Why the Sample Proportion Is Unbiased

Why Ask Whether a Proportion Is Unbiased?

A sample proportion is often used to estimate a population proportion. For example, a sample may be used to estimate the fraction of all eligible patients who experience a particular outcome. A sample proportion will vary from sample to sample, so a useful question is whether its long-run average is centered on the population value it is intended to estimate.

That question is about the sampling process, not about whether one particular sample happens to be representative. In this tutorial, we show that under random sampling, the expected sample proportion equals the population proportion. The central step is to express the sample proportion as the average of binary outcomes—one outcome for each sampled unit.

Definition: An estimator is unbiased for a population parameter if its expected value, taken over the repeated samples allowed by the sampling model, equals that parameter. The difference between an estimator's expected value and the parameter is its bias. Thus, if \(\hat{p}\) estimates \(p\), it is unbiased when \(E(\hat{p})=p\).

The expectation is taken over possible samples, not over uncertainty about a fixed population proportion. A population proportion \(p\) is fixed for the population under consideration; \(\hat{p}\) changes when the sample changes. The claim of unbiasedness says that the average value of \(\hat{p}\) across the sampling distribution is \(p\).

Representing the Sample Proportion as an Average

Suppose each sampled unit either has a characteristic of interest or does not. For the \(i\)th sampled unit, define \(Y_i=1\) if it has the characteristic and \(Y_i=0\) otherwise. These are binary outcomes: each records whether one sampled unit is a success according to the definition used in the study.

If there are \(n\) sampled units, the number of successes is \(Y_1+\cdots+Y_n\). Dividing that count by \(n\) gives the sample proportion:

\hat{p}=\frac{Y_1+Y_2+\cdots+Y_n}{n} =\frac{1}{n}\sum_{i=1}^{n}Y_i.

This expression makes the expectation calculation possible. For a binary outcome whose probability of being 1 is \(p\), its expected value is \(1\cdot p+0\cdot(1-p)=p\). Under a random sampling design in which each sampled position has probability \(p\) of being a success, each \(Y_i\) therefore has expected value \(p\).

The calculation also uses linearity of expectation: the expected value of a sum is the sum of the expected values. This rule does not require the outcomes to be independent. That matters for sampling without replacement, where the outcomes in different positions are generally dependent.

Derivation of Unbiasedness

First consider a population with \(N\) units, of which \(K\) have the characteristic. Its population proportion is \(p=K/N\). Take a simple random sample of size \(n\), meaning every possible sample of that size has the same chance of selection. The sample may be drawn without replacement. For any specified position in the sample, each population unit is equally likely to appear there, so the probability that the unit in that position is a success is \(K/N=p\). Therefore \(E(Y_i)=p\) for every position \(i\).

Now apply linearity of expectation to the average defining \(\hat{p}\). No assumption of independence between the sampled outcomes is needed for this step.

\begin{aligned} E(\hat{p}) &=E\left(\frac{1}{n}\sum_{i=1}^{n}Y_i\right)\\ &=\frac{1}{n}\sum_{i=1}^{n}E(Y_i)\\ &=\frac{1}{n}\sum_{i=1}^{n}p\\ &=\frac{1}{n}(np)\\ &=p. \end{aligned}

The same reasoning applies when the \(n\) outcomes are independent draws from a population in which each draw has success probability \(p\). In either case, each sampled position has expected outcome \(p\), so their average has expected value \(p\). Independence is not required for the expectation calculation, although it is relevant to other features of a sampling distribution.

The result depends on the sampling model. In particular, the argument requires that the chance of success at each sampled position be \(p\). If some units are more likely to enter the sample than others, an unweighted sample proportion need not be unbiased for the population proportion. Randomness alone is not a sufficient description; the way the sample is selected matters.

Theorem: If \(Y_i\) records success (1) or failure (0) for the \(i\)th sampled unit, and every sampled position has probability \(p\) of being a success, then the sample proportion \(\hat{p}=(1/n)\sum_{i=1}^{n}Y_i\) is unbiased for \(p\). This follows from \(E(Y_i)=p\) and linearity of expectation.

Worked Examples

Worked Example: Enumerating Every Sample

A hypothetical population consists of four people: two have a certain characteristic and two do not. A simple random sample of two people is drawn without replacement. Find the expected sample proportion and compare it with the population proportion.

The population proportion is \(p=2/4=1/2\). There are \(\binom{4}{2}=6\) equally likely samples. One sample contains both people with the characteristic, so its sample proportion is 1. One sample contains neither, so its sample proportion is 0. The remaining four samples each contain one person with the characteristic and one without, so each has sample proportion \(1/2\).

