Proof of the weak law of large numbers by Chebyshev's inequality

Let $X_1,X_2,X_3,...,X_n$ be a sequence of independent random variables which are defined on the same sample space $\Omega$.

Using Chebyshev’s inequality, or otherwise, prove the weak law of large numbers as it pertains to a sequence of identically distributed random variables {$X_n$} for $n=1,...,\infty$ each having finite mean=$μ$ and finite variance=$σ^2$.

I'm totally lost as to how to go about proving this question. Any help would be appreciated.

• I think the sequence also has to be independent. – kasa Jan 2 '18 at 16:20
• I've edited the question – B.K97 Jan 3 '18 at 17:46

Define $\bar X=\frac{1}{n}\sum_{i=1}^n{X_i}$. It is easy to verify that $\mathbb E\bar X=\mu$ and $\mathbb V\bar X=\sigma^2/n$. By Chebyshev's inequality, for any $k>0$ it holds that $\Pr[|\bar X-\mu|>k\sigma]\leq \frac{1}{k^2n} \to 0$ as $n\to\infty$. Thus $\bar X\overset{p}{\to} \mu$.
• Can you elaborate on why you have $\frac{1}{k^2 "n"}$ please? – hyg17 Mar 30 at 6:51