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I am trying to solve the problem shown below:

problem

To solve (a), I defined the sum of squared errors $f(\hat{\beta}) = \sum_{k = 1}^{n} (y_k - \hat{\beta})^2$. This allows us to identify the least-squares estimate of $\hat{\beta}$: $$-2 \sum_{k = 1}^{n} (y_k - \hat{\beta}) = f'(\hat{\beta}) = 0 \implies \hat{\beta} = \frac{1}{n} \sum_{k = 1}^{n} y_k.$$ This makes sense as one $\hat{\beta}$ is simply the average of all observations.

However, I don't know if my interpretation of (b) is correct. Is it essentially asking us to apply the formula $$Cov(\hat{\beta}^{(1)}, \hat{\beta}^{(2)}) = \frac{\sum_{k = 1}^{n-1} (u_k - \hat{\beta}^{(1)})(v_k - \hat{\beta}^{(2)})}{n-1}$$ where $(u_k) = (y_2, \dots, y_n)$ and $(v_k) = (y_1, y_3, \dots, y_n)$?

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1 Answer 1

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We know that the observations have uncorrelated errors with $\epsilon \sim N(0, \sigma^2)$. By properties of the covariance (see Wikipedia), we obtain $$Cov\left(\hat{\beta}^{(1)}, \hat{\beta}^{(2)}\right) = Cov\left(\frac{1}{n-1} \sum_{k \neq 1}^{n} y_k, \frac{1}{n-1} \sum_{k \neq 2}^{n} y_k\right) = \frac{1}{(n-1)^2} \sum_{k=3}^{n} Cov\left(y_k, y_k\right) = \frac{(n-2)\sigma^2}{(n-1)^2}.$$

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