本文的目的在于提供0.5，1和1.5阶强SDE数值格式的推导和内容，所有推导基于 I t o ? T a y l o r Ito-Taylor Ito?Taylor展开，由于国内外网站缺少关于强SDE数值阶的总结，笔者在此特作总结，为使用SDE数值格式的读者提供帮助，本文需要读者预先已经知道了有关Brown-motion的基本性质和随机微分的基本性质，否则格式推导会看不懂。如果想要实际应用直接看数值格式即可。
本文考虑如下SDE:
d
X
=
b
(
x
)
d
t
+
σ
(
x
)
d
W
.
dX=b(x)dt+\sigma(x)dW.
dX=b(x)dt+σ(x)dW.
使用数值SDE在时间
t
t
t下以
x
x
x为初值的近似解定义为
X
￣
t
,
x
(
t
+
h
)
\overline{X}_{t,x}(t+h)
Xt,x?(t+h),其中
t
+
h
≥
t
t+h\geq t
t+h≥t.真解为:
X
t
,
x
(
t
+
h
)
X_{t,x}(t+h)
Xt,x?(t+h).若满足如下定理:
定理1:
若存在
p
1
,
p
2
p_1,p_2
p1?,p2?,满足:
E
∣
X
t
,
x
(
t
+
h
)
?
X
￣
t
,
x
(
t
+
h
)
∣
≤
k
(
1
+
∣
x
∣
2
)
h
p
1
.
E |X_{t,x}(t+h)-\overline{X}_{t,x}(t+h)|\leq k(1+|x|^2)h^{p_1}.
E∣Xt,x?(t+h)?Xt,x?(t+h)∣≤k(1+∣x∣2)hp1?.
[
E
∣
X
t
,
x
(
t
+
h
)
?
X
￣
t
,
x
(
t
+
h
)
∣
2
]
1
2
≤
k
(
1
+
∣
x
∣
2
)
1
2
h
p
2
.
[E |X_{t,x}(t+h)-\overline{X}_{t,x}(t+h)|^2]^{\frac{1}{2}}\leq k(1+|x|^2)^{\frac{1}{2}}h^{p_2}.
[E∣Xt,x?(t+h)?Xt,x?(t+h)∣2]21?≤k(1+∣x∣2)21?hp2?.
且满足
p
1
≥
p
2
+
1
2
p_1\geq p_2+\frac{1}{2}
p1?≥p2?+21?,
p
2
≥
1
2
p_2\geq\frac{1}{2}
p2?≥21?.则可认为强收敛阶为
p
2
?
1
2
p_2-\frac{1}{2}
p2??21?.
该定理的证明笔者在此不做推导，感兴趣的读者自行阅读其他文献或教材，这一般都是有的，这里我们直接利用它进行以下的
I
t
o
?
T
a
y
l
o
r
Ito-Taylor
Ito?Taylor展开。
我们考虑:
d
x
=
b
(
x
)
d
t
+
σ
(
x
)
d
W
.
dx=b(x)dt+\sigma(x)dW.
dx=b(x)dt+σ(x)dW.
若有一个非常光滑的
f
(
x
)
f(x)
f(x),根据
I
t
o
Ito
Ito公式，会有:
d
f
(
x
)
=
[
b
(
x
)
f
x
(
x
)
+
1
2
σ
2
f
x
x
(
x
)
]
d
t
+
[
σ
(
x
)
f
x
(
x
)
]
d
W
.
df(x)=[b(x)f_x(x)+\frac{1}{2} \sigma^2 f_{xx}(x)]dt+[\sigma(x)f_x(x)]dW.
df(x)=[b(x)fx?(x)+21?σ2fxx?(x)]dt+[σ(x)fx?(x)]dW.
这两个微分算子可以被定义为:
L
0
=
b
(
x
)
d
d
x
+
1
2
σ
2
(
x
)
d
2
d
x
2
L^0=b(x)\frac{d}{dx}+\frac{1}{2}\sigma^2(x)\frac{d^2}{dx^2}
L0=b(x)dxd?+21?σ2(x)dx2d2?
L
1
=
σ
(
x
)
d
d
x
L^1=\sigma(x)\frac{d}{dx}
L1=σ(x)dxd?
则对两边取积分会有:
f
(
x
t
)
=
f
(
x
t
0
)
+
∫
t
0
t
L
0
f
(
x
s
)
d
s
+
∫
t
0
t
L
1
f
(
x
s
)
d
W
s
.
f(x_t)=f(x_{t0})+\int_{t_0}^t L^0f(x_s)ds+\int_{t_0}^tL^1f(x_s)dW_s.
f(xt?)=f(xt0?)+∫t0?t?L0f(xs?)ds+∫t0?t?L1f(xs?)dWs?.
