Binomial Variance With Different Trial Counts

Why the Response Scale Matters

A binomial model can describe either the number of successes in a fixed set of trials or the fraction of successes in those trials. These responses contain the same underlying information when the trial count is known, but they are on different scales. Their variances therefore differ: the variance of a count grows with the number of trials, while the variance of a proportion shrinks.

This distinction matters whenever binomial data are recorded in groups with different numbers of trials. A count of 4 events out of 20 and a count of 20 events out of 100 both give a proportion of 0.20, but the uncertainty of the proportions is not the same under a common binomial probability. The formulas below make the role of the trial count explicit.

Definition: Let $X_1,\ldots,X_n$ be independent Bernoulli random variables with common success probability $p$, where $n$ is a fixed positive integer and $0\leq p\leq 1$. The count of successes is $Y=\sum_{j=1}^{n}X_j$, so $Y\sim\operatorname{Bin}(n,p)$. The sample proportion of successes is $\bar{X}=Y/n$.

The requirement that the trials have a common success probability is part of the binomial model, as is their independence. The earlier material on the Bernoulli variance gives $\operatorname{Var}(X_j)=p(1-p)$. We use that fact to derive the count variance, then use the rule $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$ to obtain the proportion variance.

Deriving the Variance of the Count

Representing a binomial count as a sum of indicators makes the effect of the trial count transparent. Each indicator has variance $p(1-p)$, and independence means that the covariance terms between distinct indicators are zero.

Theorem: If $Y\sim\operatorname{Bin}(n,p)$ with fixed $n$ and $0\leq p\leq1$, then $$ \operatorname{Var}(Y)=np(1-p). $$

Proof. By the definition of the binomial count, $Y=\sum_{j=1}^{n}X_j$, where the $X_j$ are independent Bernoulli variables with success probability $p$. For a finite sum, the variance is the sum of the individual variances plus twice the sum of pairwise covariances:

\begin{aligned} \operatorname{Var}(Y) &=\operatorname{Var}\left(\sum_{j=1}^{n}X_j\right)\\ &=\sum_{j=1}^{n}\operatorname{Var}(X_j) +2\sum_{1\leq j<k\leq n}\operatorname{Cov}(X_j,X_k). \end{aligned}

For $j\ne k$, independence gives $\operatorname{Cov}(X_j,X_k)=0$. Each $X_j$ has variance $p(1-p)$, so the remaining sum has $n$ equal terms:

\operatorname{Var}(Y) =\sum_{j=1}^{n}p(1-p) =np(1-p).

This argument includes the boundary cases $p=0$ and $p=1$. In either case every trial outcome is constant, and the formula gives variance zero, as required. $\square$

The factor $n$ appears because the count accumulates the variation from $n$ independent trials. It does not mean that the variance is simply $n$ times the probability of success: the factor $1-p$ is also present. Successes are less variable when $p$ is close to either 0 or 1 than when $p$ is near the middle of its range.

Worked Example: Variance of a Success Count

Suppose 12 independent patients each have probability $p=0.25$ of experiencing a specified event. Let $Y$ be the number of patients who experience it. Find the mean and variance of the count.

The model is $Y\sim\operatorname{Bin}(12,0.25)$. The mean is $np=12(0.25)=3$. For the variance, first calculate $1-p=0.75$, then substitute:

\operatorname{Var}(Y) =12(0.25)(0.75) =3(0.75) =2.25.

Thus the expected count is 3 and the count variance is 2.25. The standard deviation of the count is $\sqrt{2.25}=1.5$ patients. The variance is measured in squared count units, whereas the standard deviation is measured in patients.

Rescaling to a Sample Proportion

The sample proportion divides the count by the fixed trial count: $\bar{X}=Y/n$. Dividing a random variable by $n$ divides its variance by $n^2$, not merely by $n$. Combining that scaling rule with the count-variance result yields the proportion variance.

Theorem: Under the same binomial assumptions, the sample proportion $\bar{X}=Y/n$ has variance $$ \operatorname{Var}(\bar{X})=\frac{p(1-p)}{n}. $$

Derivation. Since $n$ is fixed and positive, $1/n$ is a constant. Apply the variance scaling rule and then substitute the count variance:

\begin{aligned} \operatorname{Var}(\bar{X}) &=\operatorname{Var}\left(\frac{Y}{n}\right)\\ &=\frac{1}{n^2}\operatorname{Var}(Y)\\ &=\frac{1}{n^2}\bigl(np(1-p)\bigr)\\ &=\frac{p(1-p)}{n}. \end{aligned}

The derivation requires no additional distributional assumption: the proportion is the count multiplied by the constant $1/n$. When $p=0$ or $p=1$, the proportion is constant and this variance formula also gives zero. For $0<p<1$, its variance decreases as $n$ increases.

