View source for Alec's remaining probability bound

{{Stub page|grade=A**|msg=This page is a stub and just notes, it needs to be done formally and fleshed out [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 21:29, 11 January 2018 (UTC)}}{{ProbMacros}}
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==Statement==
Let {{M|X}} be a [[non-negative real random variable]], then we claim:
* {{MM|\forall\alpha\in\mathbb{R}_{>0}\left[\P{X\ge \alpha}\le \frac{\E{X} }{\alpha} \right]}}, which we may also write: {{MM|\forall\beta\in\mathbb{R}_{>0}\left[\P{X\ge \beta\E{X} }\le \frac{1 }{\beta} \right]}}<sup>{{Caveat|Unconfirmed}}</sup>
==Proof==
Recall that {{MM|\E{X}:\eq\int^\infty_0 xf(x)\mathrm{d}x}} for {{M|f(x)}} the [[probability density function]].
* Let {{M|\alpha\in\mathbb{R}_{>0} }} be given.
** Now notice that if we define:
**# {{MM|I_1:\eq\int_0^\alpha xf(x)\mathrm{d}x}} and 
**# {{MM|I_2:\eq\int_\alpha^\infty xf(x)\mathrm{d}x}}
*: that {{M|\E{X}\eq I_1+I_2}}
** Next we observe that:
**# {{MM|I_1\ge \int^\alpha_0 0f(x)\mathrm{d}x\eq 0}} {{XXX|BY WHAT THEOREM? MEASURE THEORY NEEDED!}} and,
**# {{MM|I_2\ge \int_\alpha^\infty \alpha f(x)\mathrm{d}x\eq \alpha\P{X\ge\alpha} }}
*: We see that {{M|\E{X}\ge 0+\alpha\P{X\ge\alpha} }}
** As {{M|\alpha>0}} we may divide both sides by {{M|\alpha}} to obtain:
*** {{MM|\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)