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)

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

Statement

Let $X$ be a non-negative real random variable, then we claim:

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

Proof

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

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

  that $\E{X}\eq I_1+I_2$

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

  We see that $\E{X}\ge 0+\alpha\P{X\ge\alpha}$

  • As $\alpha>0$ we may divide both sides by $\alpha$ to obtain:
    • $$\P{X\ge\alpha}\le \frac{\E{X} }{\alpha}$$

Todo

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