Complete the missing step from the lecture: for ε1 . . . , εn that are distributed i.i.d. N (0, σ2), for a pair of i ̸= j, calculate E[(εi −ε ̄)(εj −ε ̄)], where ε ̄ = 􏰆ni=1 εi/n.

Sure, here is the step-by-step solution:

Step 1: Expand the expression inside the expectation operator: E[(εi − ε ̄)(εj − ε ̄)] = E[εiεj - εiε ̄ - εjε ̄ + ε ̄^2]

Step 2: Break down the expectation of the sum into the sum of expectations (due to the linearity of expectation): = E[εiεj] - E[εiε ̄] - E[εjε ̄] + E[ε ̄^2]

Step 3: Since εi and εj are independent and identically distributed (i.i.d.) as N(0, σ^2), their expectation is 0 and their covariance is 0 for i ≠ j. Therefore, E[εiεj] = 0.

Step 4: For the terms E[εiε ̄] and E[εjε ̄], note that ε ̄ is the average of all ε's, so εi and εj each contribute 1/n to ε ̄. Therefore, E[εiε ̄] = E[εjε ̄] = σ^2/n.

Step 5: For the term E[ε ̄^2], note that the variance of the average of n i.i.d. variables is the variance of one of the variables divided by n. Therefore, E[ε ̄^2] = σ^2/n.

Step 6: Substitute the results from steps 3, 4, and 5 back into the equation from step 2: = 0 - σ^2/n - σ^2/n + σ^2/n = -σ^2/n.