Derive the MSE-optimal variance divisor

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For n independent normal draws, consider estimating the variance by \hat\sigma^2 = c\sumi (Xi - \bar X)^2 = c\,	ext{SS} for a constant c. Using \mathbb{E}[	ext{SS}] = (n-1)\sigma^2 and 	ext{Var}(	ext{SS}) = 2(n-1)\sigma^4, find the c that minimizes MSE and show it corresponds to dividing by n+1.

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Set up bias and variance. The estimator c\,	ext{SS} has 	ext{Bias} = \mathbb{E}[c\,	ext{SS}] - \sigma^2 = \big(c(n-1) - 1\big)\sigma^2, \qquad 	ext{Var} = c^2 	ext{Var}(	ext{SS}) = 2(n-1)c^2\sigma^4. Assemble the MSE. Dropping the common \sigma^4 and writing m = n-1, \frac{	ext{MSE}(c)}{\sigma^4} = 2m\,c^2 + (cm - 1)^2. Minimize. Differentiate with respect to c and set to zero: \frac{d}{dc}\Big[2m c^2 + (cm-1)^2\Big] = 4m c + 2m(cm - 1) = 0. Divide by 2m (positive): 2c + cm - 1 = 0 \implies c(m + 2) = 1 \implies c^\star = \frac{1}{m + 2} = \frac{1}{n + 1}. The second derivative is 4m +…

Derive the MSE-optimal variance divisor

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