# Excpected value of unbiased estimator.

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. :)

• That $\bar\Theta$ is unbiased means exactly that $E\bar\Theta=\Theta$. Thus, this is the definition. – Did May 28 '15 at 11:06
• Do you understand characteristic functions? One way to prove the sum of independent normal distributions is normal uses them. – Karl May 28 '15 at 11:34
• I don't know characteristic function. – user180834 May 28 '15 at 15:44
• Or a direct proof(convolution) – 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 $$E(\bar{X})=E\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n}\sum_{i=1}^nE(X_i)=E(X_1)$$

$$V\left(\bar{X}\right)=\frac{1}{n^2}\sum_{i=1}^nV(X_i)=\frac{V(X_1)}{n}$$

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.