[proofplan] Choose a positive number $b<a$ and use the two finite exponential moments $M_X(b)$ and $M_X(-b)$ to build an integrable envelope for $e^{sX}$ when $|s|<b$. This same envelope implies that all polynomial moments are finite, because powers grow more slowly than exponentials. Then choose $r<b$ and use the pointwise exponential [power series](/page/Power%20Series) together with domination by $e^{r|X|}$ to interchange expectation and infinite summation. The result is exactly the Taylor expansion of $M_X$ about $0$ on $(-r,r)$. [/proofplan]

[step:Choose a two-sided exponential envelope]Choose $b \in (0,a)$, and define the non-negative [random variable](/page/Random%20Variable) $Y: \Omega \to [0,\infty]$ by \begin{align*} Y(\omega)=e^{bX(\omega)}+e^{-bX(\omega)}. \end{align*} Since $M_X(b)<\infty$ and $M_X(-b)<\infty$, we have \begin{align*} \mathbb E[Y]=M_X(b)+M_X(-b)<\infty. \end{align*} Hence $Y$ is integrable with respect to $\mathbb P$. If $s \in (-b,b)$ and $x \in \mathbb R$, then \begin{align*} e^{sx}\le e^{bx}+e^{-bx}. \end{align*} Indeed, if $x\ge 0$, then $sx\le bx$, while if $x<0$, then $sx\le -bx$. Therefore, for every $s \in (-b,b)$, \begin{align*} e^{sX}\le Y \end{align*} pointwise on $\Omega$.[/step]

[guided]The hypothesis gives finiteness of the [moment generating function](/page/Moment%20Generating%20Function) on a two-sided interval around $0$. We need to use both sides of the interval, because $X$ may take positive and negative values. Choose $b \in (0,a)$ and define the envelope random variable $Y: \Omega \to [0,\infty]$ by \begin{align*} Y(\omega)=e^{bX(\omega)}+e^{-bX(\omega)}. \end{align*} The two terms are measurable because $X$ is measurable and the exponential function is Borel measurable. By the definition of $M_X$, \begin{align*} \mathbb E[Y]=\mathbb E[e^{bX}]+\mathbb E[e^{-bX}]=M_X(b)+M_X(-b)<\infty. \end{align*} Thus $Y$ is an integrable random variable. Now fix $s \in (-b,b)$ and $x \in \mathbb R$. If $x\ge 0$, then $s<b$ implies $sx\le bx$, so $e^{sx}\le e^{bx}$. If $x<0$, then $s>-b$ implies $sx\le -bx$, so $e^{sx}\le e^{-bx}$. In either case, \begin{align*} e^{sx}\le e^{bx}+e^{-bx}. \end{align*} Substituting $x=X(\omega)$ gives, for every $\omega \in \Omega$, \begin{align*} e^{sX(\omega)}\le Y(\omega). \end{align*} This is the basic domination estimate used throughout the proof.[/guided]

[step:Deduce finiteness of all moments]Let $n \in \mathbb N$. Define the constant \begin{align*} C_n=\sup_{u\ge 0}u^n e^{-bu}. \end{align*} The function $u \mapsto u^n e^{-bu}$ is continuous on $[0,\infty)$ and tends to $0$ as $u\to\infty$, so $C_n<\infty$. For every $x \in \mathbb R$, \begin{align*} |x|^n\le C_n e^{b|x|}. \end{align*} Also, \begin{align*} e^{b|x|}\le e^{bx}+e^{-bx}. \end{align*} Therefore, \begin{align*} |X|^n\le C_nY. \end{align*} Since $Y$ is integrable, $\mathbb E[|X|^n]<\infty$. Hence $\mathbb E[X^n]$ is well-defined and finite for every $n \in \mathbb N$. For $n=0$, $\mathbb E[X^0]=1$ because $X^0=1$ pointwise.[/step]

