Deriving the Binomial Mean and Variance

Why Derive the Mean and Variance?

A binomial model describes the number of successes in a fixed number of trials. Its probability mass function can give the probability of each possible count, but calculating a whole distribution is not necessary to find its mean or variance. The key is to examine each trial separately, then combine the results.

This approach also makes the assumptions visible. The mean comes from adding the expected contributions of the trials. The variance has an additional requirement: the trials must be independent, so their indicators have zero covariance. Without that condition, the same variance formula need not hold.

An indicator is a numerical way to record whether an event occurred. It contributes exactly 1 to the total when the event occurs and 0 otherwise. Thus, counting successes is the same as adding one indicator for each trial.

Mean: Add the Expected Contributions

For any one trial, \(I_i=1\) with probability \(p\) and \(I_i=0\) with probability \(1-p\). Its expected value is therefore

Expectation is linear: the expected value of a sum is the sum of the expected values. This property does not require independence. Since the count is the sum of the \(n\) indicators, each with expected value \(p\),

Variance of One Indicator

Variance measures how much a random variable varies around its expected value. For any random variable \(Y\) with a finite second moment, the variance can be written as

For an indicator, \(I_i\) can only be 0 or 1, so \(I_i^2=I_i\) in either case. It follows that \(E(I_i^2)=E(I_i)=p\). Substituting into the variance formula gives

The same result can be obtained by considering the two possible values directly: the squared deviation from the mean \(p\) is \((1-p)^2\) when \(I_i=1\), and \(p^2\) when \(I_i=0\). Weighting those deviations by their probabilities gives \(p(1-p)^2+(1-p)p^2=p(1-p)\).

Variance of the Count: Why Independence Matters

For a sum of random variables, variance includes both the individual variances and the covariances between pairs. For two variables \(Y\) and \(Z\), the identity is

where \(\operatorname{Cov}(Y,Z)=E[(Y-E(Y))(Z-E(Z))]\). Extending this identity to the \(n\) indicators gives

If the trials are independent, their indicators are independent too. Independent variables have covariance zero, so every pairwise covariance term disappears. Each indicator has variance \(p(1-p)\), leaving \(n\) equal terms.

Seeing the Covariance Term in a Small Model

A small example makes clear which part of the variance proof depends on independence. Suppose three hypothetical trials each have success probability \(0.40\). For each trial, \(E(I_i)=0.40\), and

Assume the three trials are independent and let \(X=I_1+I_2+I_3\). Then all three pairwise covariances are zero. The mean and variance follow by adding the individual contributions:

For a direct check of one covariance, independence gives \(P(I_1=1,I_2=1)=0.40(0.40)=0.16\). Since \(I_1I_2=1\) only when both trials succeed, \(E(I_1I_2)=0.16\), and

The other two pairs have the same calculation. This confirms that the covariance terms vanish here because the trial outcomes are independent, not merely because each indicator has the same success probability.

Common Mistake and Why the Result Matters

A frequent error is to reason, “There are \(n\) trials, each with variance \(p(1-p)\), so the count’s variance is \(np(1-p)\),” without checking independence. In general, adding variances alone is not valid for a sum of dependent variables. The missing terms are the covariances:

The correction is to check the binomial conditions before using the formula. A fixed number of trials and a common success probability give the indicator expectations and individual variances used above; independence justifies dropping the covariance terms. If outcomes tend to occur together, covariance can increase the count’s variance; if they tend to offset one another, covariance can decrease it.

The indicator method matters because it explains the formulas rather than treating them as rules to memorize. It also separates two ideas that are easy to blur: linearity of expectation supports \(E(X)=np\), while independence supports \(\operatorname{Var}(X)=np(1-p)\). These derivations apply to any setting that genuinely satisfies the binomial model, whether the successes are correct answers, functioning parts, or another clearly defined outcome.