For a sequence of i.i.d. random variable $(X_n)$ with $\Bbb{E}X=m$, $\overline{X}_n=(X_1+...+X_n)/n$, then $\overline{X}_n\stackrel{P}{\rightarrow}m$.

Since $(X_n)$ are i.i.d., $(X_n)$ induce the same probability measure $\mu$ on $\Bbb{R}$. So $$\Bbb{P}\{|\overline{X}_n-m|>\epsilon\}=\int_{\{|\frac{x_1+...+x_n}{n}-m|>\epsilon\}}\Bbb{1}(\mathrm{d}\mu)^n$$ where $(\mathrm{d}\mu)^n$ is the product measure of $\mu$. I want to know if we can prove weak law of large numbers by this representation. Any help will be appreciated.

  • $\begingroup$ You should use the Chebyshev inequality. First, you might want to compute $Var\left(\overline{X}_n\right)$. This would of course assume $X_1$ has finite variance. $\endgroup$ – Michael Apr 15 at 3:27
  • $\begingroup$ @Michael Yes, sir! I know the standard proof, first consider the case of finite variation and use Chebyshev inequality, then using truncation to approximate the general case. But I want to know if my attempt is useful. $\endgroup$ – Xin Fu Apr 15 at 3:38
  • $\begingroup$ I don't see how your integral helps to identify concentration properties. The key idea in all proofs of the law of large numbers is to compute the variance of either $\overline{X}_n$ or its truncated version. Somewhere you will want to use the fact that $E[(X_i-m)(X_j-m)]=0$ (or its truncated version $E[(\tilde{X}_i-m_i)(\tilde{X}_j-m_j)]=0$). $\endgroup$ – Michael Apr 15 at 3:41

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