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An autoregressive process of order one with independent and identically distributed random variables as innovations is an affine process when the innovation exponential moments are finite. Its one-step coefficients are $A(\theta )=a\theta$ and $B(\theta )=b\theta +\psi (\theta )$, where $\psi$ is the innovation cumulant-generating function.

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 elapsed-time $h>0$ transition distribution of a driftless square-root diffusion starting at $x$ is the compound Poisson distributionwhere all variables on the right are independent and the empty sum is zero. Here the exponential distribution parameter is its rate. The Laplace transform of this sum iswhich is the Laplace transform of the driftless square-root diffusion; uniqueness of the Laplace transform proves the distributional identity. In particular, for $0\le v<2/h$, the exponential moment isFor $x>0$ this exponential moment is infinite when $v\ge 2/h$, since the event $N=1$ has positive probability and one exponential jump already has an infinite moment there. At $x=0$ the process remains zero and every such moment is one.

The Mellin transform of call prices can be inverted along a vertical contour with real part $x_0>1$: $C(k)=\frac{1}{2\pi i}{\int }_{x_0-i\infty }^{x_0+i\infty }\mathbb{E}{e}^{z\xi }/(z(z-1)e^{(z-1)k})dz$. A finite exponential moment at $x_0$ and the quadratic denominator give absolute convergence.

Under a risk-neutral measure $Q$, let the short rate be $r_t=g(t)+\sigma W_t^Q$, with deterministic locally integrable $g$ and constant $\sigma >0$. For a zero-coupon bond paying one at $T$, the risk-neutral pricing value isIndeed, conditional on ${\mathcal{F}}_t$, the random part of the integrated future short rate is $\sigma {\int }_t^T(T-u)d{W}_u^Q$, which is Gaussian with mean zero and variance ${\sigma }^2(T-t{)}^3/3$. Its exponential moment gives the formula. At points where $g$ is continuous, the instantaneous forward rate isFor merely locally integrable $g$, the same identity holds almost everywhere. The model allows negative short rates and gives explicit bond prices and forward curves.

Suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X}<\infty$ for all $\lambda \ge 0$. Under the exponential tilting $d{\mu }^{\lambda }=e^{\lambda x}d\mu /\varphi (\lambda )$, $X$ is an integrable random variable whenever $\lambda >0$: its negative exponential tail is bounded, and its positive contribution is dominated by a larger positive exponential moment. Differentiation under the integral sign gives the displayed mean. At the endpoint $\lambda =0$, the positive part of $X$ is integrable, but the negative part need not be. The right derivative ${\varphi }_{+}^{\prime }(0)=\mathbb{E}X$ is valid as an extended value, possibly $-\infty$; a finite mean at zero requires an extra negative-tail integrability assumption.

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

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.

For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

An interest rate model must describe a whole term structure of interest rates, not just one future scalar rate. Let $P(t,T)$ be the time-$t$ price of a unit zero-coupon bond maturing at $T$. For a sufficiently regular maturity curve, define the instantaneous forward rate and short rate bySince $P(t,t)=1$, integration givesHere $B$ is the continuous-time bank account. Thus modelling the two-parameter field $f(t,T)$ determines both the bonds and the discounting account. A Gaussian random field means that every finite collection of its values is jointly Gaussian; merely making each individual forward rate normally distributed is insufficient.

A useful arbitrage-free construction works under a money-market risk-neutral measure $Q$. Take independent Brownian motions $W^j$, deterministic coefficient sequences ${\sigma }_j(t,T)$, and a deterministic initial forward curve $f(0,T)$. With the coefficients viewed as vectors in a separable Hilbert space $H$, writeFor infinitely many factors, the last term means the $L^2$ limit of ${\sum }_j{\sigma }_j(t,T)d{W}_t^j$, not an assumption that the infinite vector of Brownian motions is itself $H$-valued. Require ${\int }_0^t\Vert \sigma (s,T){\Vert }_H^{2}ds<\infty$ and the maturity/time regularity needed for stochastic integration and exchange of integrals. Deterministic drift and volatility make this a Gaussian forward-rate field.

The field covariance is determined byFor every finite set of maturities, $(c_t(T_i,T_j))$ must be positive semidefinite. A one-factor model has rank-one instantaneous covariance, while additional or infinitely many factors permit richer correlations along the yield curve. Equivalently one can drive the curve with a maturity-correlated Brownian field $Z_t(T)$ satisfying $d⟨Z(T),Z(U){⟩}_t=c(T,U)dt$. A Brownian sheet gives the exampleMultiplication by deterministic maturity-dependent amplitudes, or deterministic increasing changes of the two sheet coordinates, gives further Gaussian covariance structures. A covariance specification alone is not sufficient: the forward drift must also obey absence of arbitrage.

