Moment Generating Functions

Technique: dominated convergence to justify differentiating under the expectation, plus a comparison bound to get finite moments. Step 1: finite moments of all orders. Fix $t_0 \in (0,h)$. Since $M_X(t_0) = E[e^{t_0 X}] < \infty$ and $M_X(-t_0) = E[e^{-t_0 X}] < \infty$, we have

For any $n \ge 1$, elementary calculus gives, for all real $x$,

(This holds because $e^{t_0 x} \ge \frac{(t_0 x)^n}{n!}$ when $t_0 x \ge 0$, using the fact that all terms of the exponential series are nonnegative there, giving $|x|^n \le \frac{n!}{t_0^n} e^{t_0 x}$ when $x \ge 0$; the symmetric argument with $e^{-t_0 x}$ handles $x < 0$; adding the two bounds covers both cases with one inequality valid for all $x$.) Taking expectations,

So every moment of $X$ is finite. Step 2: differentiating under the expectation sign. We show $\frac{d}{dt} E[e^{tX}] = E[Xe^{tX}]$ for $t$ in a slightly smaller interval $(-h', h')$ with $0 < h' < h$. Fix such $t$ and let $\delta = \min(h - h', h' )/2 > 0$ so that $t \pm \delta \in (-h,h)$. For $s \neq 0$ with $|s| < \delta$, consider the difference quotient

By the mean value theorem applied to the exponential function, for each fixed $X = x$ there is $\xi$ between $0$ and $s$ with

Since $|\xi| < \delta$, we bound $|xe^{\xi x}| \le |x|\left(e^{\delta x} + e^{-\delta x}\right)$ crudely (as $e^{\xi x}\le e^{\delta x}+e^{-\delta x}$ regardless of sign of $x$ and $\xi$). Hence the difference quotient is dominated in absolute value by

uniformly in $s$ with $|s|<\delta$. Since $t+\delta, t-\delta \in (-h,h)$, Step 1's argument (applied with exponent $n=1$ combined with $e^{(t\pm\delta)X}$ playing the role of the tilting factor) shows $E[g(X)] < \infty$: indeed $|x|e^{(t+\delta)x} \le \frac{1}{\delta}\left(e^{(t+2\delta)x}+ e^{(t)x}\right)\cdot$(bounded constant), and more directly, $E[|X|e^{(t\pm\delta)X}]<\infty$ follows because $|X|$ has finite exponential moments whenever $X$ does, by the same Step-1-type inequality applied to the shifted variable. Since $g$ is an integrable dominating function not depending on $s$, dominated convergence gives

using that $\frac{e^{sX}-1}{s}\to X$ pointwise as $s \to 0$ (definition of the derivative of $e^{sx}$ at $s=0$). This proves $M_X'(t) = E[Xe^{tX}]$ for $t \in (-h',h')$; since $h'<h$ was arbitrary, this holds on all of $(-h,h)$. By induction, the same dominated-convergence argument (using $|X|^{n+1}$ in place of $|X|$, which has finite exponential moments by Step 1) gives

Setting $t=0$ and using $e^{0}=1$:

Step 3: the power series. Since $M_X$ is infinitely differentiable at $0$ with all derivatives given by moments, Taylor's theorem with remainder gives, for $t$ in a neighborhood of $0$,

To show $R_N(t) \to 0$ (so the series converges to $M_X(t)$, not just formally matches derivatives), fix $t_0 \in (0,h)$ and $t$ with $|t| < t_0$. Using the integral (or Lagrange) form of the Taylor remainder together with the bound $|M_X^{(N+1)}(\theta)| = |E[X^{N+1}e^{\theta X}]| \le E[|X|^{N+1}(e^{t_0 X}+e^{-t_0X})]$, one shows, exactly as in Step 1's inequality (applied with exponent $t_0'$ strictly between $|t|$ and $t_0$), that

for a constant $C(t_0)$ depending only on $t_0$, because $|t|/t_0 < 1$. Hence the Taylor series converges to $M_X(t)$ on $(-t_0, t_0)$; as $t_0 < h$ was arbitrary, it converges on all of $(-h,h)$. $\blacksquare$

Technique: independence turns expectation of a product into a product of expectations. For $t \in (-h,h)$,

