Equivalence of the sum of random variables and their expectation

Question

Given that $X$ is a random variable I define $$ \psi = \sum_i^{n} X_i $$ so $\psi$ is the sum of $n$ variables with the same distribution. Given that $$ \bar{X} = \frac{\sum_{i}^{n} X_i}{n} $$

I can write: $$ \sum_{i}^{n} X_i = n \bar{X} = \psi $$

At the same time if I take the expectation of $\psi$ I get: $$ E[\psi] = E\left[\sum_{i}^{n} X_i\right] = n E[X_i] = n \bar{X} $$

thus it seems that $\psi = E[\psi]$ which seems impossible. Where is the error in this reasoning?

Answer 1

I have a problem with this

$$E[\psi] = E\left[\sum_{i}^{n} X_i\right] = n E[X_i] = n \bar{X}$$

because $$E[X_i] \neq \bar{X}$$ simply because $\bar{X}$ is random whereas $E[X_i]$ is not.

Answer 2

You are confusing the 'average X', which is the average over an actual sample, and the 'average X' which is the expectation value, the ideal limit you would get from many samplings. The first is usually denoted $\overline X$ and the latter$E(X)$. $<X>$ is used for both by various authors. So the notation can be confusing, but the concepts are entirely different.