Binomial Sums Through MGFs

Why Binomial Sums Need the Right Conditions

A binomial random variable counts successes across a fixed number of trials that share one success probability. In applications, a total count may combine several groups: for example, successes in separate batches, regions, or periods. If the groups are independent and use the same success probability, their total is again binomial. If their probabilities differ, simply adding the trial counts and averaging the probabilities is generally wrong.

Moment generating functions (MGFs) make the common-probability result direct. The key is to use the product rule for independent random variables, then recognize the resulting MGF as that of a binomial distribution. The conditions matter: independence allows multiplication of MGFs, while a common success probability allows the factors to combine into one binomial MGF.

Definition: A random variable $X$ has a binomial distribution with parameters $n$ and $p$, written $X\sim\operatorname{Bin}(n,p)$, if it counts successes in $n$ independent Bernoulli trials, each with success probability $p$. Here $n$ is a nonnegative integer and $0\le p\le1$.

Deriving the Common-Probability Sum

Suppose $X_1,\ldots,X_k$ are independent and $X_j\sim\operatorname{Bin}(n_j,p)$ for every $j$. The group sizes $n_j$ may differ, but the success probability $p$ must be the same. Set $T=X_1+\cdots+X_k$. The MGF of a $\operatorname{Bin}(n,p)$ variable is $(1-p+pe^t)^n$. By the product rule for independent MGFs, the MGF of $T$ is the product of the individual MGFs.

\begin{aligned} M_T(t) &=\prod_{j=1}^{k}M_{X_j}(t)\\ &=\prod_{j=1}^{k}(1-p+pe^t)^{n_j}\\ &=(1-p+pe^t)^{n_1+\cdots+n_k}. \end{aligned}

This is the MGF of a binomial random variable with total trial count $N=n_1+\cdots+n_k$ and success probability $p$. Since the MGF identifies the distribution when it exists in a neighborhood of zero, it follows that $T\sim\operatorname{Bin}(N,p)$. The argument depends on the stated independence and common-$p$ assumptions; without them, this conclusion does not follow.

Theorem: If $X_1,\ldots,X_k$ are independent and $X_j\sim\operatorname{Bin}(n_j,p)$ for one common $p$, then $$ X_1+\cdots+X_k\sim\operatorname{Bin}\left(\sum_{j=1}^{k}n_j,p\right). $$

The same conclusion can be seen from the underlying trials: the total counts successes among all the groups’ trials. Independence means those trials are independent across groups as well as within each group, and common $p$ means every trial has the same success probability. Thus the combined collection satisfies the binomial model.

Worked Example: Combining Two Independent Groups

A hypothetical quality-control study records the number of units passing a test in two independent batches. Let $X\sim\operatorname{Bin}(4,0.30)$ and $Y\sim\operatorname{Bin}(6,0.30)$, with $X$ and $Y$ independent. Find the distribution of the total $T=X+Y$, then calculate $P(T=3)$. The success probability is $0.30$ in each trial in both batches.

Method. Use the product rule for the independent MGFs. The common success probability is $p=0.30$, and the total number of trials is $4+6=10$. Therefore,

M_T(t)=(0.70+0.30e^t)^4(0.70+0.30e^t)^6 =(0.70+0.30e^t)^{10}, \qquad T\sim\operatorname{Bin}(10,0.30).

For exactly three successes, use the binomial probability formula:

\begin{aligned} P(T=3) &=\binom{10}{3}(0.30)^3(0.70)^7\\ &=120(0.027)(0.0823543)\\ &=0.266827932\\ &\approx 0.2668. \end{aligned}

The total’s mean and variance provide a check: \[ E[T]=10(0.30)=3,\qquad \operatorname{Var}(T)=10(0.30)(0.70)=2.10. \] The separate groups give the same results: $E[X]+E[Y]=1.20+1.80=3$ and $\operatorname{Var}(X)+\operatorname{Var}(Y)=0.84+1.26=2.10$. The result is consistent. The common mistake to avoid is adding the trial counts but changing the success probability; here it remains $0.30$.

