Why the Sample Mean Is Unbiased

Why Ask Whether the Sample Mean Is Unbiased?

A sample mean is often used to estimate a population mean. But one sample can produce a mean above the population mean, and another can produce one below it. To understand whether the method is systematically off target, ask what would happen if the sampling process were repeated many times under the same conditions. The key question is whether the sample means, on average, equal the population mean.

This question concerns the sampling process, not just the observed data. Before a sample is drawn, its values are random, so the sample mean is a random variable too. Once the sample has been observed, its mean is a particular numerical estimate. Unbiasedness is a property of the rule that generates estimates over repeated samples.

Definition: An estimator is unbiased for a parameter if its expected value over the specified repeated-sampling process equals that parameter. The bias of an estimator \(T\) for a parameter \(\theta\) is \(E(T)-\theta\). Thus, \(T\) is unbiased for \(\theta\) when \(E(T)-\theta=0\).

The Expectation Rules Behind the Result

Let \(X_1,\ldots,X_n\) denote the values of a quantitative variable for the \(n\) sampled units. Before sampling, each \(X_i\) is a random variable. The sample mean is therefore the random variable

\bar{X}=\frac{X_1+\cdots+X_n}{n}.

Here \(\bar{X}\) describes the sample mean before the sample is observed. In the notation for a particular observed data set, the resulting numerical mean is written \(\bar{x}\). Let \(\mu\) be the population mean, the target quantity. The central condition is that every sampled value has expected value \(\mu\): \(E(X_i)=\mu\) for each \(i\). This condition holds when the sampling procedure gives each sampled position the same population mean.

The expectation rules needed are that multiplying a random variable by a constant multiplies its expected value by that constant, and that the expectation of a sum is the sum of the expectations. The latter rule does not require the random variables to be independent.

Theorem: If \(n\) is a fixed positive integer and each sampled value satisfies \(E(X_i)=\mu\), then the sample mean is unbiased for the population mean: \(E(\bar{X})=\mu\). Independence among the \(X_i\) is not required for this conclusion.

The result follows directly by applying those expectation rules to the definition of the sample mean:

\begin{aligned} E(\bar{X}) &=E\left(\frac{X_1+\cdots+X_n}{n}\right)\\ &=\frac{1}{n}E(X_1+\cdots+X_n)\\ &=\frac{1}{n}\left(E(X_1)+\cdots+E(X_n)\right)\\ &=\frac{1}{n}(\mu+\cdots+\mu)\\ &=\frac{1}{n}(n\mu)\\ &=\mu. \end{aligned}

The fourth line uses the condition that each sampled value has expected value \(\mu\), and the fifth line counts \(n\) copies of \(\mu\). This establishes the claim: across repeated samples generated by this process, the average of the sample means is the population mean. The derivation does not say that any one sample mean must equal \(\mu\).

Worked Examples

Worked Example: A Small Discrete Population

Suppose a variable has three possible values, 2, 5, and 11, each equally likely on an independent draw. A hypothetical sample consists of \(n=2\) such draws. Find the population mean and the expected sample mean.

Check the sampling condition. Each sampled value has the same distribution as the population variable, so each has expected value equal to the population mean. The draws are independent in this example, although the unbiasedness derivation itself does not need independence.

The population mean is \( \mu=(2+5+11)/3=18/3=6 \). For the two sampled values, \(E(X_1)=E(X_2)=6\). Applying the definition and expectation rules gives

E(\bar{X}) =E\left(\frac{X_1+X_2}{2}\right) =\frac{E(X_1)+E(X_2)}{2} =\frac{6+6}{2} =6.

Thus, the sample mean is unbiased for the population mean of 6. The sample mean need not be 6 for a particular pair of draws. For instance, the pair 2 and 5 has sample mean \((2+5)/2=3.5\); unbiasedness is about the expectation across the sampling process, not exactness in every sample.

As a direct check, the nine equally likely ordered pairs produce sample means 2, 3.5, 6.5, 3.5, 5, 8, 6.5, 8, and 11. Their average is \(54/9=6\), agreeing with the expectation calculation.

Worked Example: Mean Waiting Time in a Hypothetical Population

Imagine a population in which a randomly selected appointment has a waiting time of 0 minutes with probability 0.10, 10 minutes with probability 0.20, 20 minutes with probability 0.50, and 50 minutes with probability 0.20. A random sample of \(n=4\) appointments is taken. Find the population mean waiting time and the expected sample mean.

