# 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:

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

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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 $\mu$.

$\blacksquare$

## Sources

- 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**estimator** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**estimator**