To derive that drift restriction, putDifferentiate the moving lower endpoint in $\log P(t,T)=-{\int }_t^{T}f(t,u)du$ to obtainThe Itô formula therefore givesDiscounted bond prices must be martingales under $Q$. Their drift vanishes precisely when ${\int }_t^T\alpha (t,u)du=\frac{1}{2}\Vert \Sigma (t,T){\Vert }_H^2$. Differentiating in maturity gives the Heath-Jarrow-Morton model restrictionFor scalar amplitudes $v(t,T)$ multiplying a field with covariance kernel $c(T,U)$, this becomes $\alpha (t,T)=v(t,T){\int }_t^{T}v(t,u)c(T,u)du$. The extra maturity correlation factor cannot in general be dropped.

The complete Gaussian specification is consequentlyIts mean and covariance follow immediately from deterministic integration and the Itô isometry. The initial yield curve is an input, so fitting initial bond prices requires choosing $f(0,T)$ rather than changing a few scalar parameters. With the displayed drift, the discounted bond obeysOn finite horizons, deterministic square-integrable bond volatility gives an expectation-one stochastic exponential, hence a true discounted bond price martingale. This supplies an arbitrage-free pricing model for admissible portfolios in any finite collection of these bonds. Under a physical measure, a market price of risk vector changes the drift: if $d{W}_t^Q=d{W}_t^P+{\lambda }_{t}dt$, then the physical forward drift is $\alpha (t,T)+⟨\sigma (t,T),{\lambda }_t{⟩}_H$. Historical mean changes must therefore not be inserted in place of the risk-neutral drift restriction. Gaussianity under both measures additionally requires suitable deterministic drift changes; state-dependent risk premiums need not preserve Gaussianity under the physical measure.

There is also a useful general consistency test for a proposed Gaussian field beyond this independent-increment construction. DefineIn the filtration generated by a Gaussian field, conditional distributions of such linear functionals are Gaussian. For $s\le t\le T$, the martingale requirement is therefore exactlyThis follows by taking the conditional exponential moment and equating it to $e^{-L_s^T}$. It constrains both the deterministic mean and the covariance-induced conditional mean. Matching only unconditional discounted bond expectations does not suffice. For the Brownian-field construction, the Heath-Jarrow-Morton model restriction realizes this consistency condition through local drift cancellation.

Gaussian structure makes risk-neutral pricing particularly tractable. As a finite-factor example, take the Vasicek model under $Q$:Solving this Ornstein-Uhlenbeck process gives, for $u\ge t$,Hence, with $\tau =T-t$ and $B(\tau )=(1-e^{-a\tau })/a$, the conditional Gaussian integral $I={\int }_t^T{r}_{u}du$ hasThe exponential moment identity for a normal variable now gives the zero-coupon bond price $P(t,T)={\mathbb{E}}_Q[e^{-I}\mid {\mathcal{F}}_t]=A(\tau )e^{-B(\tau )r_t}$, whereThis outlines the derivation of the familiar exponential-affine form without assuming a deterministic future discount rate. More general forward-field models need not reduce to a finite-dimensional Markov short-rate process.

For another application, a call at time $T$ on a unit maturity-$U$ bond, with $t<T<U$ and strike $K>0$, is priced under the T-forward measure, using $P(t,T)$ as numéraire. The bond-price ratio $F_s=P(s,U)/P(s,T)$ is a martingale under that measure and has deterministic volatility $-(\Sigma (s,U)-\Sigma (s,T))$. It is consequently lognormal, with integrated log-varianceIntegrating the lognormal payoff gives the Gaussian bond-option formulaAt $V=0$ use $(P(t,U)-KP(t,T){)}^{+}$. The same idea prices other suitable Gaussian-rate derivatives through forward measures and normal exponential moments.

The attractions are exact initial-curve fitting, explicit covariance control across maturities, and tractable bond and option prices. The limitations are also structural: Gaussian short rates and instantaneous forward rates generally can be negative; finite-factor specifications restrict maturity correlations; and deterministic volatilities cannot reproduce arbitrary stochastic changes in rate dispersion. An infinite-factor model allows richer covariance but typically leaves risks that cannot be spanned by finitely many traded bonds. Thus Gaussianity is a useful modelling choice, not by itself a guarantee of positivity, market completeness, or absence of arbitrage.

For $\theta +t\in \mathcal{N}$, integrating the exponential tilt gives the moment-generating functionAt an interior natural parameter this is finite in a neighbourhood of $t=0$, so differentiation under the integral sign is justified by the nearby exponential moments. Differentiating its logarithm once and twice at zero givesAlternatively the score function is $Y-\nabla \kappa (\theta )$; its mean zero and its covariance matrix give the same exponential-family derivative identities. For any vector $v$, $v^T{\nabla }^2\kappa (\theta )v={\mathrm{Var}}_{\theta }(v^{T}Y)\ge 0$. In a minimal exponential family this variance is strictly positive whenever $v\ne 0$, proving strict convexity of the cumulant function. For a general exponential family whose sufficient statistic is $T(X)$, these identities describe the mean and covariance matrix of $T(X)$, rather than necessarily those of $X$ itself.

