Question

Question:For the conditions of Exercise 2, find an unbiased estimator of \({\left( {{\bf{E}}\left( {\bf{X}} \right)} \right)^{\bf{2}}}\). Hint: \({\left( {{\bf{E}}\left( {\bf{X}} \right)} \right)^{\bf{2}}}{\bf{ = E}}\left( {{{\bf{X}}^{\bf{2}}}} \right){\bf{ - Var}}\left( {\bf{X}} \right)\)

Short answer

\(\frac{1}{n}{\sum\limits_{i = 1}^n {{X_i}} ^2} - \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {X - \overline x } \right)}^2}} \)

Step-by-step solution

Given information

For a sample of size n, from a population with mean \(\mu \)and finite variance \({\sigma ^2}\) .

The equation of variance is given as:

\(Var\left( X \right) = E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2} \ldots \left( 1 \right)\)

Rewriting the equationand finding the terms

By rewriting the equation (1)

\({\left( {E\left( X \right)} \right)^2} = E\left( {{X^2}} \right) - Var\left( X \right) \ldots \left( 2 \right)\)

Since we are considering sample estimates, we will substitute \(\mu \) with \(\overline x \).

Now,

\(E\left( {{X^2}} \right) = \frac{1}{n}{\sum\limits_{i = 1}^n {{X_i}} ^2} \ldots \left( 3 \right)\)

We know that,

\(Var\left( X \right) = E{\left( {X - \mu } \right)^2}\)

By replacing \(\mu \) with \(\overline x \)

\(\begin{aligned}{c}Var\left( X \right) &= E{\left( {X - \overline x } \right)^2}\\ &= \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {X - \overline x } \right)}^2}} \ldots \left( 4 \right)\end{aligned}\)

Replacing the terms in the equation

By replacing (3) and (4) in (2), we get,

\(\begin{aligned}{}{\left( {E\left( X \right)} \right)^2} &= E\left( {{X^2}} \right) - Var\left( X \right)\\ &= \frac{1}{n}{\sum\limits_{i = 1}^n {{X_i}} ^2} - \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {X - \overline x } \right)}^2}} \end{aligned}\)