The Product Rule for Independent MGFs

Why Independence Makes the MGF of a Sum Useful

When a quantity is the sum of several independent random contributions, its distribution may be difficult to find directly. The moment generating function (MGF) gives a simple route: under independence, the MGF of the sum is the product of the individual MGFs. The result is useful only when the independence condition and the domains of the MGFs are handled correctly.

This tutorial proves the product rule and extends it to a finite sum. The examples focus on calculating the MGF and its domain, rather than identifying particular families of sum distributions. The methods are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: The moment generating function of a random variable $X$ is $M_X(t)=E[e^{tX}]$, for real $t$ such that this expectation is finite. Its domain is $D_X=\{t\in\mathbb{R}:E[e^{tX}]<\infty\}$. Since $e^{0X}=1$, $0\in D_X$ and $M_X(0)=1$.

The value $M_X(t)$ may be infinite outside $D_X$; it is not a finite MGF value there. For example, if $X$ has an exponential distribution with rate $\lambda>0$, its MGF is finite precisely when $t<\lambda$. A product formula for MGFs must therefore be accompanied by a domain statement.

The Product Rule and Its Domain

Theorem: If $X$ and $Y$ are independent, then $$ M_{X+Y}(t)=M_X(t)M_Y(t) $$ whenever both expectations on the right are finite. More precisely, for independent real-valued $X$ and $Y$, $M_{X+Y}(t)$ is finite exactly when both $M_X(t)$ and $M_Y(t)$ are finite. Thus the finite domain of the sum is $D_X\cap D_Y$.

The key step is independence. The random variables $e^{tX}$ and $e^{tY}$ are then independent, and $e^{t(X+Y)}=e^{tX}e^{tY}$. Independence allows the expectation of this product of nonnegative random variables to factor, including when an expectation is infinite. Because each exponential is strictly positive, its expectation cannot be zero. Consequently, if either individual expectation is infinite, the expectation of the product is infinite as well.

Why the formula holds

Independence means that the joint distribution of $(X,Y)$ is the product of the marginal distributions. Integrating the nonnegative function $e^{t(x+y)}$ against that product distribution gives

\begin{aligned} M_{X+Y}(t) &=E[e^{t(X+Y)}]\\ &=E[e^{tX}e^{tY}]\\ &=\int\!\!\int e^{tx}e^{ty}\,dF_X(x)\,dF_Y(y)\\ &=\left(\int e^{tx}\,dF_X(x)\right) \left(\int e^{ty}\,dF_Y(y)\right)\\ &=M_X(t)M_Y(t). \end{aligned}

The separation of the double integral is valid for nonnegative integrands; if the integrals are finite, it gives the usual finite product. If one is infinite, the corresponding expectation of the product is infinite because the other factor has strictly positive expectation. This also explains the domain statement: the product is finite exactly on the intersection of the two individual finite domains.

Finite-sum result: If $X_1,\ldots,X_n$ are mutually independent, then $$ M_{\sum_{j=1}^{n}X_j}(t)=\prod_{j=1}^{n}M_{X_j}(t), \qquad D_{\sum_{j=1}^{n}X_j}=\bigcap_{j=1}^{n}D_{X_j}. $$

Here, mutual independence means the joint distribution of all $n$ variables is the product of their marginal distributions; independence in pairs alone is not enough for this statement. The finite-sum formula follows by applying the two-variable result repeatedly, since a sum of some of the variables is independent of the remaining variables under mutual independence. For a finite collection, if each MGF is finite on an open interval around zero, their intersection also contains an open interval around zero.

Applying the Product Rule

Worked Example: A Bernoulli Contribution and an Exponential Contribution

Let $X\sim\operatorname{Bernoulli}(0.4)$ and let $Y$ have an exponential distribution with rate $2$, in the same time unit. Assume $X$ and $Y$ are independent. Find $M_{X+Y}(t)$, its finite domain, and its value at $t=1$. Also check its mean using the derivative at zero.

