A Lévy-driven rainfall model with applications to futures pricing

Abstract

We propose a parsimonious stochastic model for characterising the distributional and temporal properties of rainfall. The model is based on an integrated Ornstein–Uhlenbeck process driven by the Hougaard Lévy process. We derive properties of this process and propose an extended model which generalises the Ornstein–Uhlenbeck process to the class of continuous-time ARMA processes. The model is illustrated by fitting it to empirical rainfall data on both daily and hourly time scales. It is shown that the model is sufficiently flexible to capture important features of the rainfall process across locations and time scales. Finally, we study an application to the pricing of rainfall derivatives which introduces the market price of risk via the Esscher transform. We first give a result specifying the risk-neutral expectation of a general moving average process. Then we illustrate the pricing method by calculating futures prices based on empirical daily rainfall data, where the rainfall process is specified by our model.

Appendix

In the following we present the proofs of our theoretical results.

First we quote a result (Cont and Tankov 2004, Lemma 15.1) which will be used repeatedly in the following:

Lemma 1

Let $f: [0,T]\rightarrow \mathbb {R}$ be a left-continuous function and $L(t)$ a Lévy process. Then

$$\begin{aligned} E\left[ \exp \left\{ \int _0^t if(s) \mathrm{d}L(s) \right\} \right] =\exp \left\{ \int _0^t \psi (f(s)) \mathrm{d}s, \right\} , \end{aligned}$$

where $\psi (t)$ is the characteristic exponent of $L$, given by

$$\begin{aligned} \exp \{\psi (u)\}=E\left[ e^{iuL(1)}\right] . \end{aligned}$$

Characteristic function of ${\varvec{\Delta Y}}$ The characteristic function of $\Delta Y$ is given by

$$\begin{aligned} \varphi _{\Delta Y}=\exp \left\{ \int _0^{\infty }\psi (ug_1(s))\, \mathrm{d}s+\int _0^1 \psi (ug_2(s)) \, \mathrm{d}s\right\} , \end{aligned}$$

where

$$\begin{aligned} g_1(s)&=\sum _{k=1}^p \frac{e^{-\lambda _k s}-e^{-\lambda _k(1+s)}}{\lambda _k}, \\ g_2(s)&=\sum _{k=1}^p \frac{1-e^{-\lambda _k(1-s)}}{\lambda _k}. \end{aligned}$$

This follows immediately from applying Lemma 1 to the expression given in (12) and noting that because $\Delta Y$ is stationary we can set $t_{i-1}=t_0=0$, causing the second integral to vanish.

Proof of Proposition 1 By construction of $f_\delta$, we have that

$$\begin{aligned} f(x)=\frac{1}{2 \pi } \int _{\mathbb {R}} \hat{f}_\delta (\xi ) e^{(\delta +i\xi )x} \, \mathrm{d}\xi , \end{aligned}$$

and hence by the Fubini theorem, it follows that

$$\begin{aligned} E_Q[\left. f(Y(\tau _2)-Y(\tau _1)) \right| \mathcal {F}_t]=\frac{1}{2 \pi } \int _{\mathbb {R}} \hat{f}_\delta (\xi ) E_Q\left[ \left. e^{(\delta +i\xi )(Y(\tau _2)-Y(\tau _1))} \right| \mathcal {F}_t\right] \, \mathrm{d}\xi , \end{aligned}$$

similar to the proof of Proposition 8.4 in Benth and Šaltytė Benth (2013).

We now calculate the expectation involving the integrated moving average process $Y$. To this end, we first split the integrals in the expression for $Y$ as follows:

