Question
Refer to Exercise 9.18. Is the estimator of $\sigma^{2}$ given there an MVUE of $\sigma^{2} ?$
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18. The estimator of $\sigma^{2}$ is given by: \[ \hat{\sigma}^{2} = \frac{1}{2n-1} \left( \sum_{j=1}^{n} (x_{j} - \bar{x})^{2} + \sum_{j=1}^{n} (y_{j} - \bar{y})^{2} \right) \] Show more…
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Key Concepts
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