Method. Find the individual MGFs, intersect their finite domains, and multiply. The Bernoulli variable takes values 1 and 0 with probabilities 0.4 and 0.6, so $$ M_X(t)=0.6+0.4e^t,\qquad D_X=\mathbb{R}. $$ For an exponential variable with rate $2$, integration against the density $2e^{-2y}$ for $y\geq 0$ gives $$ M_Y(t)=\int_0^\infty e^{ty}2e^{-2y}\,dy =\frac{2}{2-t},\qquad t<2. $$ The integral is finite only when $2-t>0$.

Since the variables are independent, the product rule gives $$ M_{X+Y}(t)=(0.6+0.4e^t)\frac{2}{2-t},\qquad t<2. $$ The domain is $\mathbb{R}\cap(-\infty,2)=(-\infty,2)$.

At $t=1$, $$ M_{X+Y}(1)=(0.6+0.4e)\frac{2}{1} =1.2+0.8e \approx 3.3746. $$ The approximation is rounded to four decimal places.

For a check, differentiate the product at zero. The factor values at zero are $M_X(0)=M_Y(0)=1$, while $M_X'(0)=0.4$ and $M_Y'(0)=1/2$. Therefore $$ M'_{X+Y}(0)=0.4(1)+1(0.5)=0.9. $$ This agrees with $E[X]+E[Y]=0.4+1/2=0.9$. The common mistake avoided here is using an exponential MGF with the wrong sign in its denominator; the factor $e^{ty}$ in the defining expectation leads to $2-t$, not $2+t$.

Worked Example: Two Independent Uniform Contributions

Let $X$ be uniform on $[0,2]$ and $Y$ be uniform on $[1,3]$, with $X$ and $Y$ independent. Find the MGF of $X+Y$, its domain, and its value at $t=1/2$.

Method. Integrate $e^{tx}$ against each uniform density and multiply the results. For $t\ne 0$, the density of $X$ is $1/2$ on $[0,2]$, so $$ M_X(t)=\frac{1}{2}\int_0^2 e^{tx}\,dx =\frac{e^{2t}-1}{2t}. $$ Similarly, $Y$ has density $1/2$ on $[1,3]$, giving $$ M_Y(t)=\frac{1}{2}\int_1^3 e^{ty}\,dy =\frac{e^{3t}-e^t}{2t}. $$ Both MGFs are finite for every real $t$; the displayed quotients have removable values at $t=0$, where each MGF equals 1.

Independence now gives $$ M_{X+Y}(t)=\frac{(e^{2t}-1)(e^{3t}-e^t)}{4t^2},\qquad t\ne 0, $$ with $M_{X+Y}(0)=1$. Its finite domain is all real numbers, the intersection of the two individual domains.

At $t=1/2$, the denominator $2t$ in each individual MGF is 1. Hence $$ M_X(1/2)=e-1,\qquad M_Y(1/2)=e^{3/2}-e^{1/2}=e^{1/2}(e-1), $$ and $$ M_{X+Y}(1/2)=e^{1/2}(e-1)^2. $$ As a check on the formula at the origin, the individual limits as $t$ approaches zero are both 1, so the product also has limit 1. The common mistake avoided is treating the quotient expressions as undefined MGFs at zero; their defining expectations are finite there.

Worked Example: A Finite Sum of Three Exponential Variables

Suppose $X_1,X_2,X_3$ are mutually independent exponential random variables with respective rates $1$, $2$, and $4$, all measured in a consistent unit. Find the MGF and its domain for $S=X_1+X_2+X_3$, then evaluate $M_S(1/4)$.

