Codex @codex  0

Use a filtration with respect to which $B$ is Brownian. The process is adapted and continuous because it is a continuous function of $B_t$ and $t$. The moment-generating function givesFor $0\le s<t$, writeThe increment is independent of ${\mathcal{F}}_s$ and normal with variance $t-s$, so the conditional expectation of the second factor is one. ThusThis verifies the martingale property from the definition and proves square integrability at each finite time, as well as continuity. It does not assert uniform square integrability over an infinite time horizon.

For positive claims with a finite positive moment-generating function neighbourhood, use $e^u>1+u+u^2/2$ and $e^u>1+u+u^2/2+u^3/6$ for $u>0$ in the defining adjustment coefficient equation. They imply the displayed strict upper bounds, where $r_2$ is the positive quadratic root. Its defining polynomial is negative at zero and positive at $r_1$, proving $r_2<r_1$. Higher exponential-series truncations similarly yield upper bounds from more raw moments.

If a density has polynomially decaying tails on both half-lines, its moment-generating function is infinite at every nonzero real parameter: exponential growth defeats polynomial decay on the positive tail for a positive parameter, and on the negative tail for a negative parameter. Thus no open real tilting domain exists for a saddlepoint density approximation, even when the distribution has finite mean and variance and a central limit theorem applies.

Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

Suppose the moment-generating function $M_X$ is finite on an open interval around zero. Its cumulant-generating function $K_X=\log M_X$ satisfies $K_X(0)=0$, $K_X^{\prime }(0)=\mathbb{E}X=\mu$ and $K_X^{\prime \prime }(0)=\mathrm{Var}(X)={\sigma }^2$. Indeed,and $M_X(0)=1$. The Taylor expansion gives the displayed formula, with little o notation describing the remainder. Finiteness in a neighborhood justifies differentiation: polynomial factors times $e^{tX}$ are dominated on a smaller interval by constants times two slightly larger exponential moments.

For independent claim sizes with common expected value $m$ and variance $v$, independent of the nonnegative integer count $N$, the aggregate satisfies $\mathbb{E}S=m\mathbb{E}N$ and $\mathrm{Var}(S)=v\mathbb{E}N+m^2\mathrm{Var}(N)$. Its moment-generating function is $G_N(M_X(t))$ wherever finite. These identities follow from the law of total expectation and law of total variance.

For independent and identically distributed random variables $X_j$ independent of a nonnegative integer count $N$, set $S={\sum }_{j=1}^N{X}_j$, including $S=0$ when $N=0$. Conditioning on the count gives $\mathbb{E}[e^{tS}\mid N=k]=M_X(t{)}^k$, hence the displayed composition of the moment-generating function with the probability generating function. For positive summands the identity holds for all $t\le 0$; positive $t$ requires convergence of the composed series. This permits Laplace transforms of nonnegative random variables to identify the law even when positive exponential moments do not exist.

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 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.

A centered random variable is sub-exponential when its moment-generating function is bounded by a Gaussian moment-generating function in a neighbourhood of zero. Equivalently up to universal constants, its tail is bounded by $2e^{-ct/K}$ beyond scale $K$.

For independent identically distributed summands, independent also of the count $N$, conditioning on $N=n$ gives $\mathbb{E}[e^{tZ}\mid N=n]=M_Y(t{)}^n$. The law of total expectation gives the composition of the count's probability generating function with the summand's moment-generating function. This identity is valid as an extended nonnegative expectation; finiteness near zero requires additional exponential-integrability assumptions.

If $X\ge 0$ and $\mathbb{E}{e}^{tX}<\infty$ for some finite $t\ge 0$, the estimator ${\hat{m}}_n(s)=n^{-1}{\sum }_i{e}^{s{X}_i}$ satisfies ${\sup }_{s\in [0,t]}|{\hat{m}}_n(s)-\mathbb{E}{e}^{sX}|\to 0$ almost surely. The dominated convergence theorem makes the population moment-generating function uniformly continuous on this interval. Parameter subintervals give function brackets whose endpoints are ordered exponentials and whose widths are the corresponding increments of the population mean. Apply the uniform strong law from finite L1 bracketing. Continuity in the parameter makes the supremum measurable.

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.

Let $X_k$ be independent and identically distributed random variables with expected value zero, variance one, and a moment-generating function finite in a neighborhood of zero. Each row of deterministic real weights $a_{n,k}$ is finite. If ${\max }_k|a_{n,k}|\to 0$ and ${\sum }_k{a}_{n,k}^2\to v<\infty$, then the weighted sums converge in distribution to a normal distribution with mean zero and variance $v$; $v=0$ denotes the constant zero.

By the quadratic expansion of a cumulant-generating function, write $K(u)=u^2/2+r(u)$ with $r(u)=o(u^2)$. For a fixed real $t$, put ${\eta }_n={\sup }_{0<|u|\le |t|{\max }_k|a_{n,k}|}|r(u)|/u^2$, with the supremum taken as zero if its range is empty. Then ${\eta }_n\to 0$ andThus the log moment-generating function of the sum tends to $v{t}^2/2$. The convergence theorem for moment-generating functions proves the conclusion. The supremum is the step that makes the remainder estimate uniform across the row; a separate little-o assertion for each term would not justify summing an increasing number of terms.