Let $ \Theta$ is any parameter of any statistic such that $\Theta $ has an unbiased, normally distributed estimator $\overline \Theta$. Now, it is written in the book, that: $$E \overline \Theta = \Theta$$

I don't understand. I am asking for intuitive explanation and using definition of expected value.

And the second issue: It is also written If a sample $(X_1, ..., X_n)$ comes from normal distribution, them mean(X) is also Normal.

I am not sure. :)

  • 3
    $\begingroup$ That $\bar\Theta$ is unbiased means exactly that $E\bar\Theta=\Theta$. Thus, this is the definition. $\endgroup$ – Did May 28 '15 at 11:06
  • $\begingroup$ Do you understand characteristic functions? One way to prove the sum of independent normal distributions is normal uses them. $\endgroup$ – Karl May 28 '15 at 11:34
  • $\begingroup$ I don't know characteristic function. $\endgroup$ – user180834 May 28 '15 at 15:44
  • $\begingroup$ Or a direct proof(convolution) $\endgroup$ – kjetil b halvorsen May 29 '15 at 11:23

$E \overline \Theta = \Theta$ is the definition of an unbiased estimator.

If $X_1,\dots,X_n$ are i.i.d. normally distributed variables, then $\bar{X}$ is also normally distributed, as a linear combination of the components of a gaussian vector.

Thus you have $$E(\bar{X})=E\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n}\sum_{i=1}^nE(X_i)=E(X_1)$$



Perhaps what is confusing here is the fact that the definition of an unbiased estimator (for $\theta$) is that:

$$ (\forall \theta) E[\bar{\theta}-\theta] = 0 $$

Yes, what everyone means when the write $E[\bar{\theta}-\theta] = 0 $ is that $E[\bar{\theta}-\theta]$ is a function of $\theta$ equal to $0$ everywhere but on first approach it's easy to assume that $\theta$ is a random variable in $E[\bar{\theta}-\theta]$ and confuse being unbiased with Bayesian estimator concepts.


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.