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Here take $X,Y$ iid, centered with variance one, and with finite moment-generating functions at all real arguments. independence of their sum and difference gives $M(2t)=M(t{)}^{3}M(-t)$. Dividing the positive and negative argument identities shows $M(t)/M(-t)$ satisfies the symmetry step of dyadic rigidity of a moment-generating function, hence $M$ is even. The identity becomes $M(2t)=M(t{)}^4$, and the unit-variance expansion then forces $M(t)=e^{t^2/2}$. The uniqueness theorem for moment-generating functions identifies the standard normal distribution. This is a precise sufficient-hypothesis version; the displayed implication is understood with those stated assumptions.

The moment-generating function of a real random variable $X$ is the exponential transformallowing the value $+\infty$. It always satisfies $M_X(0)=1$. The important regularity condition is finiteness on an open neighbourhood of zero; finiteness only at zero provides no moment information.

Under that neighbourhood condition, differentiating under the expected value gives $M_X^{(k)}(0)=\mathbb{E}{X}^k$ for every nonnegative integer $k$. In particular the mean is $M_X^{\prime }(0)$ and the variance is $M_X^{\prime \prime }(0)-M_X^{\prime }(0{)}^2$. Exponential bounds dominate powers locally and justify the derivatives. For independent random variables, $M_{X+Y}(t)=M_X(t)M_Y(t)$ wherever finite, making addition easier to analyze.

For a Bernoulli random variable with success probability $p$, $M_X(t)=1-p+p{e}^t$. For an exponential distribution of rate $\lambda >0$, $M_X(t)=\lambda /(\lambda -t)$ for $t<\lambda$, and it is infinite for $t\ge \lambda$. A standard Cauchy random variable has infinite $M_X(t)$ at every nonzero real $t$, although its characteristic function $\mathbb{E}{e}^{itX}$ is defined for every real $t$.

Two probability distributions with equal moment-generating functions on a neighbourhood of zero are equal. The neighbourhood condition is what makes this uniqueness theorem stronger than merely listing formal moments.

For the rate-parameterized Gamma distribution with shape $r>0$ and rate $\lambda >0$, integration of its probability density giveswhen $t<\lambda$, by the gamma function integral and a change of variables. The integral is infinite for $t\ge \lambda$. In particular the moment-generating function is finite on an open interval containing zero, so the uniqueness theorem for moment-generating functions applies. Multiplication of these moment-generating functions proves additivity of independent gamma distributions with a common rate.

The independence of $X$ and $Y$ permits factoring the expectation of the product $e^{tX}{e}^{tY}$. The moment-generating function of a binomial distribution from the previous part therefore givesThis is the moment-generating function of $B(n+m,p)$. Both moment-generating functions are finite on a neighbourhood of zero, so the uniqueness theorem for moment-generating functions identifies the probability distribution:The common success probability $p$ and independence are both used in this argument.

Write $K(t)=\log \mathbb{E}[e^{tY}]$ for the cumulant-generating function in the question's $(\varphi ,\lambda )$ parametrization. Multiplying the density by $e^{ty}$ replaces $\varphi$ in the decaying part of its exponent by $\varphi -2t$. For $t<\varphi /2$, the remaining integral is the normalizing integral for an Inverse Gaussian distribution with parameters $(\varphi -2t,\lambda )$. Comparing the two normalizing constants givesIn particular $K^{\prime }(0)=\sqrt{\lambda /\varphi }$, so the mean parameter used in the more usual Inverse Gaussian distribution notation is $\mu =\sqrt{\lambda /\varphi }$; it must not be confused with the present parameter $\varphi$.

First compute the cumulant-generating function of the sample average, as requested. Independence givesThis is exactly the cumulant-generating function of an Inverse Gaussian distribution with parameters $(n\varphi ,n\lambda )$. The uniqueness theorem for moment-generating functions therefore identifies that law. Scaling $S_n=n{\overline{Y}}_n$ and including the density Jacobian $1/n$ givesEquivalently, the inverse Gaussian sum closure is $S_n\sim IG(\varphi ,n^2\lambda )$ in the parametrization of this question. Its mean is $n\sqrt{\lambda /\varphi }$ and its variance is $n\sqrt{\lambda }/{\varphi }^{3/2}$, in agreement with addition of independent means and variances.

For a general continuous iid sample with cumulant-generating function $K$, finite near zero, set $x=s/n$ and choose a real saddle $\hat{t}$ in the interior of the finite moment domain satisfying $K^{\prime }(\hat{t})=x$. With $K^{\prime \prime }(\hat{t})>0$, the saddlepoint density approximation for the sum isUnder the usual smooth nonlattice conditions, at a fixed interior $x$ the true density has this leading expression times $1+O(n^{-1})$. This is a relative density approximation; the leading expression is not automatically an exactly normalized probability density in every model.

