Alec's sample mean bound

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Needs some work, like what is a random variable for which expectation and variance are defined? Can we have complex or vector ones for example?

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Notice

It appears that this is actually 3 inequalities in one, which we shall name as follows:

1. Alec's remaining probability bound - that for $X$ a real and non-negative random variable, $$\forall \alpha\in\mathbb{R}_{>0}\left[\P{X\ge \alpha}\le\frac{\E{X} }{\alpha}\right]$$
2. Alec's deviation probability bound - that for a real random variable (possibly negative) $X$ that $$\forall\beta\in\mathbb{R}_{>0}\left[\mathbb{P}\Big[\big\vert X-\E{X}\big\vert\ge\beta\Big]\le\frac{\Var{X} }{\beta^2} \right]$$
3. Alec's sample mean bound (this page)

Inequality

Let $X_1,\ldots,X_n$ be a collection of $n$ random variables which are pairwise independent, such that:

• $\exists\mu\forall i\in\{1,\ldots,n\}\big[\mathbb{E}[X_i]\eq\mu\big]$ - all of the $X_i$ have the same expectation and
  • Alternatively: $\forall i,j\in\{1,\ldots,n\}\big[\mathbb{E}[X_i]\eq\mathbb{E}[X_j]\big]$, but note we need $\mu$ in the expression
• $\exists\sigma\forall i\in\{1,\ldots,n\}\big[\text{Var}(X_i)\eq\sigma^2\big]$ - all the $X_i$ have the same variance
  • Alternatively: $\forall i,j\in\{1,\ldots,n\}\big[\text{Var}(X_i)\eq\text{Var}(X_j)\big]$, but note again we need $\sigma$ in the expression

Then

• For all $\epsilon>0$ we have:
  • $$\mathbb{P}\left[\left\vert\frac{\sum^n_{i\eq 1}X_i}{n}-\mu\right\vert<\epsilon\right]\ge 1-\frac{\sigma^2}{\epsilon^2n}$$
    • Note that the notation here differs from that in my 2011 research journal slightly, but $\sigma$ and $\mu$ were present.
  • $$\mathbb{P}\left[\left\vert\frac{\sum^n_{i\eq 1}X_i}{n}-\mu\right\vert<\epsilon\right]\ge 1-\frac{\sigma^2}{\epsilon^2n}$$

History of form

When I "discovered" this inequality I was looking to say something like "the chance of the sample mean being within so much of the mean is at least ..."

I didn't know how to handle $\vert X-\E{X}\vert$ (what we'd now call $\Mdm{X}$) which is why I applied it to variance, and of course $\big(X-\E{X}\big)^2\ge 0$ - the only condition required for the first inequality.