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The characteristic function of a real random variable $X$ isIt determines the probability distribution of $X$.

Here $t\in \mathbb{R}$ and $i^2=-1$. By Euler's formula, its real part and imaginary part are $\mathbb{E}[\cos (tX)]$ and $\mathbb{E}[\sin (tX)]$. The complex exponential has absolute value one, so this expected value exists for every real random variable, even without finite absolute moments. Moreover,The bound uses the triangle inequality for a vector-valued integral; the last identity is complex conjugate symmetry.^[ref] Consequently the real part is an even function and the imaginary part is an odd function.

For a discrete probability distribution with values $x_j$ and probabilities $p_j$, the characteristic function is ${\sum }_j{p}_j{e}^{it{x}_j}$: a weighted average of points on the unit circle in the complex plane. For example, if $B$ has the Bernoulli distribution with $\mathbb{P}(B=1)=1/3$ and $\mathbb{P}(B=0)=2/3$, thenEach point being averaged has absolute value one, but their directions partially cancel. At $t=\pi$, the same average is $1/3$; at $t=2\pi$, it returns to one. The characteristic function is thus neither a probability density function nor a probability mass function: its argument is a frequency parameter, and its values can be complex numbers. This Bernoulli distribution has no probability density function with respect to Lebesgue measure, yet its characteristic function is defined at every frequency.

For a probability density function $f$, the same definition becomes ${\phi }_X(t)={\int }_{\mathbb{R}}{e}^{itx}f(x)dx$, the Fourier transform with the displayed sign convention. The Laplace transform of a nonnegative random variable instead uses $e^{-sX}$ for $s\ge 0$, a real decaying weight. For the Bernoulli distribution above, it is $2/3+e^{-s}/3$; neither transform is the probability density function. The uniqueness theorem for characteristic functions explains why the oscillating averages still retain the entire probability law, while the characteristic function of a sum of independent variables turns combining laws into multiplication. These two properties make the representation useful even when directly computing a sum's probability distribution is cumbersome.

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed cumulants when the relevant derivatives exist. The aggregate variance is the count rate times the raw second claim moment, not just the claim variance. Positive moment-generating function arguments need an exponential moment; finite first and second claim moments suffice for the aggregate mean and variance by conditioning or right derivatives at zero of the Laplace transform of a nonnegative random variable.

Let $Y_t={\sum }_{j=1}^{N_t}{X}_j$ for a rate-$\lambda$ Poisson process and independent and identically distributed nonnegative marks independent of that process, with $\varphi (q)=\mathbb{E}{e}^{-q{X}_1}$. The Laplace transform of a nonnegative random variable for the increment over $[s,t]$ is $\exp [\lambda (t-s)(\varphi (q)-1)]$. That increment is independent of the past of $Y$, so the displayed nonnegative process has mean one and satisfies $\mathbb{E}[Z_t\mid \sigma (Y_u:u\le s)]=Z_s$. No finite first moment of the marks is needed for this martingale when $q\ge 0$.

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 $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Conditional on $Z=k$, the summands are independent with the common Laplace transform of a nonnegative random variable $\varphi (q)$. The empty sum for $k=0$ is zero. Therefore the compound Poisson distribution hasand applying the probability generating function from part (a) givesIf $m=\mathbb{E}{X}_1<\infty$, differentiating at zero from the right gives ${\varphi }^{\prime }(0+)=-m$ and $L_Y^{\prime }(0+)=-\lambda m$. HenceThe differentiation is justified by $(1-e^{-q{X}_1})/q\le X_1$ and the dominated convergence theorem. No finite moment was assumed in the question, so the formula must also be interpreted when $m=\infty$. For any nonnegative random variable $V$, the monotone convergence theorem givesApply this to $X_1$ and $Y$, and use $1-e^{-\lambda h}\sim \lambda h$ as $h\downarrow 0$, with $h=1-\varphi (q)$; it yields the same formula with value $+\infty$ when $m=\infty$. If $m=0$, nonnegativity makes all marks zero almost surely and the result is immediate.

For the second method, condition on the count rather than differentiating a transform. For finite $m$, the tower property of conditional expectation gives $\mathbb{E}[Y\mid Z]=Zm$, and $\mathbb{E}Z=\lambda$ gives $\mathbb{E}Y=\lambda m$. More generally, Tonelli's theorem and independence giveFor $m=\infty$, already the term $j=1$ is infinite since $\mathbb{P}(Z\ge 1)>0$. Thus the two methods agree, including in the extended-valued case.

