How do Chebyshev inequality and Weak Law of Large Number translate into this?

Question

The Chebyshev inequality states that if $X$ is a random variable with mean $\mu$ and variance $\sigma^2$, then:

$$\mathbf P\bigl(|X -\mu |\geq c\bigr)\leq\frac{\sigma^2}{c^2},~~~~\textrm{for all}~c > 0.$$

The Weak Law of Large numbers states that for $n$ independent identically distributed r.v.'s $X_i$:

$${\mathbf P\bigl(|M_n -\mu |\geq \epsilon\bigr) = \mathbf P\Bigl(\Bigl|\frac{X_1+\cdots+X_n}{n}-\mu\Bigr|\geq\epsilon\Bigr)\rightarrow 0,\textrm{ as}~n\rightarrow\infty.}$$

In a sample of $n$ independent identically distributed r.v.'s $X_i$ with sample mean $M_ n=(X_1+\cdots +X_ n)/n$, it is said that the Chebyshev inequality gives us:

$$\mathbf{P}\big(|M_ n-\mu | \geq \epsilon \big)\leq\frac{\sigma^2}{n\epsilon ^2}.$$

Why is $n$ there in the denominator? Am I missing something here?

Answer 1

Let's denote $S_{n} = \frac{X_{1}+\cdots + X_{n}}{n}$ and observe that is $X_{1},\cdots X_{n}$ (independent) have the same distribution of random variable $X$ (with second moment) $m = \mathbb{E}(\frac{S_{n}}{n})$.

Using Chebyshev with $\frac{S_{n}}{n}$ we get

$$\begin{cases}\mathbb{P}(|\frac{S_{n}}{n} - m|\geq \epsilon) \leq \frac{1}{\epsilon^{2}}\cdot Var(\frac{S_{n}}{n}) \\ Var(\frac{S_{n}}{n}) = \frac{1}{n^{2}}\cdot Var(S_{n}) = \frac{1}{n^{2}}\cdot Var(X_{1}+\cdots+X_{n}) = \frac{1}{n^{2}}\cdot nVar(X) \end{cases}$$

Which explains the $n$ in the denominator.

And we conclude with $0 \leq \mathbb{P}(|\frac{S_{n}}{n} - m|\geq \epsilon) \leq \frac{1}{n}\cdot Var(X) \underset{n \to \infty}{\longmapsto} 0$

Answer 2

Well, according to page 2 this paper, and $\overline{x}$ is the sample mean in a sample of size $n$:

$$\displaystyle\sigma_{\overline{x}}^2 = \frac{\sigma^2}{n}$$

In this case, $\overline{x} = M_n$. Thus we have:

$$\mathbf{P}\big(|M_ n-\mu | \geq \epsilon \big)\leq\frac{\sigma_{\overline{x}}^2}{\epsilon ^2} = \frac{\sigma^2}{n\epsilon ^2}.$$

That explains the $n$ in the denominator.