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

Question

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 ?

Answer 1

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.