Chapter 6, Problem 31SE

An estimator θ is said to be consistent if for any ∈ > 0, $P(|\hat{\theta }\text{ - }\theta |\ge \in )\to \text{ 0}$ as n → ∞. That is, $\hat{\theta }$ is consistent if, as the sample size gets larger, it is less and less likely that $\hat{\theta }$ will be further than ∈ from the true value of θ. Show that $\bar{X}$ is a consistent estimator of μ when σ² < ∞ by using Chebyshev’s inequality from Exercise 44 of Chapter 3. [Hint: The inequality can be rewritten in the form

$P(|Y\text{ - }{\mu }_Y|\ge \in )\le {\sigma }_Y^2/\in$

Now identify Y with $\bar{X}$.]

To determine

Show that $\overline{X}$ is a consistent estimator of $\mu$ when ${\sigma }^2<\infty$, using Chebyshev’s inequality.

Explanation of Solution

Calculation:

Chebyshev’s inequality can be rewritten as:

$P\left(\left|Y-{\mu }_Y\right|\ge \in \right)\le \frac{{\sigma }_Y^2}{\in }$.

The random variable considered here is the sample mean, $\overline{X}$. The population mean is $\mu$ and the population variance is ${\sigma }^2$. It is known that the distribution of the sample mean, $\overline{X}$, for a sample of size n, has mean $\mu$ and variance $\frac{{\sigma }^2}{n}$.

The quantity $\in$ is a pre-defined, very small quantity.

Replace Y by $\overline{X}$, ${\mu }_Y$ by $\mu$, ${\sigma }_Y^2$ by $\frac{{\sigma }^2}{n}$ in Chebyshev’s inequality:

$\begin{array}{l}P\left(\left|\overline{X}-\mu \right|\ge \in \right)\le \frac{\frac{{\sigma }^2}{n}}{\in }\\ P\left(\left|\overline{X}-\mu \right|\ge \in \right)\le \frac{{\sigma }^2}{n\in }.\end{array}$

When ${\sigma }^2<\infty$, that is finite, then, the right hand side of the inequality tends to 0 as $n\to \infty$.

As a result, when $n\to \infty$, $P\left(\left|\overline{X}-\mu \right|\ge \in \right)\to 0.$

Thus, using Chebyshev’s inequality, it can be shown that $\overline{X}$ is a consistent estimator of $\mu$ when ${\sigma }^2<\infty$.