5
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 $\endgroup$ – Glen_b Dec 6 '15 at 4:39
  • 1
    $\begingroup$ 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. :) $\endgroup$ – AlaskaRon Dec 6 '15 at 9:03
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.