Binomial Additivity at a Common Success Probability

Why a Shared Success Probability Matters

A binomial count records the number of successes in a fixed number of trials that have the same success probability. If two independent studies use the same success probability, it is natural to ask whether their combined success count has a binomial distribution too. It does: the trials can be pooled, and the total number of trials is the sum of the group sizes. The shared probability is essential to this identification; merely adding two binomial counts does not guarantee a binomial result.

The probability-generating function (PGF) makes the additivity transparent. For an integer-valued random variable \(X\) taking nonnegative values, its PGF is \(G_X(z)=E[z^X]\) for \(0\le z\le 1\). The PGF of a \(\operatorname{Bin}(n,p)\) variable is \((1-p+pz)^n\). As in other transform arguments, independence turns the PGF of a sum into a product.

Definition: \(X\sim\operatorname{Bin}(n,p)\) means that \(X\) counts successes in \(n\) independent Bernoulli trials, each with success probability \(p\). Here \(n\) is a nonnegative integer and \(0\le p\le1\). Its probability mass function is \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k=0,\ldots,n\), and its PGF is \(G_X(z)=(1-p+pz)^n\).

The Additivity Theorem

Theorem: Let \(X_1,\ldots,X_m\) be mutually independent, with \(X_i\sim\operatorname{Bin}(n_i,p)\) for a common \(p\in[0,1]\) and nonnegative integers \(n_i\). Then \[ \sum_{i=1}^m X_i\sim\operatorname{Bin}\left(\sum_{i=1}^m n_i,p\right). \]

Proof. Write \(S=\sum_{i=1}^m X_i\) and \(N=\sum_{i=1}^m n_i\). Independence gives, for \(0\le z\le1\),

G_S(z)=E[z^S] =E\left[\prod_{i=1}^m z^{X_i}\right] =\prod_{i=1}^m E[z^{X_i}] =\prod_{i=1}^m(1-p+pz)^{n_i} =(1-p+pz)^N.

The last expression is the PGF of \(\operatorname{Bin}(N,p)\). PGFs uniquely determine distributions on the nonnegative integers, so \(S\) has that distribution. The argument includes \(p=0\) and \(p=1\): in those cases the sum is, respectively, identically zero or identically \(N\), as asserted. \(\square\)

The same result follows by pooling all the Bernoulli trials represented by the binomial counts. Independence across the counts, together with the independence of trials within each count, makes the pooled collection independent; every trial has probability \(p\). The PGF proof packages that reasoning into one calculation.

Worked Example: Combining Two Counts With the Same Probability

Suppose independent counts satisfy \(X\sim\operatorname{Bin}(2,0.4)\) and \(Y\sim\operatorname{Bin}(3,0.4)\). Find the distribution of \(S=X+Y\), and calculate \(P(S=2)\).

Both counts have success probability \(0.4\), so additivity gives \(S\sim\operatorname{Bin}(5,0.4)\). Therefore

P(S=2)=\binom{5}{2}(0.4)^2(0.6)^3 =10(0.16)(0.216) =0.3456.

We can check this by adding the three disjoint ways to obtain two total successes:

\begin{aligned} P(S=2) &=P(X=0)P(Y=2)+P(X=1)P(Y=1)+P(X=2)P(Y=0)\\ &=(0.36)(0.288)+(0.48)(0.432)+(0.16)(0.216)\\ &=0.10368+0.20736+0.03456\\ &=0.3456. \end{aligned}

The product of probabilities in the convolution uses independence. The agreement checks both the pooled binomial formula and the possible ways the two counts can contribute two successes.

Why Unequal Probabilities Usually Change the Distribution

For independent \(X_i\sim\operatorname{Bin}(n_i,p_i)\), the same calculation yields

G_{\sum_i X_i}(z)=\prod_{i=1}^m(1-p_i+p_i z)^{n_i}.

When the \(p_i\) differ, the factors are generally different linear polynomials. Their product is not generally the PGF of a binomial variable with \(\sum_i n_i\) trials. The following result states precisely when a binomial identification is possible if all the probabilities are strictly between zero and one.

Theorem: Suppose \(n_i\) are positive integers, \(p_i\in(0,1)\), and the \(X_i\) are independent with \(X_i\sim\operatorname{Bin}(n_i,p_i)\). Let \(N=\sum_i n_i\). Then \(\sum_i X_i\sim\operatorname{Bin}(N,q)\) for some \(q\in[0,1]\) if and only if \(p_i=q\) for every \(i\).

Proof. If every \(p_i=q\), the additivity theorem proves the claim. Conversely, suppose the sum has distribution \(\operatorname{Bin}(N,q)\). Since every \(p_i\) is strictly between zero and one, the sum has positive probability of being zero and positive probability of being \(N\). Thus \(q\) cannot be 0 or 1. Equality of the distributions implies equality of their PGF polynomials:

\prod_i(1-p_i+p_i z)^{n_i}=(1-q+qz)^N.

