Alec's remaining probability bound

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This page is a stub and just notes, it needs to be done formally and fleshed out Alec (talk) 21:29, 11 January 2018 (UTC)

[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]

[ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]

Statement

Let [ilmath]X[/ilmath] be a non-negative real random variable, then we claim:

• [math]\forall\alpha\in\mathbb{R}_{>0}\left[\P{X\ge \alpha}\le \frac{\E{X} }{\alpha} \right][/math], which we may also write: [math]\forall\beta\in\mathbb{R}_{>0}\left[\P{X\ge \beta\E{X} }\le \frac{1 }{\beta} \right][/math]^{Caveat:Unconfirmed}

Proof

Recall that [math]\E{X}:\eq\int^\infty_0 xf(x)\mathrm{d}x[/math] for [ilmath]f(x)[/ilmath] the probability density function.

• Let [ilmath]\alpha\in\mathbb{R}_{>0} [/ilmath] be given.
  • Now notice that if we define:
    1. [math]I_1:\eq\int_0^\alpha xf(x)\mathrm{d}x[/math] and
    2. [math]I_2:\eq\int_\alpha^\infty xf(x)\mathrm{d}x[/math]

  that [ilmath]\E{X}\eq I_1+I_2[/ilmath]

  • Next we observe that:
    1. [math]I_1\ge \int^\alpha_0 0f(x)\mathrm{d}x\eq 0[/math]
       TODO: BY WHAT THEOREM? MEASURE THEORY NEEDED!
       and,
    2. [math]I_2\ge \int_\alpha^\infty \alpha f(x)\mathrm{d}x\eq \alpha\P{X\ge\alpha} [/math]

  We see that [ilmath]\E{X}\ge 0+\alpha\P{X\ge\alpha} [/ilmath]

  • As [ilmath]\alpha>0[/ilmath] we may divide both sides by [ilmath]\alpha[/ilmath] to obtain:
    • [math]\P{X\ge\alpha}\le \frac{\E{X} }{\alpha} [/math]

Todo

Make this statement formal, I hate one integral for real one for natural numbers (summation)