注意到:
∫
t
0
t
L
0
f
(
x
s
)
d
s
=
L
0
f
(
x
t
0
)
(
t
?
t
0
)
+
∫
t
0
t
[
L
0
f
(
x
s
)
?
L
0
f
(
x
t
0
)
]
.
\int_{t_0}^t L^0f(x_s)ds=L^0 f(x_{t_0})(t-t_0)+\int_{t_0}^t[L^0f(x_s)-L^0f(x_{t_0})].
∫t0?t?L0f(xs?)ds=L0f(xt0??)(t?t0?)+∫t0?t?[L0f(xs?)?L0f(xt0??)].
∫
t
0
t
L
1
f
(
x
s
)
d
W
s
=
L
1
f
(
x
t
0
)
(
W
t
?
W
t
0
)
+
∫
t
0
t
[
L
1
f
(
x
s
)
?
L
1
f
(
x
t
0
)
]
d
W
s
.
\int_{t_0}^t L^1f(x_s)dWs=L^1 f(x_{t_0})(W_t-W_{t_0})+\int_{t_0}^t[L^1f(x_s)-L^1f(x_{t_0})]dWs.
∫t0?t?L1f(xs?)dWs=L1f(xt0??)(Wt??Wt0??)+∫t0?t?[L1f(xs?)?L1f(xt0??)]dWs.
注意到:
∫
t
0
t
[
L
0
f
(
x
s
)
?
L
0
f
(
x
t
0
)
]
=
∫
t
0
t
∫
t
0
s
d
[
L
0
f
(
x
s
1
)
]
d
s
\int_{t_0}^t[L^0f(x_s)-L^0f(x_{t_0})]=\int_{t_0}^t\int_{t_0}^s d[L^0f(x_{s_1})]ds
∫t0?t?[L0f(xs?)?L0f(xt0??)]=∫t0?t?∫t0?s?d[L0f(xs1??)]ds
∫
t
0
t
[
L
1
f
(
x
s
)
?
L
1
f
(
x
t
0
)
]
=
∫
t
0
t
∫
t
0
s
d
[
L
1
f
(
x
s
1
)
]
d
W
s
\int_{t_0}^t[L^1f(x_s)-L^1f(x_{t_0})]=\int_{t_0}^t\int_{t_0}^s d[L^1f(x_{s_1})]dWs
∫t0?t?[L1f(xs?)?L1f(xt0??)]=∫t0?t?∫t0?s?d[L1f(xs1??)]dWs
使用
I
t
o
Ito
Ito公式我们可以知道:
d
[
L
0
f
(
x
)
]
=
[
L
0
L
0
f
]
d
t
+
[
L
1
L
0
f
]
d
W
.
d[L^0f(x)]=[L^0L^0f]dt+[L^1L^0f]dW.
d[L0f(x)]=[L0L0f]dt+[L1L0f]dW.
d
[
L
1
f
(
x
)
]
=
[
L
0
L
1
f
]
d
t
+
[
L
1
L
1
f
]
d
W
.
d[L^1f(x)]=[L^0L^1f]dt+[L^1L^1f]dW.
d[L1f(x)]=[L0L1f]dt+[L1L1f]dW.
我们便可以得到:
f
(
x
t
)
=
f
(
x
t
0
)
+
L
0
f
(
x
t
0
)
(
t
?
t
0
)
+
L
1
f
(
x
t
0
)
(
W
t
?
