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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?

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2 Answers 2

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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.

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  • $\begingroup$ So as I understand it the population mean $\bar{X}$ is not the same as the expectation of the random variable $E[X_i]$. Can we say that they converge when the population size ($n$) is very large? $\endgroup$
    – Tom
    Commented Feb 13, 2019 at 14:24
  • $\begingroup$ What you say is true and is an application of the law of large numbers @Tom $\endgroup$
    – Ahmad Bazzi
    Commented Feb 13, 2019 at 14:25
  • $\begingroup$ Thanks, In that case $\psi=𝐸[\psi]$ in the limit where N is large right? $\endgroup$
    – Tom
    Commented Feb 13, 2019 at 14:26
  • $\begingroup$ you can say $ \frac{\psi}{n} \rightarrow E(X_i)$ as $n$ goes to infinity given iid observations. $\endgroup$
    – Ahmad Bazzi
    Commented Feb 13, 2019 at 14:27
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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.

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