7.5 Practice Problems
Problem 7.5.1. Let \(X_1,\dots ,X_n\) be independent with mean \(\mu \) and variance \(\sigma ^{2}\). Use Chebyshev’s inequality to show that \(\overline {X}\underset {P}{\longrightarrow }\mu \), and find how large \(n\) must be so that \(P\left (\left |\overline {X}-\mu \right |\geq 0.1\sigma \right )\leq 0.05\).
Show solution
Solution. Since \(\Var \left (\overline {X}\right ) = \sigma ^{2}/n\), Chebyshev gives \[P\left (\left |\overline {X}-\mu \right |\geq \varepsilon \right ) \leq \frac {\sigma ^{2}}{n\varepsilon ^{2}} \longrightarrow 0 ,\] which is the weak law. Setting \(\varepsilon = 0.1\sigma \), \[\frac {\sigma ^{2}}{n(0.1\sigma )^{2}} = \frac {100}{n} \leq 0.05 \quad \Longrightarrow \quad n \geq 2000 .\]
Chebyshev is deliberately crude. The central limit theorem gives \(1.96\sigma /\sqrt {n} = 0.1\sigma \), that is \(n\approx 385\) — about a fifth as many. The bound holds for every distribution with finite variance, which is what it is paying for.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.