W
t
0
)
+
R
￣
f(x_t)=f(x_{t0})+L^0 f(x_{t_0})(t-t_0)+L^1 f(x_{t_0})(W_t-W_{t_0})+\overline{R}
f(xt?)=f(xt0?)+L0f(xt0??)(t?t0?)+L1f(xt0??)(Wt??Wt0??)+R
其中:
R
￣
=
∫
t
0
t
∫
t
0
s
L
0
L
0
f
(
s
1
)
d
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
1
L
0
f
(
s
1
)
d
W
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
0
L
1
f
(
s
1
)
d
s
1
d
W
s
+
∫
t
0
t
∫
t
0
s
L
1
L
1
f
(
s
1
)
d
W
s
1
d
W
s
\overline{R}=\int_{t_0}^t\int_{t_0}^s L^0L^0f(s_1)ds_1ds+\int_{t_0}^t\int_{t_0}^s L^1L^0f(s_1)dWs_1ds+\int_{t_0}^t\int_{t_0}^s L^0L^1f(s_1)ds_1dWs+\int_{t_0}^t\int_{t_0}^s L^1L^1f(s_1)dWs_1dWs
R=∫t0?t?∫t0?s?L0L0f(s1?)ds1?ds+∫t0?t?∫t0?s?L1L0f(s1?)dWs1?ds+∫t0?t?∫t0?s?L0L1f(s1?)ds1?dWs+∫t0?t?∫t0?s?L1L1f(s1?)dWs1?dWs
当然可以继续展开下去。
我们在这里只关心
f
(
x
)
=
x
f(x)=x
f(x)=x的情况，因此这会导出本文的主题0.5(Euler-Maruyama), 1(Milstein), 和1.5 阶强Stochastic Differential Equation格式。
令
f
(
x
)
=
x
f(x)=x
f(x)=x,显然地,我们会得到:
L
0
f
=
L
0
x
=
b
(
x
)
L^0f=L^0x=b(x)
L0f=L0x=b(x),
L
1
f
=
L
1
x
=
σ
(
x
)
L^1f=L^1x=\sigma(x)
L1f=L1x=σ(x).
则根据：
f
(
x
t
)
=
f
(
x
t
0
)
+
L
0
f
(
x
t
0
)
(
t
?
t
0
)
+
L
1
f
(
x
t
0
)
(
W
t
?
W
t
0
)
+
R
￣
f(x_t)=f(x_{t0})+L^0 f(x_{t_0})(t-t_0)+L^1 f(x_{t_0})(W_t-W_{t_0})+\overline{R}
f(xt?)=f(xt0?)+L0f(xt0??)(t?t0?)+L1f(xt0??)(Wt??Wt0??)+R
代入
f
(
x
)
=
x
f(x)=x
f(x)=x可得:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
+
R
￣
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})+\overline{R}
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)+R
此时
R
￣
\overline{R}
R为:
R
￣
=
∫
t
0
t
∫
t
0
s
L
0
b
(
x
s
1
)
d
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
1
b
(
x
s
1
)
d
W
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
0
σ
(
x
s
1
)
d
s
1
d
W
s
+
∫
t
0
t
∫
t
0
s
L
1
σ
(
x
s
1
)
d
W
s
1
d
W
s
\overline{R}=\int_{t_0}^t\int_{t_0}^s L^0b(x_{s_1})ds_1ds+\int_{t_0}^t\int_{t_0}^s L^1b(x_{s_1})dWs_1ds+\int_{t_0}^t\int_{t_0}^s L^0\sigma(x_{s_1})ds_1dWs+\int_{t_0}^t\int_{t_0}^s L^1\sigma(x_{s_1})dWs_1dWs
R=∫t0?t?∫t0?s?L0b(xs1??)ds1?ds+∫t0?t?∫t0?s?L1b(xs1??)dWs1?ds+∫t0?t?∫t0?s?L0σ(xs1??)ds1?dWs+∫t0?t?∫t0?s?L1σ(xs1??)dWs1?dWs
称下式为Euler-Maruyama格式:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)需要用到如下随机微分性质:
E
[
d
W
s
2
]
=
d
s
E[dWs^2]=ds
E[dWs2]=ds
根据定理和随机微分的性质可知余项
R
￣
\overline{R}
R中满足:
p
1
=
2
p_1=2
p1?=2 (具有
d
W
s
dWs
dWs会使得阶数为0),
p
2
=
1
p_2=1
p2?=1(一个
d
s
ds
ds贡献一个阶,一个
d
W
s
dWs
dWs贡献0.5个阶), 满足定理格式，此时强收敛阶为
p
2
?
