k }[/math]th-moment bound?