金融衍生产品的定价问题是金融界的热门研究问题,而由利率市场变化引发的衍生产品定价问题成为目前金融理论研究和实践研究的一个热点。Deng等[1]研究了Vasicek利率模型下的零息债券和无红利股票期权的定价问题;Carr等[2]研究了Lévy跳扩散过程下的期权定价并给出定价公式;郭精军等[3]研究了Vasicek随机利率和标的资产服从次分数布朗运动环境下的欧式期权定价公式;钱晓松[4]选取不同计价单位以及相应的概率测度,简化了一些期权定价中复杂的理论,得到了连续随机利率模型下欧式期权的定价公式。由于Lévy过程具有左极右连及无限可分的特性,可以更好地刻画金融市场中随时都在发生的小幅跳跃行为及罕见的大幅波动行为,因此本文在文献[4]的基础上研究了当利率与标的资产均由带Lévy跳的随机模型给出时的欧式看涨期权的定价问题。
1 带Lévy跳的利率与标的资产模型 1.1 带Lévy跳的Vasicek利率模型设有一个概率空间(Ω,
$ \mathrm{d} r(t)=k(\varepsilon-r(t)) \mathrm{d} t+v\mathrm{d}W(t), t \geqslant 0 $ | (1) |
式中,r(t)为短期利率,W(t)为布朗运动,k、ε、v均为大于0的常数,其中k为拉力,ε为利率的长期平均水平。之后,Björk等[6]将其推广至由泊松点过程和布朗运动两个不确定因素联合驱动的利率结构模型,即利率r(t)满足如下随机微分方程
$ \begin{array}{l} \ \ \ \ \ \ \ \ \mathrm{d} r(t)=k(a-r(t)) \mathrm{d} t+\sigma_{r} \mathrm{d} W_{1}(t)+\int_{-1}^{+\infty} y\left(N_{1}\right. \\ \left.(\mathrm{d} t, \mathrm{d} y)-\lambda_{1} q(\mathrm{d} y) \mathrm{d} t\right), t \geqslant 0 \end{array} $ | (2) |
r(0)=r,式(2)即为带Lévy跳的Vasicek利率模型。其中,k、a、σr均为大于0的常数;{W1(t)}t≥0为标准布朗运动;N1(dt, dy)=
在风险中性测度Q下,考虑期权标的资产的价格S(t)为[7]
$ \begin{array}{l} \ \ \ \ \ \ \ \ \mathrm{d} S(t)=S(t)\left[r(t) \mathrm{d} t+\sigma \mathrm{d} W_{2}(t)+c \mathrm{d} \widetilde{N}_{2}(t)\right],\\ t \geqslant 0 \end{array} $ | (3) |
式中,σ是大于0的常数,{W2(t)}t≥0为标准布朗运动,r(t)为式(2)的解,{N2(t)}t≥0是强度为λ2的泊松过程,c为常数,
$ \begin{array}{l} \ \ \ \ \ \ \ \ S(t)=S(0)(1+c)^{N_{2}(t)} \exp \left\{\int_{0}^{t}(r(s)) \mathrm{d} s+\right. \\ \left.\sigma W_{2}(t)-\frac{1}{2} \sigma^{2} t-c \lambda_{2} t\right\}, t \geqslant 0 \end{array} $ | (4) |
即c>-1时,S(t)是一个严格正的价格过程。令N2(dt, dx)为由c
在给定的市场与概率空间(Ω,
引理1[7] 设在测度Q下,{Wt}t≥0为标准布朗运动,N(dt, dx)是定义在[0, T]×R上的强度为λdtυ(dx)的泊松随机测度,H(x)满足
$ \begin{aligned} &\xi_{t}=\exp \left\{\sigma W_{t}-\frac{1}{2} \sigma^{2} t+\left(\int_{R} \lg H(x) N((0, t], \mathrm{d} x)-\right.\right. \\ &\left.\left.t \int_{R}(H(x)-1) v(\mathrm{d} x)\right)\right\} \end{aligned} $ | (5) |
使得
引理2[7](Doleans-Dade指数公式) 若{X(t)}t≥0是一个跳扩散过程,那么X的Doleans-Dade指数公式有如下形式
$ \begin{aligned} Z^{X}(t) &=\exp \left\{X^{c}(t)-\frac{1}{2}\left[X^{c}, X^{c}\right](t)\right\} \prod\limits_{0 \leqslant s \leqslant t}(1+\\ \Delta X(s)), t & \geqslant 0 \end{aligned} $ | (6) |
这个过程是满足初始条件ZX(0)=1的随机微分方程dZX(t)=ZX(t-)dX(t)的解。
引理3[1] 在上述概率空间中,若利率r(t)满足式(2),则到期日为T的零息债券在时刻t的价格b(t, T)=exp{A(t, T)+G(t, T)r(t)},b(T, T)=1。其中G(t, T)=
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\partial A(t, T)}{\partial t}+\frac{\partial G(t, T)}{\partial t} r+\frac{1}{2} \sigma_{r}^{2} G^{2}(t, T)+k a G(t, \\ T)-k r G(t, T)-r+\lambda_{1}\left[\mathrm{e}^{G(t, T) y}-1-y G(t, T)\right]=0 \end{array} $ | (7) |
