Revision as of 03:28, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> If <math>X</math> is a random variable with mean <math>\mu \ne 0</math> and variance <math>\sigma^2</math>, define the ''relative deviation'' <math>D</math> of <math>X</math> from its mean by <math display="block"> D = \left| \frac {X - \mu}\mu \...")
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BBy Bot
Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

If [math]X[/math] is a random variable with mean [math]\mu \ne 0[/math] and variance [math]\sigma^2[/math],

define the relative deviation [math]D[/math] of [math]X[/math] from its mean by

[[math]] D = \left| \frac {X - \mu}\mu \right|\ . [[/math]]

  • Show that [math]P(D \geq a) \leq \sigma^2/(\mu^2a^2)[/math].
  • If [math]X[/math] is the random variable of Exercise Exercise, find an upper bound for [math]P(D \geq .2)[/math], [math]P(D \geq .5)[/math], [math]P(D \geq .9)[/math], and [math]P(D \geq 2)[/math].