For excess of loss reinsurance with retention $M$, the direct insurer caps its payment at $M$ and the reinsurer pays the excess. ThusLet $I_i=1_{\{X_i>M\}}$ and $\alpha =\mathbb{P}(X_i>M)=1-F_{X_1}(M)\in (0,1)$. These are independent Bernoulli random variables, independent of $N$. Since $N_R={\sum }_{i=1}^N{I}_i$, its conditional law given $N=n$ is a binomial distribution. Conditioning on $N$ therefore gives the Bernoulli thinning identityHere the probability generating function may be evaluated on $0\le z\le 1$, and the identity extends wherever the transforms converge.

The positive reinsurer marks have the conditional law $W\stackrel{d}{=}{X}_1-M\mid X_1>M$. HenceThe independent-mark representation is justified as follows. Conditional on any fixed count and pattern of exceedance indicators, the positive excesses are independent with this same conditional law; that law depends neither on the number of exceedances nor on their locations. Averaging over the indicator pattern thus leaves the positive mark sequence independent of $N_R$. We may extend it by additional independent copies if needed, giving $S_R={\sum }_{j=1}^{N_R}{W}_j$ with the stated independence.

Now write $q=1-p$. For exponential distribution claim sizes of mean $\mu$,Thus $W$ is exponential with the original mean $\mu$, by memorylessness of the exponential distribution. The original zero-based geometric distribution count has $G_N(z)=p/(1-qz)$. ConsequentlyThis is the geometric count under Bernoulli thinning, withFinally use the Laplace transform of a nonnegative random variable, keeping the mass at zero. For $t\ge 0$,The last expression is the transform of a mixture distribution: mass $p_R$ at zero and, with probability $q_R$, an exponential distribution of rate $p_R/\mu$. Uniqueness of the Laplace transform identifies this zero-based geometric sum of exponential variables:Equivalently, $\mathbb{P}(S_R=0)=p_R$ and its positive density is $q_R(p_R/\mu )e^{-p_{R}s/\mu }$. Therefore the requested exceedance probability isThe prefactor is essential: the aggregate is zero whenever no claim crosses the retention.

Condition on $N$. Since the amounts are independent and identically distributed random variables and are independent of $N$, their moment-generating functions multiply:Summing against the positive-support geometric distribution gives the geometric-sum moment-generating functionThe finite-transform domain is $M_X(t)<\infty$ and $(1-p)M_X(t)<1$. The terms are nonnegative for real $t$, so if the latter condition fails the sum diverges. For $t\le 0$ both conditions always hold, since the claims are positive; thus the same calculation always provides the Laplace transform of a nonnegative random variable. Positivity alone does not guarantee a finite moment-generating function on a positive neighbourhood of zero.

There is at least one strictly positive amount in the portfolio. Hence $\mathbb{P}(S=0)=0$; zero carries no atom of a measure to add to the continuous laws in the next two parts.

At the terminal time, $M_T=e^{i\theta X_T}$, so the martingale from part (c) has terminal value $e^{-{\theta }^2⟨X{⟩}_T/2}$. Its initial value is $M_0$, since $X_0=⟨X{⟩}_0=0$. Taking expectations givesHere $M_0$ may be a nonconstant ${\mathcal{F}}_0$-measurable variable; the tower property of conditional expectation still gives $\mathbb{E}{M}_0=\mathbb{E}{e}^{i\theta X_T}$. This characteristic function under conditionally symmetric martingale increments identity relates the characteristic function of the terminal martingale to the Laplace transform of a nonnegative random variable given by its quadratic variation.

The displayed density is that of a gamma distribution with shape $k+1$ and rate one. PutBy the additivity of independent gamma distributions with a common rate,Equivalently, the independence of random variables makes the Laplace transform of a nonnegative random variable of the sum equalTherefore the required probability density function is

The Laplace transform of a nonnegative random variable determines its probability law. Indeed, $U=e^{-X}$ takes values in $(0,1]$, and $\mathbb{E}[U^n]={\psi }_X(n)$. Thus the transform determines every moment of $U$; moment determinacy on a compact interval determines the law of $U$, and the injective inverse $X=-\log U$ determines the law of $X$.

If independent summands have exponential distribution of mean $\mu$ and the independent count has a zero-based geometric distribution of parameter $p$, their random sum of independent claims has an atom $p$ at zero and an exponential positive component of rate $p/\mu$. Its Laplace transform of a nonnegative random variable is $p/[1-(1-p)/(1+\mu s)]=p+(1-p)p/(p+\mu s)$. Thus $\mathbb{P}(S>x)=(1-p)e^{-px/\mu }$ for $x>0$. Unlike the geometric sum of exponential variables with a positive count, this law is not purely exponential.