$$\begin{aligned}&E_Q \left. \left[ \exp \left\{ (\delta +i \xi )(Y(\tau _2)-Y(\tau _1)) \right\} \right| \, \mathcal {F}_t \right] \nonumber \\&\quad = \exp \left\{ (\delta +i \xi )\left( A(0,\tau _2)-A(0,\tau _1) + \int _0^t \left[ g(\tau _2,v)-g(\tau _1,v) \right] \, \mathrm{d}L(v) \right) \right\} \nonumber \\&\qquad \times \underbrace{E_Q \left[ \left. \exp \left\{ (\delta +i \xi ) \int _t^{\tau _1}\left[ g(\tau _2,v)-g(\tau _1,v) \right] \, \mathrm{d}L(v) \right\} \right| \mathcal {F}_t \right] }_{\mathbf {(A)}} \nonumber \\&\qquad \times \underbrace{E_Q \left[ \left. \exp \left\{ (\delta +i \xi ) \int _{\tau _1}^{\tau _2} g(\tau _2,v) \, \mathrm{d}L(v) \right\} \right| \mathcal {F}_t \right] }_{\mathbf {(B)}}. \end{aligned}$$

(20)

By the abstract Bayes formula (see, e.g. Øksendal 2003), for the measure $Q$ such that $\mathrm{d}Q/\mathrm{d}P|_{\mathcal {F}_t}=Z(t)$, with $X$ being $\mathcal {F}_\tau$-measurable and $t<\tau$, we have that

$$\begin{aligned} E_Q(X|\mathcal {F}_t)=E\left( \left. X \frac{Z(\tau )}{Z(t)} \right| \mathcal {F}_t \right) . \end{aligned}$$

Recall that we are working with the Esscher transform, so we have

$$\begin{aligned} Z(t)=\frac{\exp \left\{ \int _0^t \theta (v) \, \mathrm{d}L(v) \right\} }{E\left[ \exp \left\{ \int _0^t \theta (v) \, \mathrm{d}L(v) \right\} \right] }. \end{aligned}$$

Applying the Esscher transform then gives

$$\begin{aligned} \mathbf {(A)} = \,&E \left[ \left. \exp \left\{ \int _t^{\tau _1} (\delta +i \xi ) \left[ g(\tau _2,v)-g(\tau _1,v) \right] \, \mathrm{d}L(v) \right\} \frac{Z(\tau _1)}{Z(t)}\right| \mathcal {F}_t \right] \\ =&\,E \left[ \exp \left\{ \int _t^{\tau _1} \left( (\delta +i \xi ) \left[ g(\tau _2,v)-g(\tau _1,v) \right] +\theta (v) \right) \, \mathrm{d}L(v) \right\} \right] \\&\times \frac{E\left[ \exp \left\{ \int _0^t \theta (v) \, \mathrm{d}L(v) \right\} \right] }{E \left[ \exp \left\{ \int _0^{\tau _1} \theta (v) \, \mathrm{d}L(v) \right\} \right] }, \end{aligned}$$

where we get an unconditional expectation due to the independent increments of $L$.

We can extend Lemma 1 to complex-valued functions to get

$$\begin{aligned} E\left[ \exp \left\{ \int _0^t (a(v)+ib(v)) \, \mathrm{d}L(v) \right\} \right] =\exp \left\{ \int _0^t \psi _1(-ia(v)+b(v)) \, \mathrm{d}v \right\} , \end{aligned}$$

(21)

where the term on the RHS equals

$$\begin{aligned} \exp \left\{ \int _0^t \ln E \left( \exp \{a(v)+ib(v)\}L(1) \right) \, \mathrm{d}v \right\} , \end{aligned}$$

and we have that

$$\begin{aligned} \left| E \left( e^{(a(v)+ib(v))L(1)} \right) \right| \le E\left( e^{a(v)L(1)}\right) , \end{aligned}$$

and so if $\sup _v a(v)<k$, then the last term is bounded by the exponential moment condition given in (17).

Applying (21) to the terms in $\mathbf {(A)}$ gives

$$\begin{aligned} \mathbf {(A)}&= \exp \Big \{\int _t^{\tau _1} \psi \Big (\xi \Big [g(\tau _2,v)-g(\tau _1,v) \Big ] -i\Big [\theta (v)\\&+\delta \Big (g(\tau _2,v)-g(\tau _1,v) \Big ) \Big ] \Big ) - \psi \Big (-i\theta (v) \Big ) \, \mathrm{d}v\Big \}, \end{aligned}$$

where the requirement $\sup _{v \in [t,\tau _2]} \left( |\theta (v)|+\delta |g(\tau _2,v)-g(\tau _1,v)| \right) <k$ ensures that the terms in the above equation are well-defined.

