MGF of an Independent Sum¶

Theorem¶

If $X$ and $Y$ are independent ( $X ⊥$ $X\perp Y$ $Y$ ), then for every $t$ at which both MGFs are finite,

\begin{aligned} M_{X+Y}(t)=M_X(t)\cdot M_Y(t). \end{aligned}

Independence is essential.

Use the Moment Generating Function definition to write $M_{X+Y}(t)=\mathbb{E}[\mathrm{e}^{tX}\mathrm{e}^{tY}]$ $E [e^{tX} e^{t Y}]$ . Independence factors this expectation into the two MGFs.

Interpretation¶

Adding independent variables corresponds to multiplying their MGFs. Since an MGF that is finite near zero determines the distribution, this often identifies the distribution of a sum directly, without working out its PMF or density value by value.

Probability bar charts of Poisson(2), Poisson(3) and their independent sum, which is Poisson(5) and lies further right with a wider spread.

The figure shows $X\sim\operatorname{Pois}(2)$ $Pois (2)$ , an independent $Y\sim\operatorname{Pois}(3)$ and their sum, which lies further right and is more spread out. Multiplying the Poisson MGFs gives $\mathrm{e}^{2(\mathrm{e}^t-1)}\mathrm{e}^{3(\mathrm{e}^t-1)}=\mathrm{e}^{5(\mathrm{e}^t-1)}$ $e^{5 (e^{t} - 1)}$ , the MGF of $\operatorname{Pois}(5)$ , so by uniqueness the sum is $\operatorname{Pois}(5)$ .

For independent $X$ and $Y$ the MGF of $X+Y$ is the product of the two curves, which identifies the distribution of the sum.

The product comes from $\mathrm{e}^{t(X+Y)}=\mathrm{e}^{tX}\mathrm{e}^{tY}$ $e^{tX} e^{t Y}$ : for independent $X$ and $Y$ the factors $\mathrm{e}^{tX}$ and $\mathrm{e}^{tY}$ are independent, and the expectation of a product of independent variables is the product of the expectations. Without independence this step fails. For $Y=X$ with $X\sim\operatorname{Bern}(\tfrac{1}{2})$ $Bern (\frac{1}{2})$ , the sum $2X$ has MGF $\tfrac{1}{2}(1+\mathrm{e}^{2t})$ $e^{2 t})$ , whereas $M_X(t)^2=\tfrac{1}{4}(1+\mathrm{e}^t)^2$ $e^{t})^{2}$ ; the difference is $(\mathrm{e}^t-1)^2/4$ $1)^{2} /4$ , which is positive for every $t\neq0$ $0$ .

For i.i.d. variables the product becomes a power. Thus $n$ independent $\operatorname{Bern}(p)$ variables give $(1-p+p\mathrm{e}^t)^n$ $p e^{t})^{n}$ , the MGF of the binomial distribution, and $r$ independent $\operatorname{Exp}(\lambda)$ variables give $(\lambda/(\lambda-t))^r$ for $t<\lambda$ $λ$ , the MGF of the gamma distribution with shape $r$ and rate $\lambda$ .

Key Results¶

For independent $X_1,\ldots,X_n$ , $\displaystyle M_{\sum_iX_i}(t)=\prod_iM_{X_i}(t)$ $i \prod M_{X_{i}} (t)$ .
For i.i.d. variables, $\displaystyle M_{\sum_iX_i}(t)=M_{X_1}(t)^n$ $M_{X_{1}} (t)^{n}$ .

Examples¶

The sum of independent Poisson variables with parameters $2$ and $3$ has MGF $\exp(5(\mathrm{e}^t-1))$ $1))$ , so it is $\operatorname{Pois}(5)$ .
The sum of two independent $\operatorname{Exp}(1)$ variables has MGF $(1-t)^{-2}$ for $t<1$ $1$ , identifying $\operatorname{Gamma}(2,1)$ in rate notation.

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 1

Aliases: MGF of a sum of independent random variables, moment generating function of an independent sum Symbols: lambda, product, sum

Last updated: 20 October 2026 19:41