The factor \(1-p_i+p_i z\) has the single root \(-\frac{1-p_i}{p_i}\), and the factor \(1-q+qz\) has the single root \(-\frac{1-q}{q}\). The right-hand side has only the latter root. Consequently every root on the left must equal it, so \(\frac{1-p_i}{p_i}=\frac{1-q}{q}\) for each \(i\). Solving this equality gives \(p_i=q\). This proves the necessity and completes the proof. \(\square\)

Worked Example: Unequal Probabilities Give a Nonbinomial Sum

Let \(X\sim\operatorname{Bin}(1,0.2)\) and \(Y\sim\operatorname{Bin}(1,0.8)\), independently. Find the probabilities for \(S=X+Y\), and check whether \(S\) could be \(\operatorname{Bin}(2,q)\).

The sum is zero only if both trials fail, and it is two only if both succeed:

P(S=0)=(0.8)(0.2)=0.16,\qquad P(S=2)=(0.2)(0.8)=0.16.

The remaining probability is \(P(S=1)=1-0.16-0.16=0.68\). If \(S\) were \(\operatorname{Bin}(2,q)\), its mean would be \(2q\). But \(E[S]=0.2+0.8=1\), so necessarily \(q=0.5\). The \(\operatorname{Bin}(2,0.5)\) probabilities are \(0.25,0.50,0.25\), not \(0.16,0.68,0.16\). Thus this unequal-probability sum is not binomial with two trials. The theorem above shows that, with both probabilities in the interior, no other \(q\) can make it \(\operatorname{Bin}(2,q)\).

Independence Is a Separate Requirement

A shared probability alone does not establish additivity. Independence is what makes the PGF of the sum the product of the individual PGFs. Without it, even the common-\(p\) conclusion can fail.

Worked Example: Common Marginals but Dependent Counts

Let \(B\sim\operatorname{Bernoulli}(0.5)\), and set \(X=B\) and \(Y=B\). Each marginal is \(\operatorname{Bin}(1,0.5)\), but \(X\) and \(Y\) are dependent because they always agree. Their sum is \(S=2B\), so

P(S=0)=0.5,\qquad P(S=1)=0,\qquad P(S=2)=0.5.

If the counts were independent, the additivity theorem would give \(\operatorname{Bin}(2,0.5)\), with probabilities \(0.25,0.50,0.25\). The actual probabilities differ. The common success probability does not replace the independence condition.

Endpoint Probabilities and Practical Interpretation

The equal-probability theorem’s converse assumed \(0<p_i<1\). This condition avoids deterministic trials. A group with \(p_i=0\) contributes zero with certainty, while a group with \(p_i=1\) contributes \(n_i\) with certainty. Such terms can be removed or treated as a constant when analyzing the random part of the sum. For example, an independent \(\operatorname{Bin}(2,0)\) count added to a \(\operatorname{Bin}(3,0.4)\) count leaves the latter distribution unchanged. This is why “unequal probabilities generally prevent a binomial identification” is more accurate than an unconditional claim that they always do.

In applications, the common-\(p\) result can combine independent counts from groups when each trial represents the same success chance—for instance, counts from separate batches under a shared, stable process. When success chances differ by group, replacing the sum with a binomial model having an averaged probability can give the wrong distribution, even if the mean is matched. The shape of the distribution depends on more than its mean.

Common Mistakes

Adding trial counts when probabilities differ. The fact that the total number of trials is \(N\) does not by itself make the success count \(\operatorname{Bin}(N,q)\). The success probabilities must also agree for the standard additivity conclusion.
Using a mean match as a distribution proof. An averaged success probability can match the sum’s mean, but different distributions can share a mean. In the unequal-probability example, the mean forces \(q=0.5\), yet the probabilities still disagree.
Forgetting independence. The product formula for the sum’s PGF requires independence. Identical marginal distributions are not enough.
Ignoring deterministic endpoint groups. Probabilities zero and one create constants, not ordinary random trials. State separately how those contributions affect the sum.

Key takeaway: Independent binomial counts with a common success probability add to a binomial count with the summed number of trials. Unequal interior probabilities generally give a different distribution, and independence must be checked separately.

Check Your Understanding

Answer each question using the conditions and reasoning developed above.

If independent \(X\sim\operatorname{Bin}(4,0.3)\) and \(Y\sim\operatorname{Bin}(6,0.3)\), what is the distribution of \(X+Y\)?
Which step in the PGF proof uses independence?
Two independent binomial counts have positive trial counts and success probabilities strictly between zero and one. If their sum is binomial with the total number of trials, what must be true of their success probabilities?
Why does matching the mean of an unequal-probability sum to the mean of a candidate binomial distribution not establish equality in distribution?
What is the sum of a \(\operatorname{Bin}(2,0)\) count and an independent \(\operatorname{Bin}(3,0.4)\) count, and why is this an endpoint exception to the equal-probability converse?