Now we consider the Lévy–Khintchine formula for subordinators, which takes the form

$$\begin{aligned} E\left( e^{iuL(t)}\right) =\exp \left\{ t \psi (u) \right\} =\exp \left\{ t \int _{\mathbb {R}_+} \left( e^{iuy}-1\right) \nu (\mathrm{d}y) \right\} , \end{aligned}$$

where $\nu (\cdot )$ is the Lévy measure associated with $L$. We can analytically continue this formula to complex arguments (Applebaum 2009, p. 338), and so we get that

$$\begin{aligned} \mathbf {(A)}&= \exp \left\{ \int _t^{\tau _1} \int _{\mathbb {R}_+} e^{\theta (v)y}\left( e^{(\delta +i \xi ) \left[ g(\tau _2,v)-g(\tau _1,v) \right] y}-1\right) \nu (\mathrm{d}y) \, \mathrm{d}v \right\} . \end{aligned}$$

We note that this expression can also be written as

$$\begin{aligned} E\left[ \exp \left\{ \int _t^{\tau _1} (\delta +i \xi ) \left[ g(\tau _2,v)-g(\tau _1,v) \right] \, \mathrm{d}L_Q(v) \right\} \right] , \end{aligned}$$

where $L_Q(v)$ is now a non-stationary stochastic process with jump measure depending on time, namely

$$\begin{aligned} \nu _\theta (\mathrm{d}v,\mathrm{d}y)=e^{\theta (v)y} \nu (\mathrm{d}y) \mathrm{d}v. \end{aligned}$$

Thus we see that conditioning with respect to the measure $Q$ has the effect of exponentially tilting the jump measure of $L$ at time $v$ according to $\theta (v)$, so the jumps of $L$ at times $v$ will be weighted more or less in the expectation depending on the sign of $\theta (v)$.

Now defining

$$\begin{aligned} \psi _\theta (v,\gamma c(\cdot )):=\int _{\mathbb {R}_+} e^{\theta (v)y} \left( e^{\gamma c(v)y}-1\right) \nu (\mathrm{d}y), \end{aligned}$$

(22)

we get that

$$\begin{aligned} \mathbf {(A)}=\exp \left\{ \int _t^{\tau _1} \psi _\theta \Big (v,(\delta +i \xi )\left[ g(\tau _2,v)-g(\tau _1,v) \right] \Big ) \, \mathrm{d}v \right\} ; \end{aligned}$$

and by similar arguments

$$\begin{aligned} \mathbf {(B)}=\exp \left\{ \int _{\tau _1}^{\tau _2} \psi _\theta \Big (v,(\delta +i \xi )g(\tau _2,v) \Big ) \, \mathrm{d}v \right\} . \end{aligned}$$

Substituting these expressions into (20) then gives the result.

Hougaard process The Hougaard process has Lévy measure given by (Grigelionis 2011)

$$\begin{aligned} \nu (\mathrm{d}y)= \left( \rho ^{\frac{1}{\kappa -1}} \Gamma \left( \frac{\kappa }{\kappa -1} \right) (\kappa -1)^{\kappa /(\kappa -1)} \right) ^{-1} y^{\frac{3-2\kappa }{\kappa -1}} \exp \left\{ - \frac{\mu ^{1-\kappa }}{\rho (\kappa -1)} y \right\} \mathrm{d}y, \end{aligned}$$

in terms of the Tweedie parameterisation. We also have that when $L$ is the Hougaard process, the function $\psi _\theta$ defined in (22) takes the form

$$\begin{aligned} \psi _\theta (v,\gamma c(v))=\frac{\mu ^{2-\kappa }}{\rho (2-\kappa )} \left[ \left( 1-\frac{\rho (\kappa -1)\left( i \gamma c(v)+\theta (s)\right) }{\mu ^{1-\kappa }} \right) ^{(\kappa -2)/(\kappa -1)}-1 \right] . \end{aligned}$$