Weight each possible sample proportion by its probability, \(1/6\):

\begin{aligned} E(\hat{p}) &=\frac{1(1)+1(0)+4(1/2)}{6}\\ &=\frac{1+0+2}{6}\\ &=\frac{3}{6}\\ &=\frac{1}{2}. \end{aligned}

Thus the expected sample proportion is \(1/2\), equal to the population proportion \(p=1/2\). Notice that some possible samples give an estimate of 0 or 1. Unbiasedness is a statement about the average over all six equally likely samples, not a promise that each sample gives the population value.

Worked Example: Random Draws With a Known Success Probability

Suppose a hypothetical process produces a success with probability \(p=0.30\) on each draw. Four independent draws are made, and \(\hat{p}\) is the fraction of successes. Find \(E(\hat{p})\).

Let \(Y_i=1\) for a success on draw \(i\), and \(Y_i=0\) otherwise. For each draw, \(E(Y_i)=1(0.30)+0(0.70)=0.30\). The sample proportion is the average of the four outcomes, so:

\begin{aligned} E(\hat{p}) &=E\left(\frac{Y_1+Y_2+Y_3+Y_4}{4}\right)\\ &=\frac{E(Y_1)+E(Y_2)+E(Y_3)+E(Y_4)}{4}\\ &=\frac{0.30+0.30+0.30+0.30}{4}\\ &=\frac{1.20}{4}\\ &=0.30. \end{aligned}

The expected number of successes is \(1.20\), which need not be a possible count in a single set of four draws. It is an average across repeated sets of draws. Dividing that expected count by four gives an expected sample proportion of 0.30, equal to the success probability.

Worked Example: A Finite Population With Unequal Numbers of Successes and Failures

A hypothetical population contains eight units, three of which have a specified characteristic. A simple random sample of two is taken without replacement. Verify the expected sample proportion by counting the possible sample types.

The population proportion is \(p=3/8=0.375\). There are \(\binom{8}{2}=28\) equally likely pairs. Of these, \(\binom{5}{2}=10\) contain no successes, \(3\cdot5=15\) contain exactly one success, and \(\binom{3}{2}=3\) contain two successes. Their sample proportions are respectively 0, \(1/2\), and 1.

Therefore the expectation is:

\begin{aligned} E(\hat{p}) &=\frac{10(0)+15(1/2)+3(1)}{28}\\ &=\frac{0+7.5+3}{28}\\ &=\frac{10.5}{28}\\ &=0.375\\ &=\frac{3}{8}. \end{aligned}

This expected value matches the population proportion. The fractional quantity 7.5 in the numerator is a weighted contribution to the expectation, not a claim that a sample can contain half a success.

What Unbiasedness Does—and Does Not—Tell Us

Unbiasedness is a property of an estimator under a specified sampling model. It does not say that every estimate is close to the parameter, nor does it say that the estimate from the sample at hand equals the population proportion. In the first example, the estimator is unbiased even though two of the six possible samples produce extreme values, 0 and 1.

Unbiasedness also does not describe how spread out the estimates are. Two estimators can both be unbiased and yet vary by very different amounts from sample to sample. The derivation here establishes the center of the sampling distribution, not its variability.

Finally, the conclusion can fail if the sample selection does not give each relevant unit the proper chance of representation. For example, suppose a survey about all patients with a condition recruits only people who visit a particular clinic, and clinic patients are more likely to have a particular outcome than other patients. The observed fraction may systematically differ from the population fraction. The algebra \(E(\hat{p})=p\) only applies when the sampling model makes \(E(Y_i)=p\) for each sampled position.

Define success.
Record each sampled unit as 1 if it has the characteristic and 0 otherwise.

Write the statistic as an average.
The sample proportion is the sum of the binary outcomes divided by \(n\).

Check the sampling model.
Confirm that each sampled position has success probability \(p\), the population proportion.

Take expectations.
Use \(E(Y_i)=p\) and linearity of expectation to obtain \(E(\hat{p})=p\).

Common Mistake

A common error is to reason, “The sample proportion is unbiased, so my sample proportion must equal the population proportion.” This confuses a long-run average with the result from one sample. The sample proportion can be above or below \(p\); unbiasedness says that neither direction occurs systematically in expectation under the stated sampling model. A second error is to reject the derivation because sampling without replacement makes outcomes dependent. Dependence does not invalidate linearity of expectation: the expected value of the sum is still the sum of the expected values.

Key takeaway: Under an appropriate random sampling model, the sample proportion is an average of binary outcomes, each with expected value \(p\). Linearity of expectation then gives \(E(\hat{p})=p\), whether or not the sampled outcomes are independent.

Check Your Understanding

Use the definition and derivation above to answer these questions.

In the derivation, what does \(Y_i=1\) represent, and what does \(Y_i=0\) represent?
A population has 12 units, of which 9 are successes. What is \(p\), and what is the expected sample proportion under simple random sampling?
Why does the expectation calculation for sampling without replacement not require the sampled outcomes to be independent?
Does an unbiased estimator have to equal the population proportion in every sample? Explain.
Give one reason a sample proportion from a nonrandomly selected group might not be unbiased for the population proportion.