Use the usual Black-Scholes model with $\sigma >0$ and the filtration generated by $W$. SetFor $u>t$, independence and the Gaussian distribution exponential moment formula give ${\mathbb{E}}_P[Z_u/Z_t\mid {\mathcal{F}}_t]=1$. Thus $Z$ is a strictly positive martingale of mean one, defining an equivalent measure $Q$ on every finite horizon by $dQ/dP{|}_{{\mathcal{F}}_t}=Z_t$.

To verify the measure change directly, put ${\hat{W}}_t=W_t+\lambda t$. For any real $a$ and $h=u-t$,The conditional characteristic function is that of $N(0,h)$ and does not depend on past information. Therefore $\hat{W}$ has independent increments with Gaussian distribution and continuous paths, and is a Brownian motion under $Q$. Substituting $dW=d\hat{W}-\lambda dt$ gives $d{S}_t=S_t(rdt+\sigma d{\hat{W}}_t)$. Hence $S_t/B_t$ is a geometric Brownian motion that is a martingale and $Q$ is an equivalent martingale measure.

The form is also forced. By the Brownian martingale representation theorem, any density process that is a martingale has $d{Z}_t=a_{t}d{W}_t$. The change of measure identity makes $Z_t{S}_t/B_t$ a local martingale under $P$. Its drift, calculated by Itô formula, is $(S_t/B_t)[Z_t(\mu -r)+\sigma a_t]dt$, which must vanish. Thus $a_t=-\lambda Z_t$. Solving $d{Z}_t=-\lambda Z_{t}d{W}_t$ with $Z_0=1$ gives exactly the displayed stochastic exponential. Equivalence here is on each finite horizon, as usual for Black-Scholes pricing.

The diffusion coefficient becomes $S\sigma (S)=\sqrt{S}$, so the stock is a driftless square-root diffusion, with zero taken as an absorbing boundary. Substitution of $V(t,S)=e^{A(t)S+B(t)}$ into the pricing equation for a local volatility model givesEquating the constant and linear terms and imposing the terminal payoff givesThus $1/A(t)=(1+t)/2$ and $B(t)=0$, and the price isThis is the exponential claim in a driftless square-root model. At the absorbing boundary its value is $1$, matching the payoff at zero.

The payoff in this part is unbounded, unlike the payoff assumed in part (i), so it is useful to verify that the proposed price is a finite conditional expectation. For $h>0$, starting from stock value $s$, defineThen $F_v+\frac{1}{2}x{F}_{xx}=0$, $F(h,x)=e^{-ux}$, and $0\le F\le 1$. By Itô formula, $F(v,S_v)$ is a bounded local martingale, hence a true martingale. It follows thatThis is also the Laplace transform of a sum of $N$ independent variables with exponential distribution of rate $2/h$, where $N$ has Poisson distribution of mean $2s/h$ and is independent of the summands: its transform is $\exp [(2s/h)((2/h)/(2/h+u)-1)]$. Uniqueness of the Laplace transform therefore gives the compound Poisson transition law of a driftless square-root diffusion. The exponential distribution jump has exponential moment $\mathbb{E}[e^Y]=(2/h)/(2/h-1)$ whenever $h<2$. Summing over the Poisson distribution yieldsFor the remaining horizon $h=1-t\le 1$, this is finite and, by the Markov property, is precisely ${\mathbb{E}}_Q[e^{S_1}\mid {\mathcal{F}}_t]=e^{2S_t/(1+t)}$. Thus the displayed process is indeed the risk-neutral pricing value and a true martingale. The replicating strategy from part (ii) has $h_t^S=[2/(1+t)]V(t,S_t)$ and $h_t^B=[1-2S_t/(1+t)]V(t,S_t)/B_t$.

Let $Q$ denote the given unique equivalent martingale measure. A zero-coupon bond paying one at $T$ has risk-neutral pricing valueWrite $h=T-t$. Split the integrated short rate into its known and future parts:By integrating the future Brownian motion increments in the opposite order, the last integral equals ${\int }_t^T(T-u)d{\hat{W}}_u$. It is independent of ${\mathcal{F}}_t$ and is a Gaussian random variable of mean zero and varianceThe Gaussian exponential moment formula $\mathbb{E}[e^{aZ}]=e^{a^2\mathrm{Var}(Z)/2}$ for centered $Z$ therefore gives the bond price in this Gaussian short-rate model with a deterministic shift:The instantaneous forward rate is the negative maturity derivative of the logarithm of the bond price. ConsequentlyIn particular, $P_t(t)=1$ and $f_t(t)=g(t)+\sigma {\hat{W}}_t=r_t$, as required. The derivative formula is pointwise for continuous $g$ and almost everywhere for locally integrable $g$.

