I am confused how $\bar X$ is used sometimes as a constant and othertimes as a random variable.

My understanding is that $\bar X$ is a random variable because it changes every time our sample changes. I also understand that $E[\bar X]= E[\frac{X_1 +X_2 +...+X_n}{n}]=\frac{ nE[X]}{n} = E[X] = \mu $

I get confused when the professor writes $\bar{X}=E[\bar X]= E[\frac{X_1 +X_2 +...+X_n}{n}]=\frac{ nE[X]}{n} = E[X] = \mu $ and when he uses $\sum_1^n \bar X = n \bar X$ to proove another equation wich is $(X_i-\bar X)^2=\sum_1^nX_i^2-n\bar X^2$.

Someone can explain me this please? thank you

  • 2
    $\begingroup$ The only problematic equation is "$\bar X = E[\bar X]$." If your professor wrote that, you should ask them what they meant by such a circular expression. Everything else is just algebraic manipulation or exploitation of the linearity of expectation. $\endgroup$ – whuber Sep 26 '18 at 17:31
  • $\begingroup$ But I think $ \sum_1^n\bar X = n \bar X $ is not just algebraic manipulation or exploitation of the linearity of expectation. It means that he considered $\bar X $ constant. Isn't it? $\endgroup$ – Youssef Sep 26 '18 at 17:35
  • 3
    $\begingroup$ No, it's pure algebra. Please read stats.stackexchange.com/questions/95993. $\endgroup$ – whuber Sep 26 '18 at 17:36
  • $\begingroup$ I red it . How $\bar X$ is always the same for i= 1, 2 ---, n if it is a random variable ? $\endgroup$ – Youssef Sep 26 '18 at 17:42
  • 2
    $\begingroup$ "$\sum_{i=1}^n \text{anything}$" always means to sum $\text{anything}$ $n$ times, giving $n$ times $\text{anything}.$ $\endgroup$ – whuber Sep 26 '18 at 18:04

Assume $X_i$ is a random variable. Then, $\bar X$, defined as $\frac{1}{n}\sum_{i=1}^n X_i$, is also a random variable. For each realisation of the sample, it will have a different value.

As stated in @whuber's comment, $\bar X$ is not equal to $E(\bar X)$ in general.

To answer the last comment/question: $\bar X$ does not vary as a function of $i$ but it varies as a function of the sample. Therefore, it is true that $\sum_{i=1}^n \bar X = n \bar X$, while $\bar X$ is still a random variable. Take a different sample of $n$ individuals, you can define $\bar X$, it will have a different value and still, $\sum_{i=1}^n \bar X = n \bar X$ will hold.


You'll find that the sample expected value of the expected value of a random variable is a constant, the population mean $\mu$.

$$\mathbf E[\mathbf E[X]]=\mathbf E[\bar X]=\mu$$

$\mathbf E[\bar X]$ is your estimation of $\mu$, but it's a random variable because it's derived from a sample estimate.

Following the same logic, $\bar X=\mathbf E[X]$ is not a constant, but also a random variable.


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