2024-09-16

Ejercicio 6

Sea $X_1,X_2,\dots ,X_n$ una m.a. de tamaño $n$ de una población $X$, con media ${\mu }_X$ y varianza ${\sigma }_X^2$. Demostrar que $\bar{X}$ es un estimador consistente de ${\sigma }_X$.

• Desigualdad de Tchevyshev $P(|{\bar{X}}_n-{\mu }_X|<k{\sigma }_{\bar{x}})\le \frac{1}{k^2}$

$$\begin{array}{r}\underset{n\to \infty }{\lim }P[{\theta }_n-\theta <\epsilon ]=1\\ 1-P(|\overline{X_n}-{\mu }_X|<k{\sigma }_{\bar{X}})\le \frac{1}{k^2}\\ P(|\overline{X_n}-{\mu }_X|<k{\sigma }_{\bar{X}})>1-\frac{1}{k^2}\\ \text{Con }\epsilon =k{\sigma }_{\bar{X}}⟹k=\frac{\epsilon }{{\sigma }_{\bar{X}}}\text{Puesto que }{\sigma }_{\bar{X}}=\frac{{\sigma }_X}{\sqrt{n}}\text{ Si }n\to \infty \\ k=\frac{\epsilon }{\frac{{\sigma }_X}{\sqrt{n}}}=\frac{\epsilon \sqrt{n}}{{\sigma }_x}⟹P(|\bar{X}-{\mu }_x|<\epsilon )>1-\frac{1}{\frac{{\epsilon }^{2}n}{{\sigma }_X}}\\ P(|\overline{X_n}-{\mu }_X|<\epsilon )>1-\frac{{\sigma }_x}{n{\epsilon }^2}\text{Tomamos el lim}\\ \underset{n\to \infty }{\lim }P(|\overline{X_n}-{\mu }_X|<\epsilon )>\underset{n\to \infty }{\lim }(1-\frac{{\sigma }_x}{n{\epsilon }^2})\text{Aplicando (1)}\\ =1⟹\bar{X}\text{ es estimador consistente.}\end{array}\text{\hspace{1em}(1)}$$

De otra manera

\begin{align} \lim_{ n \to \infty } E(\hat{\theta})=\theta \tag{1}\\ \lim_{ n \to \infty } V(\hat{\theta})=0 \tag{2}\\ \hat{\theta}=\bar{X}=\frac{1}{n} \sum_{i=1}^{n}{X_{i}}\implies_{\text{Aplicando 1 }} E(\bar{X})=E\left( \frac{1}{n} \sum_{i=1}^{n}{X_{i}} \right)=\mu_{X} \\ \lim_{ n \to \infty }{E(\bar{X})}=\lim_{ n \to \infty }{\mu_{X}}=\mu_{X} \\ \text{Como: }V(\bar{X})=V\left( \frac{1}{n} \sum_{i=1}^{n}{X_{i}} \right)=\frac{\sigma^{2}_{x}}{n} \\ \lim_{ n \to \infty } {V(\bar{X})}=\lim_{ n \to \infty }{\frac{\sigma^{2}_{x}}{n}}=0 \\ \therefore \text{Por tanto es un estimador consistente} \end{align}