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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.

Let $X_f$ have the specified probability density function $f_m$. Splitting the integral into the positive and negative half-lines, and using the Gamma integral ${\int }_0^{\infty }{x}^r{e}^{-sx}dx=r!/s^{r+1}$ for a nonnegative integer $r$ and $s>0$, givesAt $t=2$ or $t=-2$, one term becomes the divergent integral of $x^m$; beyond these endpoints it grows exponentially. Consequently, the full extended-valued moment-generating function isIn particular $M_f(0)=1$, checking normalization of a probability distribution. All absolute moments exist. Direct integration yieldsThe symmetric distribution makes every odd moment zero, so

Now let $X_g$ have probability density function $g_m$. Its moment-generating function would beFor every $t\ne 0$, one summand has exponential growth, which dominates the polynomial decay. HenceThere is no finite moment-generating function on any neighborhood of zero, but some moments still exist. Specifically,ThusFor even $k\ge m+1$, the moment is $+\infty$, so it is not a finite moment. For odd $k\ge m+1$, both the positive and negative contributions have infinite integral, so the expected value is undefined. Cancellation in a Cauchy principal value does not define an expected value. In particular, when $m=0$ there are no finite positive-order moments.

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.

The uniform distribution has probability density function $1/2$ on $(-1,1)$. Therefore, for $t\ne 0$,The value at zero follows either directly from the definition of the moment-generating function or by continuity. This answer needs no assumption about $B$.

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.

First take integers $1\le k<n$. For independent copies $X_1,\dots ,X_n$ of $X$, defineThe two blocks involve disjoint families of independent random variables, so $A$ and $B$ are independent. The assumed strictly stable distribution gives $A\stackrel{d}{=}X$, $B\stackrel{d}{=}X$, andTherefore, on any fixed probability space carrying $X$ and an independent copy $X^{\prime }$, their joint probability distribution givesNow fix an arbitrary $0<c<1$ and choose rationals $r_m=k_m/n_m\in (0,1)$ tending to $c^{\alpha }$. Continuity of the real powers gives, on that fixed probability space,The left sides all have the probability distribution of $X$. Since almost sure convergence implies convergence in distribution, their limit has that same probability distribution. HenceThis proves strict stability implies self-decomposability. The limit argument uses no moment-generating function or finite moments, which is essential for general stable distributions.

The moment-generating function is $M_X(t)=\mathbb{E}{e}^{tX}$ on the real values where this expectation is finite, normally including a neighborhood of zero. Direct substitution giveswhenever the right side is finite.

For the given Gamma distribution and $t<\lambda$,using the Gamma function after the scaling $u=(\lambda -t)x$. At $t\ge \lambda$ the positive integral diverges.

independence multiplies moment-generating functions. ThusUniqueness of a moment-generating function on a neighborhood of zero identifieswhere $\lambda$ is the rate parameter, as in the printed probability density function.

For an exponential distribution with rate $\lambda >0$, direct integration gives the moment-generating functionIt is finite in a neighborhood of zero, so its derivatives there give moments. Differentiating three times gives $M_X^{(3)}(t)=6\lambda /(\lambda -t{)}^4$, hence

Take $Y=X_1+X_2$. Independence makes the joint exponential expectation factor:This is the product rule for the moment-generating function of an independent sum.

Take $Y=a{X}_1+b$. From the definition of a moment-generating function, $\mathbb{E}{e}^{tY}=e^{bt}\mathbb{E}{e}^{at{X}_1}=e^{bt}{M}_1(at)$, for $t$ where the expectation is finite.

Take $Y=U{X}_1$, where $U$ is uniform on $(0,1)$ and independent of $X_1$. Conditioning on $U$ givesFor $t$ near zero this is finite because the moment-generating function of $X_1$ is finite on a neighborhood of zero; Tonelli's theorem justifies the conditioning integral for the nonnegative exponential.

Recognize the mixture distributionIt is a half-and-half mixture of exponential distributions with rates one and two. Uniqueness of a moment-generating function finite near zero identifies the law. Therefore

A characteristic function is useful for proving convergence in distribution, including for laws without a finite moment-generating function: the continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution to its law. For example, the Cauchy distribution has a characteristic function although its moment-generating function is not finite on any neighbourhood of zero.

A moment-generating function is especially useful for exponential tail bounds: for $t>0$ where it is finite, Markov's inequality givesOptimizing over $t$ yields a Chernoff bound. The positive real exponential in this argument permits the inequality; a unit-modulus complex exponential does not give the corresponding tail bound.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ for real $t$ where this expectation is finite. Usually one requires finiteness on a neighbourhood of zero when using it to characterize a distribution. The characteristic function isIt always exists because the integrand has absolute value one. In both definitions the expectation is taken with respect to the distribution of $X$.

At each index the two independent uniform symbols match with probability $p=k/k^2=1/k$. The match indicators are independent across indices. Consequently the waiting times between successive matches are independent geometric random variables on $1,2,\dots$ with success probability $p$, and $S_m$ is their sum.

For one waiting time $G$, summing the geometric series gives $\mathbb{E}{e}^{tG}=p{e}^t/[1-(1-p)e^t]$. Independence yieldsFor $k>1$ the moment-generating function is finite for $t<\log (k/(k-1))$. When $k=1$, every index matches, $S_m=m$ deterministically and the formula reduces to $e^{mt}$ for every real $t$.

The probabilistic form of Wick's theorem is Isserlis's theorem. If $(Y_1,\dots ,Y_k)$ is a centered random vector with a multivariate Gaussian distribution and covariance matrix $C$, its odd product moments vanish, and, for even $k$,Here ${\mathcal{P}}_2(k)$ is the set of perfect matchings of the $k$ labelled positions. The formula allows repeated coordinates and singular covariance matrices. To see the pairing rule, differentiate the moment-generating function $\exp (t^{T}Ct/2)$ once in each coordinate and set $t=0$: only terms pairing all differentiated coordinates survive, each with the displayed product of covariances.

Take $Y_a=M_{i_a{i}_{a+1}}$, with $i_{k+1}=i_1$. These variables are linear combinations of the original independent normal random variables, hence are jointly Gaussian. They have zero means, soSubstituting the entry covariance from part (a)(i) identifies each parenthesis with $\mathbb{E}[M_{i_a{i}_{a+1}}{M}_{i_b{i}_{b+1}}]$, giving the requested identity.

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore

The moment-generating function of a standard normal distribution gives $\mathbb{E}[e^{t{X}_j}]=e^{t^2/2}$. Using independence of the coordinates,The accepted draws are independent and identically distributed random variables, so their sample mean is an unbiased estimator and has varianceThis counts $n$ accepted draws, requiring $n{\sigma }^d$ proposals in expectation.