Why the MGF Identification Works

The key algebra is that the common factor $1-p+pe^t$ can be raised to powers and multiplied. Independence justifies the first equality below: for independent $X$ and $Y$, $E[e^{t(X+Y)}]=E[e^{tX}e^{tY}]=E[e^{tX}]E[e^{tY}]$. The equal success probability makes the two MGF factors have the same base.

Write each MGF.
If $X\sim\operatorname{Bin}(m,p)$ and $Y\sim\operatorname{Bin}(n,p)$, their MGFs are $(1-p+pe^t)^m$ and $(1-p+pe^t)^n$.

Apply independence.
For $T=X+Y$, the product rule gives $M_T(t)=M_X(t)M_Y(t)$.

Combine the powers.
Because the bases are identical, $M_T(t)=(1-p+pe^t)^{m+n}$.

Identify the law.
This is the MGF of $\operatorname{Bin}(m+n,p)$, so uniqueness of MGFs gives $T\sim\operatorname{Bin}(m+n,p)$.

This derivation also points to the limitation. If $X\sim\operatorname{Bin}(m,p_1)$ and $Y\sim\operatorname{Bin}(n,p_2)$, independence still gives a product, but the result is \[ M_{X+Y}(t)=(1-p_1+p_1e^t)^m(1-p_2+p_2e^t)^n. \] When $p_1\ne p_2$, the factors have different bases and generally cannot be combined into the MGF of one binomial variable.

Worked Example: Three Groups with a Shared Probability

Three independent hypothetical production lines each report their number of acceptable items. Their counts are $X_1\sim\operatorname{Bin}(3,0.25)$, $X_2\sim\operatorname{Bin}(4,0.25)$, and $X_3\sim\operatorname{Bin}(5,0.25)$. Find the distribution of the total $T=X_1+X_2+X_3$, its mean and variance, and the probability that exactly two items are acceptable. Each trial has success probability $0.25$.

Method. The three counts are independent and share $p=0.25$, so the total trial count is $3+4+5=12$. The product of the MGFs is

M_T(t)=(0.75+0.25e^t)^3(0.75+0.25e^t)^4(0.75+0.25e^t)^5 =(0.75+0.25e^t)^{12}.

Thus $T\sim\operatorname{Bin}(12,0.25)$. Its mean and variance are

E[T]=12(0.25)=3, \qquad \operatorname{Var}(T)=12(0.25)(0.75)=2.25.

For exactly two successes,

\begin{aligned} P(T=2) &=\binom{12}{2}(0.25)^2(0.75)^{10}\\ &=66(0.0625)(0.0563135147)\\ &\approx 0.2323. \end{aligned}

As a check, adding the three group means gives $0.75+1.00+1.25=3$, and adding their variances gives $0.5625+0.75+0.9375=2.25$. Both match the combined binomial model. The common mistake is to combine only two of the group sizes or to assume that independence is unnecessary; all three groups are included here, and their independence is part of the model.

When Success Probabilities Differ

For independent Bernoulli trials with probabilities $p_1,\ldots,p_N$, the total number of successes has mean $\sum_i p_i$ and variance $\sum_i p_i(1-p_i)$. If the total were $\operatorname{Bin}(N,q)$, matching the mean would require $q=(\sum_i p_i)/N$. But the binomial variance would then be $Nq(1-q)$, which is generally not the actual variance.

In fact, the actual variance is at most the variance of the binomial distribution with that matched mean. Writing $\bar p=(\sum_i p_i)/N$, the difference is

N\bar p(1-\bar p)-\sum_{i=1}^{N}p_i(1-p_i) =\sum_{i=1}^{N}(p_i-\bar p)^2.

This quantity is positive whenever the probabilities are not all equal. Therefore, for independent Bernoulli trials, unequal success probabilities rule out a binomial distribution with the same total number of trials. Such a total is often called a Poisson-binomial sum; the essential point here is that its distribution is not generally binomial.

