The MGF Product Rule for Independent Sums

Why the MGF Product Rule Matters

When a total is formed by adding independent random quantities, its distribution can be difficult to calculate directly. The moment generating function (MGF) often gives a simpler route: calculate the MGF of each quantity and multiply. This tutorial establishes exactly why that works, and the conditions that must be checked before using it.

The key condition is independence. Having the same marginal distributions as independent variables is not enough. It is also important to track where each MGF is finite: the product rule applies at values of the argument for which all the required expectations exist and are finite.

Definition: The moment generating function of a random variable $X$ is $M_X(t)=E[e^{tX}]$, for each real number $t$ at which this expectation is finite. Its domain is the set of such $t$-values. In particular, $M_X(0)=1$, since $e^{0X}=1$.

The variable $X$ may be discrete, continuous, or a mixture of discrete and continuous parts. The expectation is calculated according to its distribution, but the definition is the same. For a discrete variable, for example, $M_X(t)=\sum_x e^{tx}P(X=x)$; for a continuous variable with density $f_X$, it is $M_X(t)=\int_{-\infty}^{\infty} e^{tx}f_X(x)\,dx$, when the sum or integral is finite.

The Product Rule

Theorem: If $X$ and $Y$ are independent and $M_X(t)$ and $M_Y(t)$ are finite, then $X+Y$ has a finite MGF at $t$, and $$ M_{X+Y}(t)=M_X(t)M_Y(t). $$ More generally, if $X_1,\ldots,X_n$ are mutually independent and every $M_{X_j}(t)$ is finite, then $$ M_{\sum_{j=1}^{n}X_j}(t)=\prod_{j=1}^{n}M_{X_j}(t). $$

The condition for a particular $t$ is that every factor be finite there. Thus the product formula holds on the intersection of the individual MGF domains. If each variable has an MGF finite on an interval around zero, the sum's MGF is finite on the intersection of those intervals. Do not assume that the sum's MGF exists at every real $t$ merely because each variable has an MGF somewhere.

Why Independence Produces a Product

For any real $t$, the exponential of a sum separates algebraically: $e^{t(X+Y)}=e^{tX}e^{tY}$. Independence then allows the expectation of this product of functions of separate variables to factor. The resulting identity is the product rule.

Start with the definition.
$M_{X+Y}(t)=E[e^{t(X+Y)}]$, provided the expectation is finite.

Separate the exponential.
Since $e^{t(X+Y)}=e^{tX}e^{tY}$, this becomes $E[e^{tX}e^{tY}]$.

Use independence.
Independence of $X$ and $Y$ implies that functions of $X$ and $Y$ factor in expectation when the expectations are finite. Therefore $E[e^{tX}e^{tY}]=E[e^{tX}]E[e^{tY}]$.

Recognize the MGFs.
The two expectations are $M_X(t)$ and $M_Y(t)$, so $M_{X+Y}(t)=M_X(t)M_Y(t)$.

For $n$ mutually independent variables, the same argument gives $e^{t\sum_{j=1}^{n}X_j}=\prod_{j=1}^{n}e^{tX_j}$. Mutual independence permits the expectation of this product to factor into the product of the $n$ expectations. This is why mutual independence is the appropriate condition for the general $n$-variable rule; pairwise independence alone does not guarantee the required factorization.

Worked Example: A Sum of Two Independent Discrete Variables

Let $X$ equal 1 with probability $1/3$ and 0 with probability $2/3$. Let $Y$ equal 2 with probability $1/2$ and 0 with probability $1/2$. Suppose $X$ and $Y$ are independent. Find the MGF of $S=X+Y$, then evaluate it at $t=\ln 2$ and check the result using the possible values of $S$.

Method and assumptions. Both variables have finite support, so their MGFs are finite for every real $t$. Independence allows the product rule.

