Product MGFs for Independent Sums

Why Products of MGFs Help with Sums

When several independent random variables contribute to a total, finding the distribution of that total directly can require lengthy probability calculations. A moment generating function can turn the sum into a product: each summand contributes its own MGF, and independence makes the factorization possible. If the resulting product matches the MGF of a known distribution, a uniqueness result identifies the distribution of the sum.

The method is useful only when its conditions are met. In particular, independence is essential for the product formula, and the MGF must exist on an open interval containing zero to use the standard uniqueness theorem. The calculations below focus on these conditions and on recognizing the distribution from the resulting expression.

Definition: The moment generating function (MGF) of a random variable \(X\) is \(M_X(t)=E[e^{tX}]\), for those real values of \(t\) for which this expectation is finite. To use MGF uniqueness in the usual way, the MGF must be finite on some open interval containing zero.

For a discrete variable, the expectation is a sum over its possible values; for a continuous variable with density \(f_X\), it is an integral. An MGF may be finite for some \(t\) values and infinite for others. When working with a sum, use values of \(t\) for which the MGFs of all summands are finite.

The Product Formula and Its Conditions

Let \(X_1,\ldots,X_n\) be independent random variables, and let \(S=X_1+\cdots+X_n\). For any \(t\) at which the relevant expectations are finite, the MGF of \(S\) is the product of the individual MGFs.

Theorem: If \(X_1,\ldots,X_n\) are independent and their MGFs are finite at \(t\), then \(M_S(t)=\prod_{j=1}^{n}M_{X_j}(t)\), where \(S=\sum_{j=1}^{n}X_j\). If each MGF is finite on an open interval containing zero, their product is also finite on an open interval containing zero.

The reason is that the exponential of a sum factors into a product, and independence allows the expectation of that product to factor. For two variables, the argument is:

\begin{aligned} M_{X+Y}(t) &=E[e^{t(X+Y)}]\\ &=E[e^{tX}e^{tY}]\\ &=E[e^{tX}]E[e^{tY}]\\ &=M_X(t)M_Y(t). \end{aligned}

The third equality uses independence: for independent variables, expectations of products of suitable functions of the separate variables factor. Repeating this argument gives the formula for any finite number of independent summands. Without independence, that equality is not generally valid.

Theorem: If two random variables have MGFs that are finite and equal on an open interval containing zero, then they have the same distribution. This is the MGF uniqueness theorem. Consequently, if the product formula gives \(M_S(t)=M_Z(t)\) throughout such an interval for a known random variable \(Z\), then \(S\) and \(Z\) have the same distribution.

The uniqueness step is what turns an algebraic match into a distributional conclusion. Equality at just one value of \(t\), or a resemblance between two expressions outside a shared interval of finiteness, is not enough. First establish the product MGF on a suitable interval; then compare it with the MGF of the proposed distribution.

Worked Applications

Worked Example: Identifying a Sum of Independent Poisson Variables

Let \(X\sim\operatorname{Poisson}(1.2)\) and \(Y\sim\operatorname{Poisson}(2.3)\), independently. Identify the distribution of \(S=X+Y\) using MGFs.

Method and assumptions. The variables are independent Poisson variables with parameters \(1.2\) and \(2.3\). For \(W\sim\operatorname{Poisson}(\lambda)\), \(M_W(t)=\exp(\lambda(e^t-1))\), which is finite for every real \(t\).

Step 1: Write both MGFs.

M_X(t)=\exp(1.2(e^t-1)),\qquad M_Y(t)=\exp(2.3(e^t-1)).

Step 2: Multiply using independence.

\begin{aligned} M_S(t) &=M_X(t)M_Y(t)\\ &=\exp(1.2(e^t-1))\exp(2.3(e^t-1))\\ &=\exp((1.2+2.3)(e^t-1))\\ &=\exp(3.5(e^t-1)). \end{aligned}

Step 3: Identify the distribution. This is the MGF of a Poisson random variable with parameter \(3.5\). Both MGFs are finite around zero, so uniqueness gives \(S\sim\operatorname{Poisson}(3.5)\).

Check and common mistake avoided. The combined parameter is \(1.2+2.3=3.5\). Multiplying the parameters would give \(1.2(2.3)=2.76\), which is not the parameter in the product MGF and is incorrect. The independence assumption is what permits multiplying the MGFs.

Worked Example: Adding Gamma Variables with a Common Rate

Let \(X\sim\operatorname{Gamma}(2,\text{ rate }4)\) and \(Y\sim\operatorname{Gamma}(3,\text{ rate }4)\), independently. Identify the distribution of \(S=X+Y\). Here the gamma parameterization uses shape \(\alpha\) and rate \(\lambda\).

Method and assumptions. For a gamma variable with shape \(\alpha\) and rate \(\lambda\), its MGF is \(M(t)=(\lambda/(\lambda-t))^\alpha\) for \(t<\lambda\). The two given variables are independent and share rate \(4\), so both MGFs are finite for \(t<4\), an interval containing zero.

Step 1: Substitute each variable’s parameters.

M_X(t)=\left(\frac{4}{4-t}\right)^2,\qquad M_Y(t)=\left(\frac{4}{4-t}\right)^3,\qquad t<4.