One way to understand the statistical saddlepoint approximation is through exponential tilting. The tilted one-observation density is $f_t(y)=e^{ty-K(t)}f(y)$, with mean $K^{\prime }(t)$ and variance $K^{\prime \prime }(t)$. Undoing this tilt for the sum multiplies its tilted density at $s$ by $e^{nK(t)-ts}$. At $t=\hat{t}$, $s$ is exactly the tilted mean of the sum. A local version of the central limit theorem gives tilted density there approximately $[2\pi n{K}^{\prime \prime }(\hat{t}){]}^{-1/2}$. Multiplying these two factors gives the displayed saddlepoint density approximation. Higher local expansion terms produce the stated relative error order.

For the Inverse Gaussian distribution, differentiation givesThus for every $x>0$ the saddle exists, withAt this saddle,Substitution into the saddlepoint density approximation gives precisely the exact density already found for $S_n$. ThereforeThis exact saddlepoint density of an inverse Gaussian sum is stronger than a relative $O(n^{-1})$ claim: the leading formula is exact and needs no renormalization.

For the final distribution, both tails decay polynomially. Any positive exponential tilt makes the positive tail nonintegrable, and any negative tilt makes the negative tail nonintegrable. Its moment-generating function is therefore infinite away from zero, leaving no open interval on which to form the cumulant-generating function and solve a saddle equation. Here polynomial tails prevent exponential tilting, so the conventional saddlepoint density approximation is unavailable. A finite mean and variance, and even an ordinary central limit theorem, do not supply the missing exponential moments.

Let $S=X+Y$ and $D=X-Y$. Since they are independent, applying the product rule for moment-generating functions to $S+D=2X$ givesAll factors are finite and strictly positive for every real $t$ by the stated hypothesis. Applying the same identity to $-t$ and dividing yields $\psi (2t)=\psi (t{)}^2$, orThe centered unit-variance Taylor expansion from part (ii) gives $M(h)=1+h^2/2+o(h^2)$ and the same expansion for $M(-h)$. Consequently $\psi (h)=1+o(h^2)$.

For the first step of dyadic rigidity of a moment-generating function, take logarithms, which are allowed because $\psi >0$. For each fixed $t$ and every integer $m$,Since $\log \psi (h)=o(h^2)$, the right side tends to zero: it is $2^{m}o(t^2/4^m)$. Thus $\psi (t)=1$ for every $t$, and the moment-generating function is even.

The original identity now becomes $M(2t)=M(t{)}^4$. Iterating this time with the correct fourth-power scaling givesNear zero, $\log M(h)=h^2/2+o(h^2)$, so the right side tends to $t^2/2$. ThereforeThis is the moment-generating function of a standard normal variable. By the uniqueness theorem for moment-generating functions, each of $X$ and $Y$ has the standard normal distribution. The proof establishes the Gaussian characterization by independent sum and difference directly from the two functional identities and the first two moments.

For the exponential distribution of mean $\mu$, $M_X(t)=(1-\mu t{)}^{-1}$. Substitution into the geometric-sum moment-generating function givesThis is the moment-generating function of an exponential distribution with rate $p/\mu$. By the uniqueness theorem for moment-generating functions, the geometric sum of exponential variables therefore has probability density functionIts expected value is $\mu /p$, agreeing with $\mathbb{E}[N]\mathbb{E}[X]=\mu /p$ in the random sum of independent claims formula.

For every $\lambda \in {\mathbb{R}}^2$, the scalar ${\lambda }^{T}X$ is normal with mean ${\lambda }^T\mu$ and variance ${\lambda }^{T}V\lambda$. The moment-generating function of a normal distribution therefore givesThus $\mu$ and $V$ determine the moment-generating function on all of ${\mathbb{R}}^2$. By the uniqueness theorem for moment-generating functions, they uniquely determine the probability distribution of $X$.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ where finite. For independent $X,Y$, independence factors the expectation, giving $M_{X+Y}(t)=M_X(t)M_Y(t)$ wherever both factors are finite.

For the given probability density function and $t<1$, put $s=(1-t)x$ to obtainHere $\Gamma (n)=(n-1)!$, obtained by repeated integration by parts. For $t\ge 1$ the defining integral diverges.

Each $Y_i$ has moment-generating function $(1-t{)}^{-1}$, so independence givesThe uniqueness theorem for moment-generating functions applies when the functions are finite on a neighborhood of zero. Hence the sum has the same distribution as $X_n$, a Gamma distribution of shape $n$ and rate one, with the displayed probability density function in the source.

Independence makes the moment-generating function of the sum the product of the two moment-generating functions. For $t<\lambda$,The expectations are finite on, for example, $(-\lambda /2,\lambda /2)$, as is the moment-generating function of the candidate Gamma distribution. The uniqueness theorem for moment-generating functions therefore applies, provingThis is additivity of independent gamma distributions with a common rate; a common rate is essential to this specific formula.

