Addition of Independent Binomial Variables

Why a Common Success Probability Matters

A binomial random variable counts successes across a fixed number of independent trials that share one success probability. When two groups of trials have the same success probability, combining their counts is equivalent to counting successes across all the trials at once. The important condition is not merely independence: the success probabilities must match.

The discrete convolution formula gives the distribution of a sum of independent discrete random variables. For binomial variables with a common success probability, the terms in that convolution share the same probability factor, and a binomial-coefficient identity combines the remaining terms. If the success probabilities differ, those factors no longer align.

The Addition Theorem

For the two-variable result, independence and the discrete convolution formula give, for \(k=0,\ldots,m+n\), $$ P(X+Y=k)=\sum_{i=\max(0,k-n)}^{\min(m,k)}P(X=i)P(Y=k-i). $$ Each term in this sum is $$ \binom{m}{i}\binom{n}{k-i}p^i(1-p)^{m-i}p^{k-i}(1-p)^{n-k+i} =\binom{m}{i}\binom{n}{k-i}p^k(1-p)^{m+n-k}. $$ For \(0<p<1\), the probability factors are the same for every term, so they can be taken outside the sum. The remaining identity is Vandermonde’s identity: $$ \sum_i\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}. $$ To see why it holds, consider two disjoint groups containing \(m\) and \(n\) objects. The right side counts the ways to choose \(k\) objects from their combined \(m+n\) objects. If exactly \(i\) chosen objects come from the first group, there are \(\binom{m}{i}\binom{n}{k-i}\) such choices. Summing over all feasible \(i\) counts every selection exactly once. Therefore the convolution probability is \(\binom{m+n}{k}p^k(1-p)^{m+n-k}\), as required. If \(p=0\), both variables are zero with probability one; if \(p=1\), their sum is \(m+n\) with probability one. The result holds at those endpoints as well. \(\square\)

The finite-family result follows by applying the two-variable result repeatedly. First, \(X_1+X_2\sim\operatorname{Bin}(n_1+n_2,p)\). This sum is independent of \(X_3\), so adding \(X_3\) gives a binomial variable with \(n_1+n_2+n_3\) trials and probability \(p\). Continuing this argument gives the stated result for \(r\) variables.

What Changes When Probabilities Differ?

Suppose \(X\sim\operatorname{Bin}(m,p)\) and \(Y\sim\operatorname{Bin}(n,q)\) are independent, with different success probabilities. The convolution still computes the distribution of \(X+Y\), but the common factor used in the addition theorem is no longer available. In general, the sum is not binomial.

A variance comparison gives a precise test when asking whether this sum could be binomial with \(m+n\) trials. For independent variables, expectations add and variances add. A binomial variable with \(N\) trials and probability \(r\) has mean \(Nr\) and variance \(Nr(1-r)\).

To prove the necessary condition, let \(N=m+n\). Additivity of expectation gives $$ Nr=E[X+Y]=mp+nq, \qquad\text{so}\qquad r=\frac{mp+nq}{N}. $$ Independence gives \(\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)\), hence $$ \operatorname{Var}(X+Y)=mp(1-p)+nq(1-q). $$ If the sum is \(\operatorname{Bin}(N,r)\), its variance must also equal \(Nr(1-r)\). Subtracting the variance of the sum from this binomial variance yields $$ \begin{aligned} Nr(1-r)-\{mp(1-p)+nq(1-q)\} &=mp^2+nq^2-Nr^2\\ &=m(p-r)^2+n(q-r)^2. \end{aligned} $$ The last equality follows from \(Nr=mp+nq\): expanding the squares gives \(mp^2+nq^2-Nr^2\). The left side must be zero if the two variances are equal. Since \(m,n>0\) and both squares are nonnegative, this forces \(p=r\) and \(q=r\). Thus \(p=q=r\). The converse is the addition theorem. \(\square\)

When \(0<p,q<1\), the sum has positive probability at every integer from 0 to \(m+n\). Consequently, if its distribution is binomial at all, the binomial trial count must be \(m+n\), since that is the largest possible value in its support. The theorem then shows that the probabilities must match. Boundary cases with probabilities 0 or 1 can produce degenerate or shifted sums, so a claim about a binomial form with a different trial count requires checking those cases separately.

A Common Mistake and Why the Rule Matters

Incorrect reasoning: “The variables are independent binomial counts, so their sum must be binomial with the trial counts added.” Independence is necessary for the addition theorem, but it does not make different success probabilities equal. In the unequal-probability example, adding the trial counts would suggest a \(\operatorname{Bin}(2,r)\) law, but the mean pins down \(r=0.45\) and the resulting probabilities are wrong.

Correction: Check both independence and a common success probability before applying the theorem. If the probabilities differ, use the discrete convolution formula to calculate the sum’s probabilities, or use expectations and variances to rule out a proposed binomial model. In applications, groups can be combined under the theorem when their trials share the same response probability; matching only the total number of trials is not enough.

The addition theorem is useful because it replaces a sum of separate counts with one simpler distribution while preserving the exact law. Its conditions also clarify what the model assumes: each counted trial is independent, and every trial has the same probability of success. If a study combines groups with genuinely different response probabilities, using one binomial distribution without checking those assumptions can misstate probabilities of the total count.