Worked Example: Count and Proportion at Two Trial Counts

Compare two groups with the same success probability $p=0.40$, one with $n=8$ trials and the other with $n=32$. Calculate the variance of the count and the variance of the sample proportion in each group.

For both groups, $p(1-p)=0.40(0.60)=0.24$. With 8 trials, the count variance is $8(0.24)=1.92$, while the proportion variance is $0.24/8=0.03$.

\operatorname{Var}(Y_8)=1.92, \qquad \operatorname{Var}(\bar{X}_8)=0.03.

With 32 trials, the count variance is $32(0.24)=7.68$, while the proportion variance is $0.24/32=0.0075$.

\operatorname{Var}(Y_{32})=7.68, \qquad \operatorname{Var}(\bar{X}_{32})=0.0075.

The count variance is four times as large in the group with four times as many trials: $7.68/1.92=4$. In contrast, the proportion variance is one quarter as large: $0.0075/0.03=0.25$. The count scale accumulates successes; the proportion scale averages over more trials.

Using the Formulas When Trial Counts Differ

In grouped data, the number of trials can vary from one group to another. If group $i$ has a fixed trial count $n_i$ and a common within-group success probability $p_i$, its count $Y_i$ has variance $n_i p_i(1-p_i)$. The corresponding proportion $Y_i/n_i$ has variance $p_i(1-p_i)/n_i$. The trial count belongs in the formula for the scale actually recorded.

Worked Example: Comparing Two Clinical Groups

Suppose a clinical study models an event with probability $p=0.10$ for each participant. One clinic has $n_1=20$ participants and another has $n_2=80$. For each clinic, calculate the variance of the event count and the variance of the event proportion.

Here $p(1-p)=0.10(0.90)=0.09$. For the first clinic:

\operatorname{Var}(Y_1)=20(0.09)=1.8, \qquad \operatorname{Var}\left(\frac{Y_1}{20}\right)=\frac{0.09}{20}=0.0045.

For the second clinic:

\operatorname{Var}(Y_2)=80(0.09)=7.2, \qquad \operatorname{Var}\left(\frac{Y_2}{80}\right)=\frac{0.09}{80}=0.001125.

The second clinic has four times as many participants, so its event-count variance is four times as large: $7.2/1.8=4$. But its proportion variance is one quarter as large: $0.001125/0.0045=0.25$. The larger count variance does not imply that the event proportion is less stable; it reflects the different response scale.

Common Mistake: Forgetting Which Quantity Is Recorded

A frequent error is to use $np(1-p)$ for a sample proportion simply because the underlying observations are binomial. That formula is for the count $Y$. If the response is $Y/n$, the variance must also be divided by $n^2$, resulting in $p(1-p)/n$.

The reverse error is to use $p(1-p)/n$ for the count. That formula describes the proportion, not the number of events. A reliable check is to write the response explicitly before substituting: if it is $Y$, use the count variance; if it is $Y/n$, use the proportion variance. The two variances are related by $\operatorname{Var}(Y/n)=\operatorname{Var}(Y)/n^2$.

Key takeaway: With $Y\sim\operatorname{Bin}(n,p)$, the count has variance $np(1-p)$, while the sample proportion $Y/n$ has variance $p(1-p)/n$. Identify the response scale before choosing the formula.

Check Your Understanding

For each question, identify whether the response is a count or a proportion before calculating or interpreting its variance.

For $Y\sim\operatorname{Bin}(15,0.20)$, calculate $\operatorname{Var}(Y)$.
For the same binomial model, calculate the variance of $Y/15$. Show how it relates to the count variance.
If the number of trials is multiplied by 3 while $p$ stays fixed, what happens to the count variance and the proportion variance?
Explain why the independence of Bernoulli trials is used when deriving the binomial count variance.
Two groups share the same $p$, but one has a larger fixed trial count. Which has the larger count variance, and which has the larger proportion variance?