{W1(t)}t≥0, {W2(t)}t≥0, {N1(t)}t≥0, {N2(t)}t≥0, (Uj)j≥1是相互独立的,利率由式(2)给出,以标的资产S(t)与零息债券b(t, T)两种计价单位来对欧式看涨期权进行定价。
3 定理及证明定理 在市场利率和标的资产的价格分别由式(2)、(3)给出的双Lévy跳的市场环境中,到期时间为T,执行价格为K,欧式看涨期权在零时刻的价格
$ \begin{array}{l} \ \ \ \ \ \ \ \ C(0, S(0))=E^{Q}\left[B^{-1}(T)(S(T)-K)^{+}\right]= \\ \sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \frac{\mathrm{e}^{-\left(\lambda_{1}+\lambda_{2}\right) T}\left(\lambda_{1} T\right)^{m}\left(\lambda_{2} T\right)^{n}}{m ! n !}\left[S(0) E\left[\varPhi\left(d_{1}^{m, n}\right)\right]-\right. \\ \left.\mathrm{e}^{-\lambda_{1} U T} K b(0, T) E\left[\varPhi\left(d_{2}^{m, n}\right) \mathrm{e}^{G(t, T)} \sum\limits_{j=1}^{m} U_{j}\right]\right] \end{array} $ | (8) |
式中,E[·]为对独立同分布随机变量序列(Uj)j≥1的函数期望,Φ(y)=
$ \begin{array}{l} \ \ \ \ \ \ \ \ E\left(\widetilde{U}_{1}\right)=\int_{-1}^{+\infty}\left(\mathrm{e}^{G(t, T) y}-1\right) q(\mathrm{d} y) \\ \ \ \ \ \ \ \ \ \Delta(t)=\sqrt{\sigma^{2}+G^{2}(t, T) \sigma_{r}^{2}} \\ \ \ \ \ \ \ \ \ \widetilde{W}(t)=\int_{0}^{t} \frac{-\sigma \mathrm{d} \widetilde{W}_{2}(s)+G(s, T) \sigma_{r} \mathrm{d} \widetilde{W}_{1}(s)}{\Delta(s)} \mathrm{d} s \\ \ \ \ \ \ \ \ \ d_{1}^{m, n}=\left[\ln \frac{S(0)}{K b(0, T)}+\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t-\lambda_{1} T \cdot\right. \\ E\left(\widetilde{U}_{1}\right)-\lambda_{2}(1+c) T-\ln \left[\prod\limits_{j=0}^{m} \mathrm{e}^{G(t, T) U_{j}} \cdot\right. \\ \left.\left.\left(\frac{1}{1+c}\right)^{n}\right]\right] / \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t \end{array} $ | (9) |
$ \begin{array}{l} \ \ \ \ \ \ \ \ d_{2}^{m, n}=\left[\ln \frac{S(0)}{K b(0, T)}-\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t-\lambda_{1} T \cdot\right. \\ \left.E\left(\widetilde{U}_{1}\right)+\ln \left[\prod\limits_{j=0}^{m} \frac{1}{\mathrm{e}^{G(t, T) U_{j}}}(1+c)^{n}\right]\right] / \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t \end{array} $ | (10) |
证明 到期时间为T且执行价格为K的欧式看涨期权的风险中性价格C(0, S(0))满足
$ \begin{array}{l} \ \ \ \ \ \ \ \ C(0, S(0))=E^{Q}\left[B^{-1}(T)(S(T)-K)^{+}\right]= \\ E^{Q}\left[B^{-1}(T) S(T) I_{\{S(T) \geqslant K\}}\right]-K E^{Q}\left[B^{-1}(T) I_{\{S(T) \geqslant K\}}\right] \end{array} $ | (11) |
在此处,EQ表示在风险中性测度Q下的期望。对于EQ[B-1(T)S(T)I{S(T)≥K}],以S(t)作为计价单位,将其测度变换到测度QS;对于EQ[B-1(T)I{S(T)≥K}],以b(t, T)作为计价单位,将其测度变换到测度Qb,R-N导数分别为
