Conditional Moment Matching and Stratified Approximation for Pricing and Hedging Periodic-Premium Variable Annuities

Abstract

This paper extends the stratified approximation method using lognormal and gamma distributions - first introduced to price Asian options - to derive a close formula for pricing and hedging of periodic-premium variable annuities. We used the moment matching method to fit the lognormal and gamma distributions to the conditional distribution of the integral of the underlying asset on a time interval, given the terminal value of the underlying asset. The highly oscillating double integrals for computing an expectation about the integral of the underlying assets are simplified down to a single integral, which greatly reduces the computation time for pricing periodic-premium variable annuities. This method allowed us to construct a different delta hedging strategy, other than the one used in the existing literature for embedded option of periodic-premium variable annuities. Compared with the existing research on pricing periodic-premium variable annuities, we obtained more accurate results using the stratified approximation method than the numerical method of partial differential equations, and found that the underpricing problem with periodic-premium variable annuities is even more severe than previously stated in existing literature. We further investigated the price gap between single-premium and periodic-premium variable annuities in a variety of settings, and examined the impact that the model and product parameters had on the price gap. The robustness and accuracy of the proposed method is tested by numerical examples.

Appendices

Appendix 1

Proof of Proposition 3.1

Apply the approximated Gamma density function (3.13) into (3.7), we have

$$\begin{aligned} \begin{aligned} \Pi (u , t)&= \textrm{e}^{- r(t-u)} \int _0^\infty k \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \left[ \frac{1}{k} \left( G(t) - F_u z \right) - \Lambda _{t-u}' \right] \textrm{d} \mathbb {P}\left( \Lambda _{t-u}' \in x \big \vert S_{t-u}' = z \right) \textrm{d} \mathbb {P}(S_{t-u}' \in z) \\&\approx \textrm{e}^{- r(t-u)} \int _0^\infty k \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \left[ \frac{1}{k} \left( G(t) - F_u z \right) - x \right] \frac{\left( x/\theta _{t-u}(z)\right) ^{\nu _{t-u}(z)-1}}{\theta _{t-u}(z) \Gamma \left( \nu _{t-u}(z) \right) } \textrm{e}^{-\frac{x}{\theta _{t-u}(z)}} \textrm{d} x f_{S_{t-u}' }(z)\textrm{d}z \end{aligned} \end{aligned}$$

The inner integral can be rearranged and expressed by Gamma and incomplete upper Gamma function as

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{1}{\theta _{t-u}(z) \Gamma \big ( \nu _{t-u}(z) \big )} \left( \frac{x}{\theta _{t-u}(z)} \right) ^{\nu _{t-u}(z)-1} \textrm{e}^{-\frac{x}{\theta _{t-u}(z)}} \textrm{d} x \\&=\frac{1}{\Gamma \big ( \nu _{t-u}(z) \big )} \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \left( \frac{x}{\theta _{t-u}(z)} \right) ^{\nu _{t-u}(z)-1} \textrm{e}^{-\frac{x}{\theta _{t-u}(z)}} \textrm{d} \left( \frac{x}{\theta _{t-u}(z)} \right) \\&= \frac{\Gamma \big ( \nu _{t-u}(z) \big ) - \Gamma \bigg (\nu _{t-u}(z), \frac{1}{k} \left( G(t) - F_u z \right) \bigg )}{\Gamma \big ( \nu _{t-u}(z) \big )} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{x}{\theta _{t-u}(z) \Gamma \big ( \nu _{t-u}(z) \big )} \left( \frac{x}{\theta _{t-u}(z)} \right) ^{\nu _{t-u}(z)-1} \textrm{e}^{-\frac{x}{\theta _{t-u}(z)}} \textrm{d} x \\&=\frac{\theta _{t-u}(z)}{\Gamma \big ( \nu _{t-u}(z) \big )} \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \left( \frac{x}{\theta _{t-u}(z)} \right) ^{\nu _{t-u}(z)} \textrm{e}^{-\frac{x}{\theta _t(z)}} \textrm{d} \left( \frac{x}{\theta _t(z)} \right) \\&= \frac{\theta _{t-u}(z)}{\Gamma \big ( \nu _{t-u}(z) \big )} \left[ \Gamma \big ( \nu _{t-u}(z) + 1 \big ) - \Gamma \bigg (\nu _{t-u}(z) + 1, \frac{1}{k} \left( G(t) - F_u z \right) \bigg ) \right] , \end{aligned} \end{aligned}$$

Hence the $\Pi (u,t)$ can be simplified as a single integral as in (3.14).