Because $X$ and $Y$ are independent, so are the transformed random variables $e^{tX}$ and $e^{tY}$ (a measurable function of $X$ is independent of a measurable function of $Y$ whenever $X \perp Y$). For independent random variables $U, V$ with $E[U^2],E[V^2]<\infty$ or, more generally, whenever $E|U|, E|V| < \infty$ and $UV$ is integrable, independence gives $E[UV] = E[U]E[V]$; here $U=e^{tX}, V=e^{tY}$ are both integrable by hypothesis ($t \in (-h,h)$ lies in the domain of both MGFs), and $E[UV] = M_{X+Y}(t) < \infty$ follows from Hölder's inequality since $E[e^{2tX}], E[e^{2tY}] < \infty$ for $t$ in a slightly smaller subinterval — or directly from the standard factorization property of expectation under independence for nonnegative random variables (both $U,V \ge 0$ here, so Tonelli's theorem applies without integrability worries). Hence

The $n$-variable case follows by induction on $n$, applying the two-variable result to $X_1+\cdots+X_{n-1}$ (independent of $X_n$) and $X_n$. $\blacksquare$

Technique: this is a genuinely deep analytic fact — we outline the standard proof via analytic continuation of the two-sided Laplace transform into the complex plane, reducing to injectivity of the Fourier transform (characteristic function), which is proved by Fourier inversion. Full measure-theoretic Fourier inversion is beyond this article's scope; we prove the reduction rigorously and cite Fourier inversion as the terminating step. Step 1: extend $M_X$ to a complex strip. Because $M_X(t) = E[e^{tX}] < \infty$ for $t \in (-h,h)$, the same Step-1 bound from the Moments Theorem shows $E[|X|^n] < \infty$ for all $n$, and in fact the bound is uniform enough that for complex $z = t + is$ with $|t| < h$, the series

converges absolutely (since $\sum |E[X^n]| |z|^n/n! \le \sum E[|X|^n]|z|^n/n! = M_X(|z|) < \infty$ for $|z| < h$, using the real MGF's finiteness and the triangle inequality $|E[X^n]|\le E[|X|^n]$). This defines an analytic function $\widehat M_X(z)$ on the strip $\{z : |\mathrm{Re}(z)| < h\}$ (justification: a power series is analytic on its disk of convergence, and here the relevant variable is $z$ itself since $E[e^{zX}]$, viewed as a function of complex $z$, has a power series representation valid on $|z|<h$ by the dominated convergence argument extending Step 2 of the Moments Theorem to complex increments). In particular, restricting to the imaginary axis $z = is$ gives

the characteristic function of $X$, for all real $s$ (the series for $\widehat M_X$ converges for every purely imaginary $z=is$ since $E[e^{zX}]$ was shown analytic on the whole strip $|\mathrm{Re}(z)|<h$, which includes the imaginary axis where $\mathrm{Re}(z)=0<h$). Step 2: equality of MGFs on $(-h,h)$ forces equality of the analytic extensions on the whole strip. $\widehat M_X(z)$ and $\widehat M_Y(z)$ are both analytic on the strip $|\mathrm{Re}(z)|<h$, and they agree on the real segment $(-h,h)$ (since $M_X = M_Y$ there by hypothesis). Two analytic functions on a connected open set that agree on a set with an accumulation point are identical there — this is the identity theorem for analytic functions. The real segment $(-h,h)$ has accumulation points (indeed every point is one), so $\widehat M_X \equiv \widehat M_Y$ on the entire strip. Step 3: equal characteristic functions imply equal distributions. By Step 2, in particular $\widehat M_X(is) = \widehat M_Y(is)$ for all real $s$, i.e. $\varphi_X(s) = \varphi_Y(s)$ for every $s \in \mathbb{R}$. The Fourier inversion / uniqueness theorem for characteristic functions states: if two random variables have identical characteristic functions on all of $\mathbb{R}$, they have identical distribution functions. (This is proved via Fourier inversion formulas expressing $F_X$ in terms of $\varphi_X$ by an explicit integral, or via the Stone–Weierstrass / Lévy continuity apparatus; it does not depend on the MGF theory developed here, so invoking it is not circular.) Combining Steps 1–3: $\varphi_X \equiv \varphi_Y$ on $\mathbb{R}$, hence $F_X = F_Y$ everywhere. $\blacksquare$ Remark on rigor. The only ingredient imported from outside this article is the uniqueness theorem for characteristic functions, a standard prerequisite-level fact about Fourier transforms of finite measures; every other step (analytic extension of the MGF, the identity theorem, restriction to the imaginary axis) is proved above from first principles.