exercise:06c2817872: Difference between revisions

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\usepackage{pgfplots}
(‘naive exponential estimate’) Let <math>Y_1,\dots,Y_d</math> be independent random variables and assume that <math>|\E(Y_i^k)|\leqslant k!</math> holds for all <math>k\geqslant0</math> and <math>i=1,\dots,d</math>.
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\label{PROBL-EXP-CHERNOFF} (‘naive exponential estimate’) Let <math>Y_1,\dots,Y_d</math> be independent random variables and assume that <math>|\E(Y_i^k)|\leqslant k!</math> holds for all <math>k\geqslant0</math> and <math>i=1,\dots,d</math>.
<ul style="list-style-type:lower-roman"><li> Use the series expansion of <math>\exp(\cdot)</math> and the assumption to get
<ul style="list-style-type:lower-roman"><li> Use the series expansion of <math>\exp(\cdot)</math> and the assumption to get


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</li>
</li>
<li> Compare the bound in (iii) with the bound of [[#MTB-THM |Theorem]].
<li> Compare the bound in (iii) with the bound of [[guide:B846f441d7#MTB-THM |Theorem]].
</li>
</li>
</ul>
</ul>

Latest revision as of 02:17, 2 June 2024

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(‘naive exponential estimate’) Let [math]Y_1,\dots,Y_d[/math] be independent random variables and assume that [math]|\E(Y_i^k)|\leqslant k![/math] holds for all [math]k\geqslant0[/math] and [math]i=1,\dots,d[/math].

  • Use the series expansion of [math]\exp(\cdot)[/math] and the assumption to get
    [[math]] \E(\exp(tY_i))\leqslant \Bigsum{k=0}{\infty}t^k=\left\{\begin{array}{cl}\textfrac{1}{1-t} &\text{ for } t\in(-1,1),\\\infty & \text{otherwise.}\end{array}\right. [[/math]]
  • Show by means of calculus that
    [[math]] \inf_{t\in(0,1)}\exp(-ta)\prod_{i=1}^d\smallfrac{1}{1-t}=\left\{\begin{array}{cl}(\textfrac{a}{d})^d\exp(d-a) &\text{ if } a \gt d,\\ 1 & \text{otherwise.}\end{array}\right. [[/math]]
  • Derive an estimate for [math]\P\bigl[|Y_1+\cdots+Y_d|\geqslant a\bigr][/math] from the above.
  • Compare the bound in (iii) with the bound of Theorem.