$\square$

Appendix 2

Proofy of Proposition 3.2

Apply the approximated conditional lognormal density function (3.14) to (3.7), we have

$$\begin{aligned} \begin{aligned} \Pi (u , t)&= \textrm{e}^{- r(t-u)} \int _0^\infty k \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \left[ \frac{1}{k} \left( G(t) - F_u z \right) - \Lambda _{t-u}' \right] f_{\Lambda '_{t-u}|S'_{t-u}=z} (x)\textrm{d} x f_{S'_{t-u}}(z) \textrm{d} z \\&\approx \textrm{e}^{- r(t-u)} \int _0^\infty k \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{ \left( G(t) - F_u z \right) /k - x }{x \sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + \log x \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} x f_{S'_{t-u}}(z) \textrm{d} z \end{aligned} \end{aligned}$$

The inner integral can be derived in close form by

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{1}{x \sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + \log x \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} x. \\&=\int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{1}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + \log x \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} \log x\\&= \int _0^{\log \left[ \frac{1}{k} \left( G(t) - F_u z \right) \right] } \frac{1}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + y \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} y \\&= \Phi \left( \frac{\log \left[ \frac{1}{k} \left( G(t) - F_u(z) \right) \right] + \mu _{t-u}(z) \sigma _{t-u}^2(z)(t-u) / 2}{\sigma _{t-u}(z) \sqrt{t-u}} \right) , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{x}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + \log x \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \frac{1}{x} \textrm{d} x. \\&= \int _0^{\frac{1}{k} \left( G(t) - F_u z \right) } \frac{x}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + \log x \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} \log x \\&= \int _0^{\log \left[ \frac{1}{k} \left( G(t) - F_u z \right) \right] } \frac{\textrm{e}^y}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}}\textrm{e}^{-\frac{\left( \mu _{t-u}(z) \sigma _{t-u}^2(z) (t-u) / 2 + y \right) ^2}{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} y \\&= \int _0^{\log \left[ \frac{1}{k} \left( G(t) - F_u z \right) \right] } \frac{1}{\sigma _{t-u}(z) \sqrt{2 \pi (t-u)}} \textrm{e}^{- \frac{\left( y + (\mu _{t-u}(z) - 2) \sigma _{t-u}^2(z) (t-u) / 2\right) ^2 + \left( \mu _{t-u}(z) - 1 \right) \left( \sigma _{t-u}^2(z) (t-u) \right) ^2 }{2 \sigma _{t-u}^2(z) (t-u)}} \textrm{d} y \\&= \textrm{e}^{-\frac{(\mu _{t-u}(z) - 1) \sigma _{t-u}^2(z) (t-u)}{2}} \Phi \left( \frac{\log \left[ \frac{1}{k} \left( G(t) - F_u z \right) \right] + \left( \mu _{t-u}(z) - 2 \right) \sigma _{t-u}^2(z) (t-u) / 2}{\sigma _{t-u}(z) \sqrt{t-u}} \right) . \end{aligned} \end{aligned}$$

Hence the calculation of $\Pi (u,t)$ can be simplified as a single integrals as in (3.15).

$\square$

Appendix 3

Proof of Proposition 4.1

Take derivative with respect to $F_u$ in both sides of (3.14), we have

$$\begin{aligned} \begin{aligned} \frac{{\partial \Pi (u,t)}}{\partial F_u} \approx&\ \textrm{e}^{-r(t-u)} \int _0^{\infty } \left[ (-z) \frac{\Gamma (\nu _{t-u}(z))-\Gamma (\nu _{t-u}(z), \frac{G(t)-F_u z}{k}) }{\Gamma (\nu _{t-u}(z))}\right. \\&- \frac{G(t)- F_u z}{\Gamma (\nu _{t-u}(z))}\frac{\partial \Gamma (\nu _{t-u}(z), \frac{G(t)-F_u z}{k} )}{\partial F_u}\\&+ \left. \frac{k\theta _{t-u}(z)}{\Gamma (\nu _{t-u}(z))} \frac{\partial \Gamma (\nu _{t-u}(z)+1, \frac{G(t)-F_u z}{k}) }{\partial F_u} \right] f_{S'_{t-u}}(z)\textrm{d} z.\\ \end{aligned} \end{aligned}$$