0.5
=
0.5
p_2-0.5=0.5
p2??0.5=0.5阶,则该格式具有0.5阶强收敛性。
注意到余项:
R
￣
=
∫
t
0
t
∫
t
0
s
L
0
b
(
x
s
1
)
d
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
1
b
(
x
s
1
)
d
W
s
1
d
s
+
∫
t
0
t
∫
t
0
s
L
0
σ
(
x
s
1
)
d
s
1
d
W
s
+
∫
t
0
t
∫
t
0
s
L
1
σ
(
x
s
1
)
d
W
s
1
d
W
s
\overline{R}=\int_{t_0}^t\int_{t_0}^s L^0b(x_{s_1})ds_1ds+\int_{t_0}^t\int_{t_0}^s L^1b(x_{s_1})dWs_1ds+\int_{t_0}^t\int_{t_0}^s L^0\sigma(x_{s_1})ds_1dWs+\int_{t_0}^t\int_{t_0}^s L^1\sigma(x_{s_1})dWs_1dWs
R=∫t0?t?∫t0?s?L0b(xs1??)ds1?ds+∫t0?t?∫t0?s?L1b(xs1??)dWs1?ds+∫t0?t?∫t0?s?L0σ(xs1??)ds1?dWs+∫t0?t?∫t0?s?L1σ(xs1??)dWs1?dWs
我们想要提升强格式阶，那么显然此时若能够提升
p
2
p_2
p2?的阶就可以了，注意到限制了
p
2
p_2
p2?的阶为
R
￣
\overline{R}
R中这一项:
G
=
∫
t
0
t
∫
t
0
s
L
1
σ
(
x
s
1
)
d
W
s
1
d
W
s
G=\int_{t_0}^t\int_{t_0}^s L^1\sigma(x_{s_1})dWs_1dWs
G=∫t0?t?∫t0?s?L1σ(xs1??)dWs1?dWs那么，我们需要对其进行再一次的展开，此时根据
I
t
o
Ito
Ito公式，我们可以将它展开为:
G
=
L
1
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
W
s
1
d
W
s
+
∫
t
0
t
∫
t
0
s
∫
t
0
s
1
L
0
L
1
σ
(
x
s
2
)
d
s
2
d
W
s
1
d
W
s
+
∫
t
0
t
∫
t
0
s
∫
t
0
s
1
L
1
L
1
σ
(
x
s
2
)
d
W
s
2
d
W
s
1
d
W
s
G=L^1\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s dWs_1dWs+\int_{t_0}^t\int_{t_0}^s\int _{t_0}^{s_1}L^0L^1\sigma(x_{s_2})d{s_2}dWs_1dWs+\int_{t_0}^t\int_{t_0}^s\int _{t_0}^{s_1}L^1L^1\sigma(x_{s_2})d{Ws_2}dWs_1dWs
G=L1σ(xt0??)∫t0?t?∫t0?s?dWs1?dWs+∫t0?t?∫t0?s?∫t0?s1??L0L1σ(xs2??)ds2?dWs1?dWs+∫t0?t?∫t0?s?∫t0?s1??L1L1σ(xs2??)dWs2?dWs1?dWs
此时可以得到Milstein格式为:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
+
L
1
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
W
s
1
d
W
s
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})+L^1\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s dWs_1dWs
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)+L1σ(xt0??)∫t0?t?∫t0?s?dWs1?dWs这即:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
+
σ
′
(
x
t
0
)
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
W
s
1
d
W
s
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})+\sigma^{'}(x_{t_0})\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s dWs_1dWs
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)+σ′(xt0??)σ(xt0??)∫t0?t?∫t0?s?dWs1?dWs其中:
∫
t
0
t
∫
t
0
s
d
W
s
1
d
W
s
=
1
2
(
W
t
?
W
t
0
)
2
?
1
2
(
t
?
t
0
)
\int_{t_0}^t\int_{t_0}^s dWs_1dWs=\frac{1}{2}(W_t-W_{t_0})^2-\frac{1}{2}(t-t_0)
∫t0?t?∫t0?s?dWs1?dWs=21?(Wt??Wt0??)2?21?(t?t0?)这即可以得到最终格式为:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
+
1
2
σ
′
(
x
t
0
)
σ
(
x
t
0
)
[
(
W
t
?
W
t
0
)
2
?
(
t
?
t
0
)
]
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})+\frac{1}{2}\sigma^{'}(x_{t_0})\sigma(x_{t_0})[(W_t-W_{t_0})^2-(t-t_0)]
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)+21?σ′(xt0??)σ(xt0??)[(Wt??Wt0??)2?(t?t0?)]
