[proofplan] We first prove that the [moment generating function](/page/Moment%20Generating%20Function) of the sum factors as the product of the two moment generating functions. The factorization follows from independence after applying it to the Borel functions $x \mapsto e^{tx}$ and $y \mapsto e^{ty}$. Since moment generating functions are finite and strictly positive on the interval, taking logarithms gives the identity for cumulant generating functions. Finally, smoothness near $0$ follows from the local finiteness of the moment generating functions, and differentiating the identity gives additivity of cumulants. [/proofplan]

[step:Factor the moment generating function using independence]Fix $t \in (-a,a)$. Define the non-negative random variables $U_t,V_t: (\Omega,\mathcal F) \to ([0,\infty),\mathcal B([0,\infty)))$ by \begin{align*} U_t(\omega)=e^{tX(\omega)} \end{align*} and \begin{align*} V_t(\omega)=e^{tY(\omega)}. \end{align*} The maps $x \mapsto e^{tx}$ and $y \mapsto e^{ty}$ are Borel measurable, so $U_t$ and $V_t$ are measurable. Since $X$ and $Y$ are independent, the random variables $U_t$ and $V_t$ are independent. Equivalently, if $\mu_X=\mathbb P\circ X^{-1}$ and $\mu_Y=\mathbb P\circ Y^{-1}$ are the laws of $X$ and $Y$, then the law of $(X,Y)$ is $\mu_X\otimes \mu_Y$. Using Tonelli's theorem for the non-negative Borel function $(x,y)\mapsto e^{t(x+y)}$, we obtain \begin{align*} M_{X+Y}(t)=\int_{\mathbb R^2} e^{t(x+y)}\,d(\mu_X\otimes\mu_Y)(x,y). \end{align*} Because $e^{t(x+y)}=e^{tx}e^{ty}$, Tonelli's theorem and the product measure identity give \begin{align*} M_{X+Y}(t)=\left(\int_{\mathbb R} e^{tx}\,d\mu_X(x)\right)\left(\int_{\mathbb R} e^{ty}\,d\mu_Y(y)\right). \end{align*} Thus \begin{align*} M_{X+Y}(t)=M_X(t)M_Y(t). \end{align*}[/step]

[guided]Fix $t \in (-a,a)$. The goal is to rewrite the exponential of the sum as a product and then use independence. Define $U_t,V_t: (\Omega,\mathcal F) \to ([0,\infty),\mathcal B([0,\infty)))$ by \begin{align*} U_t(\omega)=e^{tX(\omega)} \end{align*} and \begin{align*} V_t(\omega)=e^{tY(\omega)}. \end{align*} These are measurable because $X$ and $Y$ are measurable and the exponential maps $x\mapsto e^{tx}$ and $y\mapsto e^{ty}$ are Borel measurable. Since independence is preserved under measurable transformations, $U_t$ and $V_t$ are independent. A precise way to express this is through the laws. Let $\mu_X=\mathbb P\circ X^{-1}$ and $\mu_Y=\mathbb P\circ Y^{-1}$. Independence of $X$ and $Y$ means that the joint law of $(X,Y)$ is the product measure $\mu_X\otimes\mu_Y$ on $(\mathbb R^2,\mathcal B(\mathbb R^2))$. Therefore, \begin{align*} M_{X+Y}(t)=\mathbb E[e^{t(X+Y)}]=\int_{\mathbb R^2} e^{t(x+y)}\,d(\mu_X\otimes\mu_Y)(x,y). \end{align*} The integrand is non-negative, so Tonelli's theorem applies without any prior integrability assumption. Since $e^{t(x+y)}=e^{tx}e^{ty}$, Tonelli's theorem for product measures gives \begin{align*} \int_{\mathbb R^2} e^{t(x+y)}\,d(\mu_X\otimes\mu_Y)(x,y)=\left(\int_{\mathbb R} e^{tx}\,d\mu_X(x)\right)\left(\int_{\mathbb R} e^{ty}\,d\mu_Y(y)\right). \end{align*} The two one-dimensional integrals are exactly $M_X(t)$ and $M_Y(t)$. Hence \begin{align*} M_{X+Y}(t)=M_X(t)M_Y(t). \end{align*} This is the only place where independence is used.[/guided]

[step:Take logarithms after verifying finiteness and positivity] For every $t \in (-a,a)$, the hypotheses give $M_X(t)<\infty$ and $M_Y(t)<\infty$. Since $e^{tX}>0$ and $e^{tY}>0$ everywhere, we also have \begin{align*} M_X(t)>0 \end{align*} and \begin{align*} M_Y(t)>0. \end{align*} The factorization from the previous step therefore gives \begin{align*} 0<M_{X+Y}(t)=M_X(t)M_Y(t)<\infty. \end{align*} Thus $K_{X+Y}(t)=\log M_{X+Y}(t)$ is defined for every $t \in (-a,a)$, and \begin{align*} K_{X+Y}(t)=\log(M_X(t)M_Y(t)). \end{align*} Using $\log(ab)=\log a+\log b$ for $a,b>0$, we get \begin{align*} K_{X+Y}(t)=K_X(t)+K_Y(t). \end{align*} [/step]

[step:Differentiate the cumulant generating function identity] By [citetheorem:9544], the functions $M_X$ and $M_Y$ are infinitely differentiable on $(-a,a)$. Since $M_X$ and $M_Y$ are strictly positive on $(-a,a)$ and $\log:(0,\infty)\to\mathbb R$ is infinitely differentiable, the chain rule implies that $K_X=\log\circ M_X$ and $K_Y=\log\circ M_Y$ are infinitely differentiable on $(-a,a)$. The identity \begin{align*} K_{X+Y}(t)=K_X(t)+K_Y(t) \end{align*} then shows that $K_{X+Y}$ is also infinitely differentiable on $(-a,a)$. Let $n\in\mathbb N$. Applying linearity of the $n$th derivative to the preceding identity and evaluating at $t=0$ gives \begin{align*} K_{X+Y}^{(n)}(0)=K_X^{(n)}(0)+K_Y^{(n)}(0). \end{align*} By the definition $\kappa_n(Z)=K_Z^{(n)}(0)$ of the $n$th cumulant, this is exactly \begin{align*} \kappa_n(X+Y)=\kappa_n(X)+\kappa_n(Y). \end{align*} This proves the claimed additivity of cumulants. [/step]