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I have a simple question. $X$ is a random variable with mean $μ$, and there is a sample of size $n$: $X_1, X_2, \cdots, X_n$. Then what is the expected value of the sample mean $\overline{X}$?

This is what I thought:

$$\overline{X} = \frac{1}{n}(X_1+X_2+X_3+\cdots+X_n),$$

thus $\overline{X}$ is a certain value (constant), therefore $E(\overline{X}) = \overline{X}$. Similarly, $\overline{X}^2$is also a certain value (constant) and $E(\overline{X}^2) = \overline{X}^2$.

But it turns out there's something wrong. Could someone please explain it to me? Thanks in advance.

Ѕᴀᴀᴅ
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Matheoo
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    The sample mean is not a constant; it's a sum of random variables so it's another random variable. But it still has mean $\mu$ by linearity of expectation. – Qiaochu Yuan Jan 07 '18 at 01:30
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    Assuming the $X_i$ are independent each with mean $\mu$ and variance $\sigma^2$, then their sum has mean $n\mu$ and variance $n\sigma^2$ and their average has mean $\mu$ and variance $\frac1n \sigma^2$ so $E\left(\overline{X}^2\right)=\mu^2+\frac1n \sigma^2$ – Henry Jan 07 '18 at 09:42

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The problem in your reasoning is that the random variables $X_1,X_2,\ldots,X_n$ are actually NOT constants and so the sum $\frac{1}{n}(X_1+\ldots X_n)$ is not a constant either. As mentioned by Qiaochu, the expected value of the sample mean is the population mean (often denoted by $\mu$).

David Reed
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