MSE of estimator for normal distribution

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Assume $X \sim N(\mu, \sigma)$. Find $c$ for which $ T = c \sum_i (X_i - \bar{X})$ has minimum MSE.

We know that $ \operatorname{MSE}(T)= \operatorname{Var}(T)+ [E(T) - \theta]^2 $

Then:

$$ E(T) = c(n-1)\sigma^2$$

$$ \operatorname{Var}(T) = 2(n-1)c^2\sigma^4$$

$$ \operatorname{MSE}(T) = 2c^2(n-1)\sigma^4 + [(n-1)c\sigma^2 - \sigma^2]^2 = 2c^2(n-1)\sigma^4 + \sigma^4 [(n-1)c \mathbf{{}-1}]^2$$

However I dont get where the highlighted -1 comes form ?

Also I am not sure how to calculate the minimum. Could someone break this problem down to me? I have almost no mathematical background, please state the obvious also.

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The highlighted -1 comes from the factorisation of $\sigma^2$ in the last term: $$\bigl[(n-1)c\sigma^2 - \sigma^2\bigr]^2=\bigl[\sigma^2\bigl( (n-1)c- 1\bigr)\bigr]^2=\sigma^4\bigl[ (n-1)c- 1\bigr]^2.$$