Check the sampling condition. Each sampled appointment is selected from the stated population distribution, so the expected waiting time at each sampled position is the same population mean. No assumption about the shape of the distribution is needed for the calculation.

The population mean is the probability-weighted average of the possible waiting times:

\mu=(0)(0.10)+(10)(0.20)+(20)(0.50)+(50)(0.20) =0+2+10+10 =22\text{ minutes}.

Each of \(X_1,X_2,X_3,X_4\) therefore has expected value 22 minutes. By linearity of expectation,

E(\bar{X}) =\frac{E(X_1)+E(X_2)+E(X_3)+E(X_4)}{4} =\frac{22+22+22+22}{4} =22\text{ minutes}.

So the sample mean waiting time is unbiased for the population mean waiting time. This conclusion does not require that every sample of four appointments have a mean of 22 minutes; it says the mean of the sample means over repeated sampling is 22 minutes.

Worked Example: When Unequal Selection Makes the Mean Biased

Consider a finite population with three units whose values are 10, 20, and 30. The population mean is \(\mu=(10+20+30)/3=20\). Suppose a sample of one unit is selected, with probabilities 0.10, 0.30, and 0.60 for the units with values 10, 20, and 30, respectively. Find the expected sample mean and its bias.

Check the sampling condition. Because the sample contains one unit, its sample mean is that unit's value. But the selection probabilities are unequal, so the expected value of the sampled value need not equal the equally weighted population mean.

The expected sample mean is

E(\bar{X}) =(10)(0.10)+(20)(0.30)+(30)(0.60) =1+6+18 =25.

The bias for estimating the population mean is \(E(\bar{X})-\mu=25-20=5\). Thus, this sample mean has an upward bias of 5 units under the specified selection process. There is no contradiction with the theorem: the condition \(E(X_i)=\mu\) fails here. The theorem does not guarantee unbiasedness when the sampling procedure favors some population values.

What Unbiasedness Does—and Does Not—Tell Us

Unbiasedness is a statement about long-run centering. If a sampling procedure is repeated many times, the expected value of the resulting sample means equals the target population mean. It does not guarantee that the observed mean in one sample is close to the population mean, nor does it tell us how much sample means vary. Those are separate questions about variability.

The result is also not limited to normally distributed populations. The derivation uses only the expectation rules and the condition \(E(X_i)=\mu\). The population can be skewed, discrete, or continuous, as long as the relevant expectations exist and the sampling process gives each sampled value the target mean.

Independence is often part of a convenient random-sampling model, but it is not needed for this proof. Even if sampled values are dependent, linearity still gives \(E(X_1+\cdots+X_n)=E(X_1)+\cdots+E(X_n)\). What matters for unbiasedness is that the expected value of each sampled observation is \(\mu\), not that the observations are independent.

Common Mistakes

Claiming every sample mean equals the population mean. The equation \(E(\bar{X})=\mu\) concerns an average over repeated samples. Individual sample means can differ from \(\mu\).
Assuming unbiased means precise. Unbiasedness says the estimator is centered correctly in expectation. It does not describe the spread of its possible values.
Requiring independence for the expectation calculation. Independence is not needed to add expectations. The key condition here is \(E(X_i)=\mu\) for every sampled position.
Ignoring how units are selected. If the sampling procedure gives some values a systematically different chance of selection, the expected sampled value may not equal the population mean.
Treating the parameter as random in this setting. The population mean \(\mu\) is the fixed target; the sample mean \(\bar{X}\) varies from sample to sample.

Key takeaway: When each sampled observation has expected value \(\mu\), linearity of expectation gives \(E(\bar{X})=\mu\). The sample mean is then unbiased for the population mean, regardless of whether the sampled observations are independent.

Check Your Understanding

Use the definition and expectation rules to answer these questions.

In one sentence, what does it mean for the sample mean to be unbiased for \(\mu\)?
If \(E(X_1)=E(X_2)=E(X_3)=12\), what is \(E((X_1+X_2+X_3)/3)\)? Show the expectation rule you use.
Does the derivation of \(E(\bar{X})=\mu\) require the sampled values to be independent? Explain.
Can an unbiased sample mean be different from the population mean in a particular sample? Explain what unbiasedness says instead.
A sampling procedure gives larger population values greater selection probabilities. Which condition in the unbiasedness argument should be checked before concluding that the sample mean is unbiased?