# Alec's remaining probability bound

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
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)

## Statement

Let [ilmath]X[/ilmath] 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 [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. $I_1:\eq\int_0^\alpha xf(x)\mathrm{d}x$ and
2. $I_2:\eq\int_\alpha^\infty xf(x)\mathrm{d}x$
that [ilmath]\E{X}\eq I_1+I_2[/ilmath]
• 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 [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:
• $\P{X\ge\alpha}\le \frac{\E{X} }{\alpha}$

## Todo

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