Step 2: Multiply and combine powers.

\begin{aligned} M_S(t) &=M_X(t)M_Y(t)\\ &=\left(\frac{4}{4-t}\right)^2 \left(\frac{4}{4-t}\right)^3\\ &=\left(\frac{4}{4-t}\right)^5,\qquad t<4. \end{aligned}

Step 3: Match and apply uniqueness. The result is the MGF of a gamma variable with shape \(5\) and rate \(4\). Since the MGFs agree on the interval \(t<4\), which contains zero, \(S\sim\operatorname{Gamma}(5,\text{ rate }4)\).

Check and common mistake avoided. The shape parameters add, \(2+3=5\), while the common rate remains \(4\). The expression depends on having the same rate: if the rates differed, the factors would have different denominators and this particular gamma MGF match would not follow.

Worked Example: Finding the Distribution of a Finite-Support Sum

Let \(X_1,X_2,X_3\) be independent Bernoulli variables with success probabilities \(0.20\), \(0.60\), and \(0.75\), respectively. Thus each variable is \(1\) on success and \(0\) otherwise. Find the distribution of \(S=X_1+X_2+X_3\).

Method and assumptions. For a Bernoulli variable with success probability \(p\), its MGF is \(M(t)=1-p+pe^t\), since it takes values \(0\) and \(1\). Independence lets us multiply these three MGFs. Set \(z=e^t\) to make the coefficients of the product easy to read.

Step 1: Form the product.

M_S(t) =(0.80+0.20z)(0.40+0.60z)(0.25+0.75z), \qquad z=e^t.

Step 2: Multiply the first two factors.

$$ (0.80+0.20z)(0.40+0.60z) =0.32+0.56z+0.12z^2. $$

Step 3: Multiply by the third factor and collect powers.

\begin{aligned} M_S(t) &=(0.32+0.56z+0.12z^2)(0.25+0.75z)\\ &=0.08+0.38z+0.45z^2+0.09z^3. \end{aligned}

For a variable supported on \(0,1,2,3\), its MGF has the form \(\sum_{k=0}^{3}P(S=k)e^{kt}\). Since \(z=e^t\), the coefficients therefore give \(P(S=0)=0.08\), \(P(S=1)=0.38\), \(P(S=2)=0.45\), and \(P(S=3)=0.09\).

Check and common mistake avoided. The probabilities add to \(0.08+0.38+0.45+0.09=1.00\), as required. The sum is not generally binomial because the three success probabilities are not equal. Treating it as a binomial variable would incorrectly impose one common success probability.

When the Product Method Does Not Apply

The product formula requires independence, not merely zero covariance or similar-looking marginal distributions. To see the distinction, let \(B\) be Bernoulli with success probability \(0.40\), and set \(X=B\), \(Y=B\). These variables are dependent: they are always equal. Their sum is \(2B\), so

M_{X+Y}(t)=E[e^{2tB}]=0.60+0.40e^{2t}.

But multiplying the two marginal MGFs would give

$$ M_X(t)M_Y(t)=(0.60+0.40e^t)^2, $$

which is not the MGF of \(2B\). For example, at \(t=\ln 2\), the true MGF is \(0.60+0.40(4)=2.20\), while the product of the marginal MGFs is \((0.60+0.40(2))^2=1.96\). This confirms that the factorization fails here.

Common exam trap: Do not multiply marginal MGFs just because you know each summand’s distribution. Verify independence first. Also check that the proposed MGF match holds on an open interval containing zero; without that condition, the standard MGF uniqueness conclusion is not justified.

In practice, the method is a short chain of reasoning: write each individual MGF with its parameterization, use independence to multiply, simplify the product, and compare it with a known MGF on a common interval around zero. If no standard form matches, a finite-support product may still reveal probabilities through its coefficients, as in the Bernoulli example.

Check Your Understanding

For each problem, identify whether independence and the MGF conditions permit the product method, then use the product or explain why the proposed identification fails.

Independent \(U\sim\operatorname{Poisson}(0.8)\) and \(V\sim\operatorname{Poisson}(1.7)\) are added. Identify the distribution of \(U+V\). (Answer: \(\operatorname{Poisson}(2.5)\).)
Independent gamma variables have shapes \(1.5\) and \(2.5\), both with rate \(3\). Identify their sum’s distribution. (Answer: \(\operatorname{Gamma}(4,\text{ rate }3)\).)
Independent Bernoulli variables have success probabilities \(0.30\) and \(0.40\). Find the probabilities that their sum is \(0\), \(1\), and \(2\). (Answer: \(P(S=0)=0.42,\ P(S=1)=0.46,\ P(S=2)=0.12\).)
Independent \(R\sim\operatorname{Gamma}(2,\text{ rate }2)\) and \(T\sim\operatorname{Gamma}(3,\text{ rate }5)\) are added. Does their MGF product identify the sum as a gamma variable by the common-rate matching used above? (Answer: No; the product has different denominators, \( (2/(2-t))^2(5/(5-t))^3 \), so it does not match a single gamma MGF by that argument.)
Let \(C\sim\operatorname{Bernoulli}(0.25)\) and define \(D=1-C\). Is it valid to multiply their marginal MGFs to obtain the MGF of \(C+D\)? Give the actual sum distribution. (Answer: No; they are dependent, and \(C+D=1\) with probability \(1\).)

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