Why is the sample mean an unbiased estimator of the population mean?

What Does It Mean That the Sample Mean Is Unbiased?

Formally, if \(X_1, X_2, \ldots, X_n\) are i.i.d. with mean \( \mu \), then:

\[ \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i \]

The unbiasedness condition is:

\[ E[\bar{X}] = \mu \]

Why the Sample Mean Is Unbiased

1. Expectation is linear

The expectation of a sum is the sum of expectations, regardless of independence.

2. Each observation has the same mean

Since all \(X_i\) have mean \( \mu \), their average also has mean \( \mu \).

3. Dividing by n preserves the mean

Averaging does not change the expected value — it only reduces variance.

4. No systematic over‑ or under‑estimation

The sample mean fluctuates around \( \mu \), but its long‑run average equals \( \mu \).

5. Works for any distribution with finite mean

No normality assumption is required — unbiasedness holds universally.

How to Show the Sample Mean Is Unbiased (Step by Step)

Step 1: Write the sample mean

\[ \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i \]

Step 2: Apply expectation

\[ E[\bar{X}] = E\left[\frac{1}{n}\sum_{i=1}^n X_i\right] \]

Step 3: Pull out constants

\[ = \frac{1}{n} \sum_{i=1}^n E[X_i] \]

Step 4: Use the fact that each \(X_i\) has mean \( \mu \)

\[ = \frac{1}{n} \cdot n\mu \]

Step 5: Simplify

\[ E[\bar{X}] = \mu \]

This proves the sample mean is unbiased.

Numerical Example

Suppose the population mean is \( \mu = 10 \), and we draw samples of size 3:

• Sample 1: {8, 12, 10} → mean = 10
• Sample 2: {9, 11, 7} → mean = 9
• Sample 3: {13, 10, 8} → mean = 10.33

Averaging many such sample means will converge to 10 — the true population mean.