$ \begin{array}{l} \ \ \ \ \ \ \ \ C(0, S(0)) =S(0) Q^{S}(S(T) \geqslant K)-K b(0, T) \\ Q^{b}(S(T) \geqslant K) =S(0) Q^{S}\left(\frac{b(T, T)}{S(T)} \leqslant \frac{1}{K}\right)-K b(0,\\ T) Q^{b}\left(\frac{S(T)}{b(T, T)} \geqslant K\right) \end{array} $ | (12) |
定义Y(t)=b(t, T)/S(t)和Z(t)=S(t)/b(t, T),先分别求出Y(t)、Z(t)在测度QS和Qb下的表达式, 再分别计算QS(S(T)≥K)与Qb(S(T)≥K)。
引理4 由Girsanov's定理,经由测度变换
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Q^{S}}{\mathrm{d} Q} \mid \mathscr{F}_{t}=\frac{S(t)}{S(0) B(t)}=(1+c)^{N_{2}(t)} \exp \left\{\sigma W_{2}(t)-\right. \\ \left.\frac{1}{2} \sigma^{2} t-c \lambda_{2} t\right\} \end{array} $ | (13) |
在测度QS下,
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Q^{b}}{\mathrm{d} Q} \mid \mathscr{F}_{t}=\frac{b(t, T)}{b(0, T) B(t)}=\int_{0}^{t} \int_{-1}^{\infty} \mathrm{e}^{G(s, T) y} N_{1}(\mathrm{d} s, \\ \mathrm{d} y) \exp \left\{\int_{0}^{t} G(s, T) \sigma_{r} \mathrm{d} W_{1}(s)-\frac{1}{2} \int_{0}^{t} G^{2}(s, T) \sigma_{r}^{2} \mathrm{d} s+\right. \\ \left.\int_{0}^{t} \int_{-1}^{+\infty}\left(\mathrm{e}^{G(s, T) y}-1\right) \lambda_{1} q(\mathrm{d} y) \mathrm{d} s\right\} \end{array} $ | (14) |
在测度Qb下,
引理5 在测度QS下,Y(T)具有如下表达式
$ \begin{array}{l} \ \ \ \ \ \ \ \ Y(T)=\frac{b(0, T)}{S(0)}\left(\frac{1}{1+c}\right)^{N_{2}(T)} \prod\limits_{j=0}^{N_{1}(T)} \mathrm{e}^{G(t, T) U_{j}}\cdot \\ \exp \left\{\lambda_{1} T E\left(\widetilde{U}_{1}\right)-\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t+\lambda_{2}(1+c) T+\right. \\ \left.\int_{0}^{t} \Delta(t) \mathrm{d} \widetilde{W}(t)\right\} \end{array} $ | (15) |
证明 由Itô公式,对于b(t, T)和1/S(t)分别有
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} b(t, T)}{b(t, T)}=r(t) \mathrm{d} t+G(t, T) \sigma_{r} \mathrm{d} W_{1}(t)+ \\ \int_{-1}^{+\infty}\left(\mathrm{e}^{G(t, T) y}-1\right) \widetilde{N}_{1}(\mathrm{d} t, \mathrm{d} y) \end{array} $ | (16) |
又在测度QS下有
$ \begin{array}{l} \ \ \ \ \ \ \ \ \widetilde{N}_{1}(\mathrm{d} t, \mathrm{d} y)=N_{1}(\mathrm{d} t, \mathrm{d} y)-\lambda_{1} q(\mathrm{d} y) \mathrm{d} t \\ \ \ \ \ \ \ \ \ \frac{\mathrm{d}(1 / S(t))}{1 / S(t)}=-r(t) \mathrm{d} t-\sigma \mathrm{d} W_{2}(t)+\sigma^{2} \mathrm{d} t+ \\ c \lambda_{2} \mathrm{d} t+\left(\frac{-c}{1+c}\right) \mathrm{d} N_{2}(t) \end{array} $ | (17) |
且Y(t)=b(t, T)/S(t),再由Itô公式可得
