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Chapter 8: Q16E (page 514)

Question:Reconsider the conditions of Exercise 3. Suppose that n = 2, and we observe\({{\bf{X}}_{\bf{1}}}{\bf{ = 2}}\,\,{\bf{and}}\,\,{{\bf{X}}_{\bf{2}}}{\bf{ = - 1}}\). Compute the value of the unbiased estimator of\({\left[ {{\bf{E}}\left( {\bf{X}} \right)} \right]^{\bf{2}}}\) found in Exercise 3. Describe a flaw that you have discovered in the estimator.

Short Answer

Expert verified

The value of an unbiased estimate of \({\left[ {E\left( X \right)} \right]^2}\) is -2. But this is unacceptable because \({\left[ {E\left( X \right)} \right]^2} \ge 0\)we should demand an estimate that is also nonnegative.

Step by step solution

01

Given information

Referring to exercise 3,

Suppose that \(n = 2\) and observe that, \({X_1} = 2\,\,and\,\,{X_2} = - 1\)

02

Finding the value of an unbiased estimator

\({\delta _1} = \frac{1}{n}\sum\nolimits_{i = 1}^n {X_i^2} \)is an unbiased estimator of\(E\left( {{X^2}} \right)\)

Also,

\({\delta _2} = \frac{1}{{n - 1}}\({\left[ {E\left( X \right)} \right]^2}\)\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} \)is an unbiased estimator of\(Var\left( X \right)\)

So,

\({\delta _1} - {\delta _2}\)will be an unbiased estimator of

Then,

\({\left[ {E\left( X \right)} \right]^2} = \frac{1}{n}\sum\nolimits_{i = 1}^n {X_i^2} - \frac{1}{{n - 1}}\sum\nolimits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} \)

For the observed values\({X_1} = 2\,\,and\,\,{X_2} = - 1\)

Obtain the value -2 for the estimate.

This is unacceptable because, \({\left[ {E\left( X \right)} \right]^2} \ge 0\), we should demand an estimate that is also nonnegative.

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