Method. An exponential variable with rate $\lambda$ has MGF $\lambda/(\lambda-t)$ for $t<\lambda$. Thus $$ M_{X_1}(t)=\frac{1}{1-t}\quad(t<1),\qquad M_{X_2}(t)=\frac{2}{2-t}\quad(t<2),\qquad M_{X_3}(t)=\frac{4}{4-t}\quad(t<4). $$ The domain of the sum is the intersection of these three domains, which is $t<1$. Mutual independence gives $$ M_S(t)=\frac{1}{1-t}\frac{2}{2-t}\frac{4}{4-t},\qquad t<1. $$

At $t=1/4$, the three factors are $$ \frac{1}{1-1/4}=\frac{4}{3},\qquad \frac{2}{2-1/4}=\frac{8}{7},\qquad \frac{4}{4-1/4}=\frac{16}{15}. $$ Therefore $$ M_S(1/4)=\frac{4}{3}\frac{8}{7}\frac{16}{15} =\frac{512}{315} \approx 1.6254. $$ The approximation is rounded to four decimal places. At $t=0$, each factor is 1, so the product gives $M_S(0)=1$, as required. The common mistake avoided is using the smallest rate only as a label: it determines the upper boundary of the intersection, so the sum MGF is finite for $t<1$, not merely for $t<4$.

Common Exam Traps and Practical Checks

The product rule is short, but its assumptions and domain are part of the result. Check each of the following before using it:

Check mutual independence. The product formula is justified by independence. Do not apply it just because the variables have individual MGFs.
Intersect the domains. For a finite independent sum, $t$ must belong to every individual MGF domain. A single factor that is infinite makes the product MGF infinite.
Keep the exponential sign consistent. The MGF uses $e^{tX}$. In particular, integrating an exponential density produces a denominator $\lambda-t$, with finite domain $t<\lambda$.
Handle $t=0$ correctly. A quotient formula may appear to have zero in its denominator at $t=0$, but the defining expectation gives $M_X(0)=1$. Use the limit of the quotient when appropriate.
Distinguish a product of MGFs from a product of random variables. The theorem concerns the MGF of a sum. It does not say that the MGF of $XY$ is $M_X(t)M_Y(t)$.

Key takeaway: For mutually independent random variables, the MGF of their sum is the product of their MGFs, and its finite domain is the intersection of the individual finite domains. Independence justifies the factorization; checking the domain completes the result.

Check Your Understanding

For each problem, use the product rule only after stating the independence and domain conditions.

Let $X\sim\operatorname{Bernoulli}(0.25)$ and $Y\sim\operatorname{Bernoulli}(0.60)$ be independent. Find $M_{X+Y}(t)$ and its value at $t=\ln 2$. (Answer: $M_{X+Y}(t)=(0.75+0.25e^t)(0.40+0.60e^t)$, domain $\mathbb{R}$, and $M_{X+Y}(\ln 2)=2$.)
Let independent $X$ and $Y$ be exponential with rates $5$ and $2$. Find the finite domain of the sum MGF and evaluate it at $t=1$. (Answer: domain $t<2$, and $M_{X+Y}(1)=(5/4)(2)=2.5$.)
Let independent $X$ and $Y$ be uniform on $[-1,1]$ and $[0,2]$, respectively. Write the sum MGF for $t\ne0$, give its value at zero, and state its domain. (Answer: $M_{X+Y}(t)=\frac{e^t-e^{-t}}{2t}\frac{e^{2t}-1}{2t}$ for $t\ne0$, $M_{X+Y}(0)=1$, domain $\mathbb{R}$.)
Let $X_1,X_2,X_3$ be mutually independent exponential variables with rates $1,3,6$, respectively. Find the sum MGF domain and evaluate the MGF at $t=1/2$. (Answer: domain $t<1$, and $M(1/2)=2\cdot(6/5)\cdot(12/11)=144/55$.)
Three mutually independent random variables have MGFs finite precisely on $(-2,3)$, $(-1,4)$, and $(-0.5,2)$, respectively. State the finite domain of the MGF of their sum and its value at zero. (Answer: domain $(-0.5,2)$, and the MGF at zero is $1$.)

What You'll Learn