高级算法 (Fall 2022)/Problem Set 2: Difference between revisions
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== Problem 1 == | == Problem 1 == | ||
Let <math>X</math> be a random variable with expectation <math>0</math> such that moment generating function <math>\mathbf{E}[\exp(t|X|)]</math> is finite for some <math> t > 0 </math>. We can use the following two kinds of tail inequalities for <math> X </math>. | |||
'''''Chernoff Bound''''' | |||
:<math> | |||
\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>k</math>th-Moment Bound''''' | |||
:<math> | |||
\begin{align} | |||
\mathbf{Pr}[|X| \geq \delta] \leq \frac{\mathbf{E}[|X|^k]}{\delta^k} | |||
\end{align} | |||
</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''''': 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>k</math>-th moment bound? |
Revision as of 03:06, 31 October 2022
Problem 1
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?