# Sample Mean is Unbiased Estimator of Population Mean

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## Theorem

Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.

Then:

- $\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$

is an unbiased estimator of $\mu$.

## Proof

If $\bar X$ be an unbiased estimator of $\mu$, then:

- $\expect {\bar X} = \mu$

This article, or a section of it, needs explaining.In particular: The structure of this proof is confusing. It starts with what looks like an assertion of what we are trying to prove.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

We have:

\(\ds \expect {\bar X}\) | \(=\) | \(\ds \expect {\frac 1 n \sum_{i \mathop = 1}^n X_i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 n \sum_{i \mathop = 1}^n \expect {X_i}\) | Expectation is Linear | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 n \sum_{i \mathop = 1}^n \mu\) | as $\expect {X_i} = \mu$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac n n \mu\) | as $\ds \sum_{i \mathop = 1}^n 1 = n$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \mu\) |

So $\bar X$ is an unbiased estimator of $\mu$.

$\blacksquare$