高级算法 (Fall 2022)/Problem Set 2

Problem 1

Suppose we want to estimate the value of [math]\displaystyle{ Z }[/math]. Let [math]\displaystyle{ \mathcal{A} }[/math] be an algorithm that outputs [math]\displaystyle{ \widehat{Z} }[/math] satisfying [math]\displaystyle{ \Pr[ (1-\epsilon)Z \leq \widehat{Z} \leq (1+\epsilon )Z] \geq \frac{3}{4} . }[/math]

We run [math]\displaystyle{ \mathcal{A} }[/math] independently for [math]\displaystyle{ s }[/math] times, and obtain the outputs [math]\displaystyle{ \widehat{Z}_1,\widehat{Z}_2,\ldots,\widehat{Z}_s }[/math].

Let [math]\displaystyle{ X }[/math] be the median (中位数) of [math]\displaystyle{ \widehat{Z}_1,\widehat{Z}_2,\ldots,\widehat{Z}_s }[/math]. Find the number [math]\displaystyle{ s }[/math] such that [math]\displaystyle{ \Pr[ (1-\epsilon)Z \leq X \leq (1+\epsilon )Z] \geq 1 - \delta . }[/math]

Express [math]\displaystyle{ s }[/math] as a function of [math]\displaystyle{ \delta }[/math]. Make [math]\displaystyle{ s }[/math] as small as possible.

Remark: in this problem, we boost the probability of success from [math]\displaystyle{ \frac{3}{4} }[/math] to [math]\displaystyle{ 1-\delta }[/math]. This method is called the median trick.

Hint: Chernoff bound.

Problem 2

Let [math]\displaystyle{ X }[/math] be a random variable with expectation [math]\displaystyle{ 0 }[/math] such that moment generating function [math]\displaystyle{ \mathbf{E}[\exp(t|X|)] }[/math] is finite for some [math]\displaystyle{ t \gt 0 }[/math]. We can use the following two kinds of tail inequalities for [math]\displaystyle{ X }[/math].

Chernoff Bound

[math]\displaystyle{ \begin{align} \mathbf{Pr}[|X| \geq \delta] \leq \min_{t \geq 0} \frac{\mathbf{E}[e^{t|X|}]}{e^{t\delta}} \end{align} }[/math]

[math]\displaystyle{ k }[/math]th-Moment Bound

[math]\displaystyle{ \begin{align} \mathbf{Pr}[|X| \geq \delta] \leq \frac{\mathbf{E}[|X|^k]}{\delta^k} \end{align} }[/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: Consider the Taylor expansion of the moment generating function and apply the probabilistic method.

• Why would we still prefer the Chernoff bound to the (seemingly) stronger [math]\displaystyle{ k }[/math]-th moment bound?