[guided]We next justify why the coefficients in the Taylor series are meaningful. Fix $n \in \mathbb N$. Define \begin{align*} C_n=\sup_{u\ge 0}u^n e^{-bu}. \end{align*} The function $u \mapsto u^n e^{-bu}$ is continuous on $[0,\infty)$, equals $0$ at $u=0$, and tends to $0$ as $u\to\infty$ because exponential decay dominates polynomial growth. Therefore the supremum is finite, so $C_n<\infty$. For every $x \in \mathbb R$, substitute $u=|x|$ into the definition of $C_n$. This gives \begin{align*} |x|^n e^{-b|x|}\le C_n, \end{align*} and multiplying by $e^{b|x|}>0$ yields \begin{align*} |x|^n\le C_n e^{b|x|}. \end{align*} The two-sided exponential envelope controls $e^{b|x|}$ because, if $x\ge 0$, then $e^{b|x|}=e^{bx}$, while if $x<0$, then $e^{b|x|}=e^{-bx}$. Hence in both cases \begin{align*} e^{b|x|}\le e^{bx}+e^{-bx}. \end{align*} Putting $x=X(\omega)$ gives, for every $\omega \in \Omega$, \begin{align*} |X(\omega)|^n\le C_nY(\omega). \end{align*} Since $Y$ is integrable with respect to $\mathbb P$ and $C_n<\infty$, the random variable $C_nY$ is integrable. Therefore $|X|^n$ is integrable, so \begin{align*} \mathbb E[|X|^n]<\infty. \end{align*} This implies that $\mathbb E[X^n]$ is well-defined and finite for every $n\in\mathbb N$. For $n=0$, the convention $X^0=1$ pointwise gives $\mathbb E[X^0]=1$.[/guided]

[step:Dominate the exponential power series on a smaller interval]Choose \begin{align*} r=\frac{b}{2}. \end{align*} For every $t \in (-r,r)$ and every $x \in \mathbb R$, \begin{align*} e^{|t||x|}\le e^{r|x|}\le e^{b|x|}\le e^{bx}+e^{-bx}. \end{align*} Thus, for every $t \in (-r,r)$, \begin{align*} e^{|t||X|}\le Y \end{align*} pointwise on $\Omega$, and so \begin{align*} \mathbb E[e^{|t||X|}]<\infty. \end{align*}[/step]

[guided]Now we choose the neighbourhood on which the Taylor expansion will be justified. Define \begin{align*} r=\frac{b}{2}. \end{align*} Then $r>0$ and $r<b$. Fix $t\in(-r,r)$ and $x\in\mathbb R$. Since $|t|<r$, we have $|t||x|\le r|x|$, and monotonicity of the exponential function gives \begin{align*} e^{|t||x|}\le e^{r|x|}. \end{align*} Because $r<b$, we also have $r|x|\le b|x|$, hence \begin{align*} e^{r|x|}\le e^{b|x|}. \end{align*} As in the moment estimate, the two-sided exponential expression dominates $e^{b|x|}$: \begin{align*} e^{b|x|}\le e^{bx}+e^{-bx}. \end{align*} Substituting $x=X(\omega)$ gives, for every $\omega\in\Omega$, \begin{align*} e^{|t||X(\omega)|}\le Y(\omega). \end{align*} The right-hand side is integrable with respect to $\mathbb P$, so $e^{|t||X|}$ is integrable and \begin{align*} \mathbb E[e^{|t||X|}]<\infty. \end{align*} This is the domination needed for the finite partial sums of the exponential series.[/guided]

[step:Exchange the exponential series with expectation]Fix $t \in (-r,r)$. For each $N \in \mathbb N$, define the partial-sum random variable $S_N: \Omega \to \mathbb R$ by \begin{align*} S_N(\omega)=\sum_{n=0}^{N}\frac{(tX(\omega))^n}{n!}. \end{align*} The power series expansion of the exponential function gives pointwise convergence \begin{align*} \lim_{N\to\infty}S_N(\omega)=e^{tX(\omega)} \end{align*} for every $\omega \in \Omega$. Moreover, \begin{align*} |S_N(\omega)|\le \sum_{n=0}^{N}\frac{|tX(\omega)|^n}{n!}\le e^{|t||X(\omega)|}\le Y(\omega). \end{align*} Since $Y$ is integrable, the [Dominated Convergence Theorem](/theorems/4) applies to the sequence $(S_N)_{N\in\mathbb N}$ on the [probability space](/page/Probability%20Space) $(\Omega,\mathcal F,\mathbb P)$ and gives \begin{align*} M_X(t)=\mathbb E[e^{tX}]=\lim_{N\to\infty}\mathbb E[S_N]. \end{align*} By linearity of expectation over finite sums, \begin{align*} \mathbb E[S_N]=\sum_{n=0}^{N}\frac{\mathbb E[X^n]}{n!}t^n. \end{align*} Therefore, \begin{align*} M_X(t)=\sum_{n=0}^{\infty}\frac{\mathbb E[X^n]}{n!}t^n. \end{align*}[/step]

