# How to show that the $E[x_i] = \bar{X}$ [duplicate]

I know that $E[x_i] = \sum X_iP_i= \bar{X}$

But I can't quite figure out the probability in the middle step. I can't find any material online to help clarify this. Any suggestions ?

• Usually, the mean is just an estimator for the expectation of a random variable. Are you sure you don't have to prove the mean is an unbiased estimator of the expectation? Apr 5, 2014 at 19:25
• I need to show that the sample mean $\bar{x}$ is an unbiased estimator of $\bar{X}$. I have $E[\bar{x}] = \frac{1}{n} \sum E[x_i]$ Now, I need to evauluate $E[x_i]$ Apr 5, 2014 at 19:27
• Okay, so you are given that $E[x_i]=\overline{X}$ and have tow show the sample mean $\overline{x}$ is unbiased estimator. So, please have a look on my answer! Apr 5, 2014 at 20:05

Usually, in basic statistics one has to prove that the sample mean $\overline{x}$ is an unbiased estimator of the true expectation of a random variable $x_i$. Lets call this true expectation $\mu$. So, you are given $E[x_i]=\mu$ and the sample mean $\overline{x}=\frac{1}{n} \sum_{i=1}^{n} x_i$ as an estimator of the expectation.
In order to show an estimator is unbiased, the expectation of the estimator must be equal to the true expectation. Thus, one wants to prove $E[\overline{x}]=\mu$ and can do this in the following way:
$E[\overline{x}]= E[\frac{1}{n} \sum_{i=1}^{n} x_i] = \frac{1}{n} \sum_{i=1}^{n} E[x_i] = \frac{1}{n} \sum_{i=1}^{n} \mu= \frac{1}{n} n \mu=\frac{n}{n} \mu= \mu$.
At first you plug out the fraction and the sum since they are non-stochastic. Then, you know the expectation of $x_i$ was given as $\mu$. Since $\mu$ is a constant, you can replace the sum by $n$. Eventually, the $n$'s cancel out and you are left with $\mu$. Now, one has shown, the sample mean is an unbiased estimator of the expectation of a random variable.