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Y(t)}{Y(t)}=G(t, T) \sigma_{r} \mathrm{d} W_{1}(t)+\sigma^{2} \mathrm{d} t-\sigma \mathrm{d} W_{2}(t)+\\ c \lambda_{2} \mathrm{d} t+\left(\frac{-c}{1+c}\right) \mathrm{d} N_{2}(t)+\int_{-1}^{+\infty}\left(\mathrm{e}^{G(t, T) y}-1\right) \widetilde{N}_{1}(\mathrm{d} t,\\ \mathrm{d} y) \end{array} $ | (18) |
下面给出在测度QS下Y(T)的表达式。记
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Y(t)}{Y(t)}=G(t, T) \sigma_{r} \mathrm{d} \widetilde{W}_{1}(t)-\sigma \mathrm{d} \widetilde{W}_{2}(t)+ \\ \int_{-1}^{+\infty}\left(\mathrm{e}^{G(t, T) y}-1\right) \widetilde{N}_{1}(\mathrm{d} t, \mathrm{d} y)+\left(\frac{-c}{1+c}\right) \mathrm{d} \widetilde{N}_{2}(t) \end{array} $ | (19) |
令
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Y(t)}{Y(t)}=\Delta(t) \mathrm{d} \widetilde{W}(t)+\int_{-1}^{+\infty}\left(\mathrm{e}^{G(t, T) y}-1\right) \widetilde{N}_{1}(\mathrm{d} t,\\ \mathrm{d} y)+\left(\frac{-c}{1+c}\right) \mathrm{d} \widetilde{N}_{2}(t) \end{array} $ | (20) |
再由引理2可得到Y(T)即为式(15)。
引理6 在测度Qb下,Z(T)表达式为
$ \begin{array}{l} \ \ \ \ \ \ \ \ Z(T)=\frac{S(0)}{b(0, T)}(1+c)^{N_{2}(T)} \prod\limits_{j=1}^{N_{1}(T)} \frac{1}{\mathrm{e}^{G(t, T) U_{j}}} \cdot \\ \exp \left\{-\lambda_{1} T E\left(\widetilde{U}_{1}\right)-\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t-\int_{0}^{T} \Delta(t) \mathrm{d} \widetilde{W}(t)\right\} \end{array} $ | (21) |
证明 类似地,对Z(t)=S(t)/b(t, T)有
$ \begin{aligned} &\ \ \ \ \ \ \ \ \frac{\mathrm{d} Z(t)}{Z(t)}=\sigma \mathrm{d} W_{2}(t)-G(t, T) \sigma_{r} \mathrm{d} W_{1}(t)+ \\ &\int_{-1}^{+\infty} \frac{1-\mathrm{e}^{G(t, T) y}}{\mathrm{e}^{G(t, T) y}}\left(N_{1}(\mathrm{d} t, \mathrm{d} y)-\mathrm{e}^{G(t, T) y} \lambda_{1} q(\mathrm{d} y) \mathrm{d} t\right)+ \\ &G^{2}(t, T) \sigma_{r}^{2} \mathrm{d} t+c \mathrm{d} N_{2}(t) \end{aligned} $ | (22) |
以下给出在测度Qb下Z(T)的表达式。记
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Z(t)}{Z(t)}=\sigma \mathrm{d} \widetilde{W}_{2}(t)-G(t, T) \sigma_{r} \mathrm{d} \widetilde{W}_{1}(t)+c \mathrm{d} N_{2}(t)+ \\ \int_{-1}^{+\infty} \frac{1-\mathrm{e}^{G(t, T) y}}{\mathrm{e}^{G(t, T) y}} \widetilde{N}_{1}(\mathrm{d} t, \mathrm{d} y) \end{array} $ | (23) |
令
$ \begin{array}{l} \ \ \ \ \ \ \ \ \frac{\mathrm{d} Z(t)}{Z(t)}=-\Delta(t) \mathrm{d} \widetilde{W}(t)+c \mathrm{d} N_{2}(t)+ \\ \int_{-1}^{+\infty} \frac{1-\mathrm{e}^{G(t, T) y}}{\mathrm{e}^{G(t, T) y}} \widetilde{N}_{1}(\mathrm{d} t, \mathrm{d} y) \end{array} $ | (24) |
