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The adjustment coefficient is a positive root of $\lambda (M_X(r)-1)=cr$ in the classical risk model. When the moment-generating function is finite at $R$, the process $\exp (R[{\sum }_{j=1}^{N_t}{X}_j-ct])$ is a continuous-time martingale. It yields the Lundberg inequality and, under the relevant tilted integrability, the Cramér–Lundberg ruin asymptotic.

If each claim includes an independent expense $A$, replace its payment law by the convolution of independent random variables $X+A$. Its moment-generating function is $M_X(r)M_A(r)$ and its expected value is $\mathbb{E}X+\mathbb{E}A$. Keeping the relative safety loading $\theta$ fixed therefore changes the premium rate as well. The new coefficient solves $M_X(R)M_A(R)-1=(1+\theta )(\mathbb{E}X+\mathbb{E}A)R$, within the common finite-transform domain.

An affine process is a time-homogeneous Markov process whose conditional moment-generating function is exponential-affine in its starting state. In discrete time one writes $\log {\mathbb{E}}_x{e}^{\theta X_1}=A(\theta )x+B(\theta )$ on the finite exponential moment domain. Continuous-time definitions allow the functions to depend also on elapsed time.

The Bennett rate function has Taylor expansion $\varphi (u)=u^2/2-u^3/6+O(u^4)$ near zero. It is the optimized exponent in the Chernoff bound based on the moment-generating function estimate for the Bennett inequality.

Suppose independent centered sub-exponential random variables satisfy $\log \mathbb{E}{e}^{\lambda X_i}\le v_i{\lambda }^2/(2(1-b|\lambda |))$ for $|\lambda |<1/b$, and put $v={\sum }_i{v}_i$. Independence adds the log moment-generating functions. The Chernoff bound with $\lambda =t/(v+bt)$ gives the displayed two-sided tail. If each variable has common scale $K$, one may take $v_i\le C{K}^2$ and $b\le CK$, giving $\mathbb{P}(|n^{-1}{\sum }_i{X}_i|>t)\le 2e^{-cn\min \{t^2/K^2,t/K\}}$. This form of Bernstein's inequality allows unbounded variables with factorial moment control, including the centered square of a sub-Gaussian random variable.

For every $\lambda >0$, Markov inequality applied to $e^{\lambda X}$ givesOptimizing over $\lambda$ gives the Chernoff bound in terms of the moment-generating function of $X$.

The law of a sum of independent identically distributed claims with an independent Poisson distribution count of parameter $\lambda$. If the claim moment-generating function is $M_X$, the aggregate transform is $\exp (\lambda (M_X(r)-1))$ wherever finite. It has an atom at zero when the claims are positive. This is the fixed-time law of a Compound Poisson process.

The entropy method combines a coordinatewise certificate with tensorization of entropy to bound the moment-generating function of a certifiable random quantity and hence its tails.

For a random walk with independent and identically distributed random variables as increments and finite moment-generating function at $\theta$, put $\kappa (\theta )=\log \mathbb{E}{e}^{\theta X_1}$. Then $e^{\theta S_n-n\kappa (\theta )}$ is a martingale by conditioning on the next independent increment. In particular, if $r>0$ and $\mathbb{E}{r}^{X_1}=1$, then $r^{S_n}$ is a martingale. Stopping at a finite interval exit can make this process bounded, allowing the dominated convergence theorem to justify passage from bounded stopping times to the exit time.

An exponential moment of a random variable $X$ is an expectation $\mathbb{E}[e^{\theta X}]$ or $\mathbb{E}[e^{\theta |X|}]$ at a specified parameter $\theta$. Its finiteness controls tail decay and the domain of the moment-generating function.