With the derivatives of the incomplete upper Gamma function

$$\begin{aligned} \begin{aligned} \frac{\partial \Gamma (\nu _{t-u}(z), \frac{G(t)-F_u z}{k}) }{\partial F_u}&= \left( \frac{G(t)-F_u z}{k}\right) ^{\nu _{t-u}(z)-1} \textrm{e}^{-\frac{G(t)-F_u z}{k}} \frac{z}{k},\\ \frac{\partial \Gamma (\nu _{t-u}(z)+1, \frac{G(t)-F_u z}{k}) }{\partial F_u}&= \left( \frac{G(t)-F_u z}{k}\right) ^{\nu _{t-u}(z)} \textrm{e}^{-\frac{G(t)-F_u z}{k}} \frac{z}{k}, \end{aligned} \end{aligned}$$

the result of the proposition can be easily obtained by some rearrangement. $\square$

Appendix 4

Proof of Proposition 4.2

By (3.15), we have

$$\begin{aligned} \begin{aligned} \frac{{\partial \Pi (u,t)}}{\partial F_u} \approx&\ \textrm{e}^{-r(t-u)} \int _0^{\infty } \left[ (-z) \Phi \left( \frac{\log \left( \frac{G(t)-F_u z}{k}\right) + \mu _{t-u}(z)\sigma _{t-u}^2(z)(t-u)/2}{\sigma _{t-u}(z)\sqrt{t-u}}\right) \right. \\&+ (G(t)- F_u z) \frac{\partial \Phi \left( \frac{\log \left( \frac{G(t)-F_u z}{k}\right) + \mu _{t-u}(z)\sigma _{t-u}^2(z)(t-u)/2}{\sigma _{t-u}(z)\sqrt{t-u}}\right) }{\partial F_u}\\&-\left. k \text{ e}^{-\frac{(\mu _{t-u}(z)-1)\sigma ^2_{t-u}(z)(t-u)}{2}} \frac{\partial \Phi \left( \frac{\log \left( \frac{G(t)-F_u z}{k}\right) + (\mu _{t-u}(z)-2)\sigma _{t-u}^2(z)(t-u)/2}{\sigma _{t-u}(z)\sqrt{t-u}}\right) }{\partial F_u}\right] f_{S'_{t-u}}(z) \textrm{d}z, \end{aligned} \end{aligned}$$

(D.1)

with the following partial derivatives

$$\begin{aligned} \begin{aligned}&\frac{\partial \Phi \left( \frac{\log \left( \frac{G(t)-F_u z}{k}\right) + \frac{\mu _{t-u}(z)\sigma _{t-u}^2(z)(t-u)}{2}}{\sigma _{t-u}(z)\sqrt{t-u}}\right) }{\partial F_u}\\&= \frac{1}{\sqrt{2\pi }} \exp \left\{ - \frac{\left[ \log \left( \frac{G(t)-F_u z}{k}\right) + \frac{\mu _{t-u}(z)\sigma _{t-u}^2(z)(t-u)}{2}\right] ^2}{2\sigma ^2_{t-u}(z)(t-u)}\ \right\} \frac{1}{\sigma _{t-u}(z)\sqrt{t-u}} \frac{-z}{G(t)-F_u z} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\frac{\partial \Phi \left( \frac{\log \left( \frac{G(t)-F_u z}{k}\right) + \frac{(\mu _{t-u}(z)-2)\sigma _{t-u}^2(z)(t-u)}{2}}{\sigma _{t-u}(z)\sqrt{t-u}}\right) }{\partial F_u}\\&= \frac{1}{\sqrt{2\pi }} \exp \left\{ - \frac{\left[ \log \left( \frac{G(t)-F_u z}{k}\right) +\frac{ (\mu _{t-u}(z)-2)\sigma _{t-u}^2(z)(t-u)}{2}\right] ^2}{2\sigma ^2_{t-u}(z)(t-u)}\ \right\} \frac{1}{\sigma _{t-u}(z)\sqrt{t-u}} \frac{-z}{G(t)-F_u z}, \end{aligned} \end{aligned}$$

substituting into (D.1) yields the result of proposition. $\square$