再由引理2可得Z(T)即为式(21)。
在得到Y(T)和Z(T)的表达式后,对式(12)中QS(S(T)≥K)进行如下计算。
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{S}(S(T) \geqslant K)=Q^{S}\left(\frac{b(T, T)}{S(T)} \leqslant \frac{1}{K}\right)=Q^{S}(K \cdot \\ Y(T) \leqslant 1)=Q^{S}(\ln K Y(T) \leqslant 0)=Q^{S}(\ln K+\ln Y(T) \leqslant\\ 0 ) \end{array} $ | (25) |
将Y(T)的表达式(式(15))代入式(25),整理得到
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{S}(S(T) \geqslant K)=Q^{S}\left(\int_{0}^{T} \Delta(t) \mathrm{d} \widetilde{W}(t) \leqslant\right. \\ \ln \frac{S(0)}{K b(0, T)}+\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t-\lambda_{1} \operatorname{TE}\left(\widetilde{U}_{1}\right)-\lambda_{2}(1+c) T- \\ \ln \left[\prod\limits_{j=0}^{N_{1}(T)} \mathrm{e}^{G(t, T) U_{j}}\left(\frac{1}{1+c}\right)^{N_{2}(T)}\right] \end{array} $ | (26) |
注意到在测度Q下随机积分
由于(Uj)j≥1, {N1(t)}t≥0, {N2(t)}t≥0相互独立,在测度QS下,
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{S}(S(T) \geqslant K)=\sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \frac{\mathrm{e}^{-\lambda_{1} T}\left(\lambda_{1} T\right)^{m}}{m !} \cdot \\ \frac{\mathrm{e}^{-\lambda_{2} T}\left(\lambda_{2} T\right)^{n}}{n !} E^{Q^{S}}\left[\varPhi\left(d_{1}^{m, n}\right)\right]=\sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \\ \frac{\mathrm{e}^{-\left(\lambda_{1}+\lambda_{2}\right) T}\left(\lambda_{1} T\right)^{m}\left(\lambda_{2} T\right)^{n}}{m ! n !} E\left[\varPhi\left(d_{1}^{m, n}\right)\right] \end{array} $ | (27) |
其中d1m, n即为式(9)。
同理,对于式(12)中Qb(S(T)≥K)有
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{b}(S(T) \geqslant K)=Q^{b}\left(\frac{S(T)}{b(T, T)} \geqslant K\right)=Q^{b}(\ln (Z(T) / \\ K) \geqslant 0)=Q^{b}(\ln Z(T)-\ln K \geqslant 0) \end{array} $ | (28) |
将Z(T)的表达式(式(21))代入式(28),整理得到
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{b}(S(T) \geqslant K)=Q^{b}\left(\int_{0}^{T} \varDelta(t) \mathrm{d} \widetilde{W}(t) \leqslant\right. \\ \ln \frac{S(0)}{K b(0, T)}-\frac{1}{2} \int_{0}^{T} \varDelta^{2}(t) \mathrm{d} t-\lambda_{1} T E\left(\widetilde{U}_{1}\right)+ \\ \left.\ln \left(\prod\limits_{j=1}^{N_{1}(T)} \frac{1}{\mathrm{e}^{G(t, T) U_{j}}}(1+c)^{N_{2}(T)}\right)\right) \end{array} $ | (29) |
又因为在测度Qb下,