The uniqueness theorem for moment-generating functions states: if two real random variables $V,W$ have finite moment-generating functions on a common open interval $(-\delta ,\delta )$ with $\delta >0$, andthen $\boxed{V\text{ and }W\text{ have the same probability distribution}.}$ Finiteness throughout a neighborhood of zero is a hypothesis; the identity $M_V(0)=M_W(0)=1$ alone carries no information.

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence

It suffices that the moment-generating function be finite on some open interval containing zero:Equivalently, some two-sided exponential moment exists: $\mathbb{E}{e}^{r|Y|}<\infty$ for some $r>0$. For the forward implication, choose $0<r<\delta$ and use $e^{r|Y|}\le e^{rY}+e^{-rY}$; the converse follows from $e^{tY}\le e^{r|Y|}$ for $|t|\le r$.

Under this weaker hypothesis, all moments exist. For $|t|<r$, dominated convergence justifies the Taylor series argument in (i). Vanishing odd moments gives $M_Y(t)=M_{-Y}(t)$ near zero, and the local uniqueness theorem for moment-generating functions gives the symmetric distribution. The reverse implication follows directly from equality of the two probability distributions.

More generally, it suffices that all moments exist and the probability distribution have moment determinacy: $Y$ and $-Y$ then have identical moments and hence identical probability distributions. Existence of all moments by itself does not provide that uniqueness.

The uniqueness theorem for moment-generating functions states that if $M_X$ and $M_Y$ are finite and equal on an open real interval containing zero, thenFor the convergence theorem for moment-generating functions, suppose that the moment-generating functions $M_{X_n}$ are finite on a common open real interval $(-a,a)$, $a>0$, and $M_{X_n}(t)\to M_X(t)$ for every $|t|<a$, where $M_X$ is finite there. ThenOne may equivalently assume convergence to a finite function $M$ on that interval: the theorem then asserts existence of a unique limiting probability distribution with moment-generating function $M$ as well as convergence in distribution.

For $X\sim \mathrm{Exp}(\lambda )$, integration gives $M_X(t)=\lambda /(\lambda -t)$ for $t<\lambda$. For $0<c<1$, the candidate remainder moment-generating function isThis is the moment-generating function of the mixture distributionConstruct it by taking an exponential random variable $E$ of rate $\lambda$ and a Bernoulli random variable $J$ with $\mathbb{P}(J=1)=1-c$, with $E,J,X$ mutually independent, and setting $R_c=JE$. Then $R_c$ is independent of $X$, andon a neighborhood of zero. By the uniqueness theorem for moment-generating functions, $cX+R_c\stackrel{d}{=}X$, so the exponential distribution is a self-decomposable distribution.

Fix $|t|<t_0$. Since $|ct|<t_0$, $0<M_X(ct)<\infty$. Using independence of random variables and the Tonelli theorem for the positive exponential integrands gives the extended-real identityThe left side is finite and the first factor is finite and strictly positive. Consequently $M_R(t)$ is finite andThe same reasoning applies to $R^{\prime }$, so their moment-generating functions are finite and equal throughout $(-t_0,t_0)$. The uniqueness theorem for moment-generating functions yieldsThis proves uniqueness of a self-decomposition remainder while also checking the required finiteness; no independence between $R$ and $R^{\prime }$ is needed.

A probability distribution is self-decomposable if, for every $0<c<1$, an independent remainder $R_c$ can be chosen so that the displayed equality in probability distribution holds. A normal distribution $\mathcal{N}(\mu ,{\sigma }^2)$ admits a normal distribution remainder with mean $(1-c)\mu$ and variance $(1-c^2){\sigma }^2$. For an exponential distribution of rate $\lambda$, choose $R_c=0$ with probability $c$ and otherwise an independent exponential distribution of rate $\lambda$. Its moment-generating function is $c+(1-c)\lambda /(\lambda -t)=(\lambda -ct)/(\lambda -t)$; multiplication by $M_X(ct)$ proves the claim using the uniqueness theorem for moment-generating functions.

Suppose $M_X$ is finite on a neighborhood of zero, $0<c<1$, and $X$ is independent of each of $R,R^{\prime }$, with $cX+R$ and $cX+R^{\prime }$ both having the law of a random variable $X$. independence and the Tonelli theorem first give $M_X(t)=M_X(ct)M_R(t)$ as an extended nonnegative integral. Since $0<M_X(ct)<\infty$ and $M_X(t)<\infty$, both remainder moment-generating functions are finite and equal to the displayed ratio on the same neighborhood. The uniqueness theorem for moment-generating functions gives equality of their probability distributions. No independence between $R$ and $R^{\prime }$ is required.