First calculate the individual MGFs: $$ M_X(t)=\frac{2}{3}e^{0t}+\frac{1}{3}e^t=\frac{2+e^t}{3}, \qquad M_Y(t)=\frac{1}{2}e^{0t}+\frac{1}{2}e^{2t}=\frac{1+e^{2t}}{2}. $$ Thus $$ M_S(t)=M_X(t)M_Y(t)=\frac{(2+e^t)(1+e^{2t})}{6}. $$

At $t=\ln 2$, $e^t=2$ and $e^{2t}=4$. Therefore $$ M_S(\ln 2)=\frac{(2+2)(1+4)}{6}=\frac{4\cdot5}{6}=\frac{10}{3}. $$

Check by listing the sum values. Independence gives $P(S=0)=(2/3)(1/2)=1/3$, $P(S=1)=(1/3)(1/2)=1/6$, $P(S=2)=(2/3)(1/2)=1/3$, and $P(S=3)=(1/3)(1/2)=1/6$. These probabilities sum to $1$. Directly, $$ M_S(\ln 2)=\frac{1}{3}+\frac{1}{6}(2)+\frac{1}{3}(4)+\frac{1}{6}(8) =\frac{1}{3}+\frac{1}{3}+\frac{4}{3}+\frac{4}{3} =\frac{10}{3}. $$ The direct calculation agrees with the product rule. The common mistake avoided here is to multiply the probabilities for $X$ and $Y$ without first checking that the variables are independent.

Worked Example: Combining a Continuous and a Uniform Variable

Let $X$ have an exponential distribution with rate parameter 2, so its density is $f_X(x)=2e^{-2x}$ for $x\geq 0$. Let $Y$ be uniform on $[0,1]$, independently of $X$. Find the MGF of $S=X+Y$, state its domain, and evaluate it at $t=1$, to four decimal places.

Method and assumptions. The exponential rate is 2, not its mean; the mean is $1/2$. Independence is given. Calculate each MGF from its defining integral.

For $X$, $$ M_X(t)=\int_0^\infty e^{tx}2e^{-2x}\,dx =2\int_0^\infty e^{-(2-t)x}\,dx =\frac{2}{2-t},\qquad t<2. $$ The integral converges only when $2-t>0$. For $Y$, $$ M_Y(t)=\int_0^1 e^{ty}\,dy =\frac{e^t-1}{t},\qquad t\ne 0, $$ and $M_Y(0)=1$. This MGF is finite for every real $t$. By independence, the sum MGF is $$ M_S(t)=\frac{2}{2-t}\frac{e^t-1}{t},\qquad t<2,\ t\ne 0, $$ with $M_S(0)=1$ by the definition of the MGF (or by taking the limit as $t$ approaches zero). The domain is $t<2$, the intersection of the two individual domains.

At $t=1$, the exponential factor is $2/(2-1)=2$, and the uniform factor is $(e-1)/1$. Hence $$ M_S(1)=2(e-1)=2(2.7182818\ldots-1)=3.4365636\ldots. $$ Rounded only at the final step, $M_S(1)=3.4366$ to four decimal places. The check is that $1<2$, so the exponential MGF is finite at the requested value. Using $2/(2+t)$ instead would correspond to a different expression and would not follow from the defining integral.

Worked Example: Applying the Rule to Three Independent Variables

Suppose $X,Y,Z$ are mutually independent. Let $X$ equal 1 with probability $1/4$ and 0 otherwise; let $Y$ equal $-1$ or 1, each with probability $1/2$; and let $Z$ be exponential with rate 3. Find $M_{X+Y+Z}(t)$, its domain, and its value at $t=\ln 2$, to four decimal places.

Method and assumptions. The first two variables have finite support and hence MGFs finite for all real $t$. The exponential variable has a finite MGF only for $t<3$. Mutual independence allows all three factors to be multiplied.

The individual MGFs are $$ M_X(t)=\frac{3}{4}+\frac{1}{4}e^t,\qquad M_Y(t)=\frac{e^{-t}+e^t}{2},\qquad M_Z(t)=\frac{3}{3-t}\quad(t<3). $$ Therefore $$ M_{X+Y+Z}(t)= \left(\frac{3}{4}+\frac{1}{4}e^t\right) \left(\frac{e^{-t}+e^t}{2}\right) \frac{3}{3-t},\qquad t<3. $$ The domain is $t<3$, since that is the only restriction from the three individual MGFs.

