# why is $1/n \sum_{i} (X_i -10)^2$ unbiased

Let $\{ X_1,X_2,...,X_n \}$ be n observations randomly drawn from normal distribution with mean $10$ and unknown variance. Prove that the estimator $1/n \sum_{i} (X_i -10)^2$ is unbiased. Why is this estimator unbiased? Isn't we've proved that only $1/(n-1) \sum_{i} (X_i -10)^2$ is unbiased?

• No, you proved that $\frac{1}{n-1}\sum_i (X_i-\bar{x})^2$ is unbiased. $\bar{x}$ is closer to the data than $\mu$ is, so you need to divide by a smaller quantity – Glen_b -Reinstate Monica Dec 6 '15 at 4:39
• It's clearly an unbiased statistic for the population parameter $E((1/n)\sum (X_i-10)^2)$. They didn't ask if it was an unbiased estimator of $\sigma^2$. On the other hand, it isn't an unbiased estimator of lots of other parameters,for instance, $\sigma$, kurtosis, etc. :) – AlaskaRon Dec 6 '15 at 9:03

If $X_i \sim \mathcal N(10, \sigma^2)$, then $Y_i := X_i - 10 \sim \mathcal N(0, \sigma^2)$. Thus $$\mathbb E[(X_i - 10)^2] = \mathbb E[Y_i^2] = \mathrm{Var}(Y_i) + \mathbb E[ Y_i]^2 = \sigma^2,$$ so by linearity of expectation $$\mathbb E\left[\frac1N \sum_{i=1}^N (X_i - 10)^2\right] = \frac1N \sum_{i=1}^N \sigma^2 = \sigma^2.$$
Because you know the mean, you don't have to do the Bessel correction of dividing by $N-1$, and in fact that would bias your estimator.