$ \begin{array}{l} \ \ \ \ \ \ \ \ Q^{b}(S(T) \geqslant K)=\sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \frac{\mathrm{e}^{-\lambda b T}\left(\lambda^{b} T\right)^{m}}{m !} \frac{\mathrm{e}^{-\lambda_{2} T}\left(\lambda_{2} T\right)^{n}}{n !} \\ E^{Q^{b}}\left[\varPhi\left(d_{2}^{m, n}\right)\right]=\sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \frac{\mathrm{e}^{-\lambda_{1}(1+U) T}\left(\lambda_{1}\left(1+E\left(\widetilde{U}_{1}\right)\right) T\right)^{m}}{m !} \cdot \\ \frac{\mathrm{e}^{-\lambda_{2} T}\left(\lambda_{2} T\right)^{n}}{n !}\left[\varPhi\left(d_{2}^{m, n}\right) \frac{\prod\limits_{j=1}^{m} \mathrm{e}^{G(t, T) U_{j}}}{\left(1+E\left(\widetilde{U}_{1}\right)\right)^{m}}\right]=\sum\limits_{m=0}^{\infty} \sum\limits_{n=0}^{\infty} \\ \frac{\mathrm{e}^{-\left(\lambda_{1}+\lambda_{2}\right) T}\left(\lambda_{1} T\right)^{m}\left(\lambda_{2} T\right)^{n}}{m ! n !} \mathrm{e}^{-\lambda_{1} T E\left(\tilde{U}_{1}\right)} E\left[\varPhi\left(d_{2}^{m, n}\right) \mathrm{e}^{G(t, T) \sum\limits_{j=1}^{m} U_{j}}\right] \end{array} $ | (30) |
其中d2m, n即为式(10)。
综合式(11)、(27)、(30),即可整理得到欧式看涨期权的价格公式(8)。
4 结束语本文研究利率和资产均服从Lévy跳扩散模型下的欧式看涨期权的定价问题。在计算过程中利用带跳的测度变换公式完成测度变换,利用Lévy-Itô型积分公式与Doleans-Dade指数公式完成计价单位转换的计算。该模型可以更好地捕捉市场中的利率波动以及适应利率市场环境,并且使用计价单位转化原理使得定价问题变得简便。
[1] |
DENG G H, XI H. Pricing reset option in a fractional Brownian motion market[C]//Proceedings of the 30th Chinese Control Conference. Yantai, 2011: 5727-5731.
|
[2] |
CARR P, WU L R. The finite moment log stable process and option pricing[J]. The Journal of Finance, 2003, 58(2): 753-777. DOI:10.1111/1540-6261.00544 |
[3] |
郭精军, 张亚芳. 次分数Vasicek随机利率模型下的欧式期权定价[J]. 应用数学, 2017, 30(3): 503-511. GUO J J, ZHANG Y F. European option pricing under subfractional Vasicek stochastic interest rate model[J]. Mathematica Applicata, 2017, 30(3): 503-511. (in Chinese) |
[4] |
钱晓松. 跳扩散模型中的测度变换与期权定价[J]. 应用概率统计, 2004, 20(1): 91-99. QIAN X S. Changes of probability measure and options pricing in jump-diffusion models[J]. Chinese Journal of Applied Probability and Statistics, 2004, 20(1): 91-99. (in Chinese) |
[5] |
VASICEK O. An equilibrium characterization of the term structure[J]. Journal of Financial Economics, 1977, 5(2): 177-188. DOI:10.1016/0304-405X(77)90016-2 |
[6] |
BJÖRK T, KABANOV Y, RUNGGALDIER W. Bond market structure in the presence of marked point processes[J]. Mathematical Finance, 1997, 7(2): 211-223. DOI:10.1111/1467-9965.00031 |
[7] |
NIU L Q. Some stability results of optimal investment in a simple Lévy market[J]. Mathematics and Economics, 2008, 42: 445-452. DOI:10.1016/j.insmatheco.2007.05.005 |