At $t=\ln 2$, $e^t=2$ and $e^{-t}=1/2$. The three factors are $$ M_X(\ln 2)=\frac{3}{4}+\frac{2}{4}=\frac{5}{4},\qquad M_Y(\ln 2)=\frac{1/2+2}{2}=\frac{5}{4}, $$ $$ M_Z(\ln 2)=\frac{3}{3-\ln 2}. $$ Their product is $$ M_{X+Y+Z}(\ln 2)=\frac{5}{4}\cdot\frac{5}{4}\cdot\frac{3}{3-\ln 2} =\frac{75}{16(3-\ln 2)} \approx 2.0320. $$ The final value is rounded to four decimal places. Since $\ln 2\approx0.6931<3$, the evaluation is within the MGF domain. A common error in a three-variable problem is to use pairwise independence as if it necessarily implied mutual independence; here mutual independence is explicitly assumed.

Where the Rule Fails

The algebraic identity $e^{t(X+Y)}=e^{tX}e^{tY}$ is always true. What may fail is the next step: factoring the expectation. That factorization requires independence. If $X$ and $Y$ are dependent, the MGF of their sum is still $E[e^{tX}e^{tY}]$, but it need not equal $M_X(t)M_Y(t)$.

Worked Example: A Dependence Trap

Let $X$ be 0 or 1, each with probability $1/2$, and define $Y=X$. Find the actual MGF of $S=X+Y$, then compare it with the product of the marginal MGFs at $t=\ln 2$.

Method and common mistake. The variables have identical marginal MGFs, but $Y=X$, so they are not independent. Do not apply the product rule.

Since $S=2X$, it is 0 or 2, each with probability $1/2$. Thus $$ M_S(t)=\frac{1}{2}e^0+\frac{1}{2}e^{2t} =\frac{1+e^{2t}}{2}. $$ Each marginal MGF is $M_X(t)=M_Y(t)=(1+e^t)/2$. The incorrect independent-product calculation at $t=\ln2$ would give $$ M_X(\ln2)M_Y(\ln2)=\left(\frac{1+2}{2}\right)^2=\frac{9}{4}=2.25. $$ The actual value is $$ M_S(\ln2)=\frac{1+4}{2}=\frac{5}{2}=2.5. $$ The discrepancy confirms that multiplying marginal MGFs is not justified for these dependent variables.

Practical Checks Before Multiplying

A reliable calculation follows a short sequence. First identify the sum and the individual variables. Next establish mutual independence (or independence for a two-variable sum). Then calculate each MGF from its definition, noting its domain. Only after those checks should the factors be multiplied. Finally, verify that the requested value of $t$ lies in every individual domain.

This method is aligned with the SOA Exam P (Probability) syllabus because it turns a statement about independence into a precise calculation for a sum. It does not by itself identify the distribution of the sum; the product rule supplies its MGF wherever finite. Domain checks and the independence condition remain essential whether the variables are discrete or continuous.

Key takeaway: For mutually independent variables, the MGF of their sum is the product of their MGFs at every $t$ where all factors are finite. The factorization comes from independence, not merely from the algebra of exponentials.

Check Your Understanding

For each problem, state any needed domain restriction as well as the requested MGF value or expression.

$X$ and $Y$ are independent Bernoulli variables with probabilities of 1 equal to 0.2 and 0.5, respectively. Find $M_{X+Y}(\ln 2)$. (Answer: $1.8$)
$X$ and $Y$ are independent exponential variables with rates 4 and 6. Write $M_{X+Y}(t)$ and its domain. (Answer: $M_{X+Y}(t)=\frac{4}{4-t}\frac{6}{6-t}$, $t<4$)
$X$ is uniform on $[-1,1]$, and independently $Y=0$ with probability $3/4$ or $Y=2$ with probability $1/4$. Write $M_{X+Y}(t)$, including the value at zero. (Answer: $M_{X+Y}(t)=\frac{\sinh(t)}{t}\left(\frac34+\frac14e^{2t}\right)$ for $t\ne0$, and $M_{X+Y}(0)=1$; finite for all real $t$)
$X_1,X_2,X_3$ are mutually independent Bernoulli variables with probabilities of 1 equal to 0.1, 0.4, and 0.7. Find $M_{X_1+X_2+X_3}(\ln2)$. (Answer: $1.1(1.4)(1.7)=2.618$)
$X$ is Bernoulli with probability of 1 equal to $1/2$, and $Y=X$. Give the actual MGF of $X+Y$ and the value at $t=\ln2$. (Answer: $M_{X+Y}(t)=\frac{1+e^{2t}}{2}$, so $M_{X+Y}(\ln2)=2.5$; the product of marginal MGFs would incorrectly give $2.25$)

What You'll Learn