Worked Example: Different Probabilities Do Not Combine into a Binomial

Two independent hypothetical inspection groups have counts $X\sim\operatorname{Bin}(2,0.20)$ and $Y\sim\operatorname{Bin}(3,0.60)$. Let $T=X+Y$. Determine whether $T$ can be $\operatorname{Bin}(5,q)$ for some $q$, using the mean and variance. The trial probabilities are $0.20$ for the first group and $0.60$ for the second.

Method. First calculate the mean and variance of the sum using independence:

\begin{aligned} E[T]&=2(0.20)+3(0.60)=0.40+1.80=2.20,\\ \operatorname{Var}(T)&=2(0.20)(0.80)+3(0.60)(0.40)\\ &=0.32+0.72=1.04. \end{aligned}

If $T$ were $\operatorname{Bin}(5,q)$, its mean would be $5q$. Matching the mean forces $q=2.20/5=0.44$. That candidate binomial distribution would have variance

$$ 5(0.44)(0.56)=1.232, $$

which differs from the actual variance $1.04$. Therefore $T$ is not binomial. As a second check, the actual probability of zero successes is \[ P(T=0)=(0.80)^2(0.40)^3=0.04096, \] whereas the candidate $\operatorname{Bin}(5,0.44)$ would give $P(T=0)=(0.56)^5=0.0550731776$. These probabilities also disagree.

The common mistake is to average the two group probabilities and declare the total binomial. The weighted average $0.44$ matches the mean, but it does not match the variance or the probability of zero successes. Matching one summary is not enough to establish a distribution.

Checks and Exam Traps

Before identifying a sum as binomial, check the model rather than relying on the fact that the summands are binomial. A reliable checklist is:

Independence: the group counts must be independent for the product rule to apply.
Common probability: each group must use the same success probability for the binomial factors to combine.
Total trial count: add the group sizes; do not average them.
Parameter check: a common-$p$ sum has mean $(\sum n_j)p$ and variance $(\sum n_j)p(1-p)$.
Different probabilities: do not replace them by an unweighted average. Even a weighted average that matches the mean does not establish a binomial law.

If independence fails, the MGF product rule cannot be used as written, and the sum need not be binomial even when the marginal distributions have the same parameters. If probabilities differ, the MGF product is still valid when the group counts are independent, but it generally does not simplify to a single binomial MGF. These conditions are distinct and should be checked separately.

Key takeaway: Independent binomial counts with a shared success probability combine by adding their trial counts and keeping that probability unchanged. Different probabilities generally produce a non-binomial sum, even if a proposed binomial model matches the total mean.

Check Your Understanding

For each problem, state the model conditions you use before applying a binomial-sum result.

Independent counts satisfy $X\sim\operatorname{Bin}(2,0.40)$ and $Y\sim\operatorname{Bin}(3,0.40)$. Find the distribution, mean, and variance of $X+Y$. (Answer: $\operatorname{Bin}(5,0.40)$, mean $2$, variance $1.2$.)
Independent counts satisfy $X\sim\operatorname{Bin}(4,0.30)$ and $Y\sim\operatorname{Bin}(6,0.30)$. Find $P(X+Y=2)$, rounded to four decimals. (Answer: $0.2335$.)
Three independent counts each have distribution $\operatorname{Bin}(2,0.50)$. Find the probability that their total is at least four. (Answer: $0.3438$.)
Three independent Bernoulli trials have success probabilities $0.20$, $0.50$, and $0.80$. Find the probability of no successes and the variance of the total number of successes. Is the total binomial? (Answer: Probability of no successes $0.0800$, variance $0.57$, and the total is not binomial because the probabilities differ.)
Independent counts satisfy $X\sim\operatorname{Bin}(3,0.20)$ and $Y\sim\operatorname{Bin}(2,0.70)$. Find the total’s mean and variance, then decide whether it is binomial with five trials. (Answer: Mean $2$, variance $0.90$, and it is not binomial; matching the mean gives $q=0.40$, whose binomial variance would be $1.20$.)

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