For independent and identically distributed random variables with law $N(-1,1)$, $e^{2S_n}$ is a nonnegative martingale, since the increment's moment-generating function equals one at parameter two. The Doob maximal inequality gives $\mathbb{P}(Z>t)\le e^{-2t}$ for the all-time maximum $Z\ge 0$. Integrating the tail gives the displayed bound. For $b\le 0$, $e^{bZ}\le 1$. The strong law of large numbers guarantees finiteness and attainment of the maximum because $S_n\to -\infty$.

If $\Lambda$ has gamma distribution with shape $2$ and rate $p/q$, $N\mid \Lambda$ has Poisson distribution with intensity $\Lambda$, and the independent claim sizes have exponential distribution of mean $\mu$, where $p+q=1$, the aggregate law is $p^2{\delta }_0+2pq\mathrm{Exp}(p/\mu )+q^2\mathrm{Gamma}(2,\text{rate }p/\mu )$. This follows by expanding its moment-generating function as $[p+qp/(p-\mu t){]}^2$.

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.

For a normal distribution $N(\lambda ,{\sigma }^2)$, the logarithm of its moment-generating function is $s\lambda +s^2{\sigma }^2/2$. Dividing by $s>0$ gives its effective bandwidth. Independent demands add these bandwidths. For total mean $m$ and positive total variance $v$, optimizing a Chernoff bound with target $e^{-\gamma }$, $\gamma >0$, yields the sufficient margin $C\ge m+\sqrt{2\gamma v}$.

For $N$ variables with the standard normal distribution, the expected maximum of their absolute values is at most $\sqrt{2\log (2N)}$, even when they are dependent. Bound the exponential of the maximum by the sum of the $2N$ signed exponentials, use the normal moment-generating function, then apply Jensen inequality and minimize over the exponential parameter. The resulting logarithmic growth is useful in simultaneous estimation.

For a geometric distribution count on the positive integers, independent of independent and identically distributed random variables $X_i$, the moment-generating function of $S={\sum }_{i=1}^N{X}_i$ is the displayed expression. Condition on the count and sum a geometric series. For real $t$, the finite domain requires $M_X(t)<\infty$ and $(1-p)M_X(t)<1$. Positive summands always permit $t\le 0$, even when no positive exponential moment exists; this gives a Laplace transform of a nonnegative random variable. The aggregate has no mass at zero when all summands are strictly positive.

Let the summands have gamma distribution with shape two and rate $\beta$, and let the independent positive-support geometric distribution count have parameter $p$. Put $a=\beta (1-\sqrt{1-p})$ and $b=\beta (1+\sqrt{1-p})$. The aggregate moment-generating function factors as $ab/((a-t)(b-t))$, so it is the sum of two independent exponential distributions of rates $a,b$. Its probability density function is $ab(e^{-as}-e^{-bs})/(b-a)$ for $s>0$, nonnegative and integrating to one. This contrasts with the geometric sum of exponential variables, which is itself exponential.

A finite mixture distribution of exponential distributions with rates $r_i>0$ and nonnegative weights summing to one. Its expected value is ${\sum }_i{w}_i/r_i$ and its moment-generating function is ${\sum }_i{w}_i{r}_i/(r_i-t)$ on the common finite domain. The mixture chooses one component for each observation; it is not a sum of independent exponential variables.

When the adjustment coefficient lies inside the finite domain of the claim moment-generating function, the tilted kernel has finite mean $(\lambda M^{\prime }(R)-c)/(cR)$. The forcing integral is $(c-\lambda \mu )/(cR)$. Their ratio is the constant $C$ in $\psi (u)\sim C{e}^{-Ru}$, by the key renewal theorem. An extra exponential moment controls the forcing tails and proves direct Riemann integrability.

For $C(k)=\mathbb{E}(e^{\xi }-e^k{)}^{+}$ and $\mathrm{Re}z>1$ in the exponential moment domain, the moment-generating function satisfies $\mathbb{E}{e}^{z\xi }={\int }_{\mathbb{R}}C(k)z(z-1)e^{(z-1)k}dk$.