Write $\lambda ={\sum }_i{\lambda }_i$, and suppose first that $\lambda >0$. Let $T_i$ be policy $i$'s aggregate, and $M_i$ the moment-generating function of its claim sizes. The Poisson distribution count has probability generating function $G_i(z)=\exp ({\lambda }_i(z-1))$, so the random-sum transform identity and independence of the policies giveThe expression in the weighted sum is the moment-generating function of a mixture distribution with claim cumulative distribution functionThus the aggregate has a compound Poisson distribution with intensity $\lambda$ and severity law $F_{\ast }$. This identification remains valid without positive exponential moments, using the same calculation for Laplace transforms of nonnegative random variables and their uniqueness. It is the Poisson superposition of insurance portfolios.

For common unit-rate exponential distribution severities, the merged count $K$ is Poisson distributed with mean $\lambda$. Because each severity is strictly positive, $T=0$ exactly when $K=0$, givingFor $k\ge 1$, the sum of $k$ unit-rate exponential distribution claims has Erlang distribution density $x^{k-1}{e}^{-x}/(k-1)!$. This density follows by induction: convolving the $k$-claim density with $e^{-x}$ gives $e^{-x}{\int }_0^x{y}^{k-1}dy/(k-1)!=e^{-x}{x}^k/k!$. Conditional on a positive aggregate, the count weights areTherefore the hurdle decomposition of a positive random sum has positive-component probability density functionEach density integrates to one and the weights sum to one. Hence $F_T(x)=a+(1-a){\tilde{F}}_T(x)$ for $x\ge 0$, where ${\tilde{F}}_T$ is the conditional cumulative distribution function given $T>0$. If $\lambda =0$, the aggregate is identically zero, and the positive component is undefined; the requested $a\in (0,1)$ presupposes a positive total intensity.

Condition first on the count in the random sum of independent claims. For $N=k$, independence of the summands and independence from $N$ giveThe empty product is one when $k=0$. Taking expected values proves the random-sum transform identity:For positive claim sizes this composition is always finite for $u\le 0$, and is an identity of Laplace transforms of nonnegative random variables after writing $u=-s$. For positive $u$, it holds wherever the composed series is finite; no positive exponential moment is implied merely by positivity of the summands.

Let $P(z)$, $F(z)$ and $G(z)$ be the probability generating functions of $N$, one severity, and $S$, respectively. Positivity of the severities gives $F(0)=0$ and $g_0=p_0$; the random-sum transform identity gives $G=P\circ F$. The count recurrence in the Panjer claim-count class impliesUse the chain rule and substitute $F(z)$ to obtainEquating the coefficient of $z^{r-1}$ givesThe factor multiplying $f_j{g}_{r-j}$ after rearrangement is $a(r-j)+(a+b)j=ar+bj$. Therefore the Panjer recursion isThese power-series calculations are valid inside the convergence discs of the probability generating functions, and hence justify the coefficient identities without assumptions about positive exponential moments.

For a Poisson distribution count, $p_n/p_{n-1}=\lambda /n$, so $a=0$ and $b=\lambda$. The aggregate recursion becomesTo obtain a moment recursion, write ${\mu }_j=\mathbb{E}{X}^j$, $m_j=\mathbb{E}{S}^j$, and $m_0=1$. For nonnegative measurable $h$, conditioning on $N=n$ and using exchangeability of the severities giveswhere $X$ on the right is independent of $S$. The second equality uses $n{p}_n=\lambda p_{n-1}$, the Poisson size-bias identity. Choosing $h(s)=s^{k-1}$, applying the binomial theorem, and using independence proves the raw moment recursion for a compound Poisson distribution:For any particular $k$, a finite severity $k$th moment suffices: $({\sum }_{i=1}^n{X}_i{)}^k\le n^{k-1}{\sum }_{i=1}^n{X}_i^k$, and the Poisson distribution has finite moments of every order. No positive moment-generating function domain is required.

The first three iterations giveSubtracting the appropriate powers of the expected value yieldsIn particular, the aggregate variance and third central moment involve the raw second and third severity moments, rather than their centered counterparts.

Assume the innovations are independent of the previous history and have finite exponential moments at every real argument, as required by the finite-valued affine definition. Conditioning on $X_{t-1}=x$ givesHence the autoregressive process of order one is an affine process, withHere and in the remaining parts, an affine process uses a time-homogeneous Markov process: the one-step transform is the same at every time. The displayed definition at time $1$ alone would not determine later transitions of an arbitrary time-inhomogeneous Markov process.