[guided]Fix $t \in (-r,r)$. The goal is not merely to write the pointwise identity \begin{align*} e^{tX(\omega)}=\sum_{n=0}^{\infty}\frac{(tX(\omega))^n}{n!}. \end{align*} We must justify taking expectations term by term. Define, for each $N \in \mathbb N$, the measurable partial-sum random variable $S_N: \Omega \to \mathbb R$ by \begin{align*} S_N(\omega)=\sum_{n=0}^{N}\frac{(tX(\omega))^n}{n!}. \end{align*} The usual real power series for the exponential function gives \begin{align*} \lim_{N\to\infty}S_N(\omega)=e^{tX(\omega)} \end{align*} for every $\omega \in \Omega$. To pass the limit through expectation, we need an integrable dominating random variable independent of $N$. Taking absolute values and using the triangle inequality for finite sums, \begin{align*} |S_N(\omega)|\le \sum_{n=0}^{N}\frac{|tX(\omega)|^n}{n!}. \end{align*} The finite sum is bounded above by the full exponential series: \begin{align*} \sum_{n=0}^{N}\frac{|tX(\omega)|^n}{n!}\le e^{|t||X(\omega)|}. \end{align*} Because $|t|<r$, the envelope estimate from the previous step gives \begin{align*} e^{|t||X(\omega)|}\le Y(\omega). \end{align*} Thus \begin{align*} |S_N|\le Y \end{align*} pointwise, and $Y$ is integrable. The [Dominated Convergence Theorem](/theorems/4) therefore applies to the sequence $(S_N)_{N\in\mathbb N}$ on the probability space $(\Omega,\mathcal F,\mathbb P)$: the sequence converges pointwise to $e^{tX}$, and the single integrable random variable $Y$ dominates $|S_N|$ for every $N$. Hence \begin{align*} \mathbb E[e^{tX}]=\lim_{N\to\infty}\mathbb E[S_N]. \end{align*} For each fixed $N$, linearity of expectation gives \begin{align*} \mathbb E[S_N]=\sum_{n=0}^{N}\frac{t^n}{n!}\mathbb E[X^n]. \end{align*} The moments are finite by the earlier step, so every term is well-defined. Taking the limit in $N$ gives \begin{align*} M_X(t)=\mathbb E[e^{tX}]=\sum_{n=0}^{\infty}\frac{\mathbb E[X^n]}{n!}t^n. \end{align*}[/guided]

[step:Conclude the expansion holds on a nontrivial neighbourhood of zero]The number $r=b/2$ is positive because $b>0$. The preceding step proves the desired identity for every $t \in (-r,r)$. Hence there exists $r>0$ such that \begin{align*} M_X(t)=\sum_{n=0}^{\infty}\frac{\mathbb E[X^n]}{n!}t^n \end{align*} for every $t \in (-r,r)$, completing the proof.[/step]

[guided]The construction produced a specific radius, \begin{align*} r=\frac{b}{2}, \end{align*} where $b\in(0,a)$. Since $b>0$, this $r$ is positive. For an arbitrary $t\in(-r,r)$, the previous step proved \begin{align*} M_X(t)=\mathbb E[e^{tX}]=\sum_{n=0}^{\infty}\frac{\mathbb E[X^n]}{n!}t^n. \end{align*} The moment-finiteness step also proved that each coefficient $\mathbb E[X^n]$ is finite. Therefore the moment [generating function](/page/Generating%20Function) agrees with its Taylor series about $0$ throughout the nontrivial neighbourhood $(-r,r)$, which is the asserted conclusion.[/guided]