此时注意到
p
1
=
2
p_1=2
p1?=2不变，因为提升了
p
2
=
3
2
p_2=\frac{3}{2}
p2?=23?，满足定理要求，因此该格式收敛阶为1.下面继续提升，注意到此时若想提升强收敛阶，很明显提升
p
2
p_2
p2?是不够的了，因为此时
p
1
p_1
p1?也需要被提升，但是很明显的是在
R
￣
\overline{R}
R中有
p
2
=
3
2
p_2=\frac{3}{2}
p2?=23?的项两个:
∫
t
0
t
∫
t
0
s
L
0
σ
(
x
s
1
)
d
s
1
d
W
s
\int_{t_0}^t\int_{t_0}^s L^0\sigma(x_{s_1})ds_1dWs
∫t0?t?∫t0?s?L0σ(xs1??)ds1?dWs和
∫
t
0
t
∫
t
0
s
L
1
b
(
x
s
1
)
d
W
s
1
d
s
\int_{t_0}^t\int_{t_0}^s L^1b(x_{s_1})dWs_1ds
∫t0?t?∫t0?s?L1b(xs1??)dWs1?ds则他们也需要被包含在内才可以提升该格式的收敛阶，因此在
M
i
l
s
t
e
i
n
Milstein
Milstein的基础上，我们继续延拓格式
T
T
T为:
T
=
M
i
l
s
t
e
i
n
+
K
T=Milstein+K
T=Milstein+K
K
=
L
0
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
W
s
+
L
1
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
W
s
1
d
s
+
L
0
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
s
+
(
L
1
)
2
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
∫
t
0
s
1
d
W
s
2
d
W
s
1
d
W
s
K=L^0\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^sds_1dWs+L^1b(x_{t_0})\int_{t_0}^t\int_{t_0}^s dWs_1ds+L^0b(x_{t_0})\int_{t_0}^t\int_{t_0}^s ds_1ds+(L^1)^2\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s\int _{t_0}^{s_1}d{Ws_2}dWs_1dWs
K=L0σ(xt0??)∫t0?t?∫t0?s?ds1?dWs+L1b(xt0??)∫t0?t?∫t0?s?dWs1?ds+L0b(xt0??)∫t0?t?∫t0?s?ds1?ds+(L1)2σ(xt0??)∫t0?t?∫t0?s?∫t0?s1??dWs2?dWs1?dWs我们仅考虑从
n
n
n到
n
+
1
n+1
n+1会发生什么，令
△
t
=
t
n
+
1
?
t
n
\triangle t=t_{n+1}-t_n
△t=tn+1??tn?,
△
W
=
W
t
n
+
1
?
W
t
n
\triangle W=W_{t_{n+1}}-W_{t_n}
△W=Wtn+1???Wtn??
(
L
1
)
2
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
∫
t
0
s
1
d
W
s
2
d
W
s
1
d
W
s
=
(
L
1
)
2
σ
(
x
t
0
)
[
1
6
△
W
2
?
1
2
△
t
]
△
W
(L^1)^2\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s\int _{t_0}^{s_1}d{Ws_2}dWs_1dWs=(L^1)^2\sigma(x_{t_0})[\frac{1}{6}\triangle W^2-\frac{1}{2}\triangle t]\triangle W
(L1)2σ(xt0??)∫t0?t?∫t0?s?∫t0?s1??dWs2?dWs1?dWs=(L1)2σ(xt0??)[61?△W2?21?△t]△W这一点使用
I
t
o
Ito
Ito公式很容易证明。
另一方面令:
△
Z
=
∫
t
n
t
n
+
1
∫
t
n
s
d
W
s
1
d
s
=
∫
t
n
t
n
+
1
[
t
n
+
1
?
s
]
d
W
s
\triangle Z=\int_{t_n}^{t_{n+1}}\int_{t_n}^s dWs_1ds=\int_{t_n}^{t_{n+1}}[t_{n+1}-s] dW_s
△Z=∫tn?tn+1??∫tn?s?dWs1?ds=∫tn?tn+1??[tn+1??s]dWs?根据随机微分的性质,显然
△
Z
～
N
(
0
,
∫
t
n
t
n
+
1
[
t
n
+
1
?
s
]
2
d
s
)
=
N
(
0
,
1
3
△
t
3
)
\triangle Z \sim N(0,\int_{t_n}^{t_{n+1}}[t_{n+1}-s]^2 ds)=N(0,\frac{1}{3} \triangle t^3)
△Z～N(0,∫tn?tn+1??[tn+1??s]2ds)=N(0,31?△t3)注意到:
E
[
△
Z
△
W
]
=
∫
t
n
t
n
+
1
[
t
n
+
1
?
s
]
d
s
=
1
2
△
t
2
E[\triangle Z \triangle W]=\int_{t_n}^{t_{n+1}}[t_{n+1}-s] ds=\frac{1}{2} \triangle t^2
E[△Z△W]=∫tn?tn+1??[tn+1??s]ds=21?△t2因此:
E
[
(
△
Z
?
1
2
△
W
)
△
W
]
=
0
E[(\triangle Z-\frac{1}{2} \triangle W)\triangle W]=0
E[(△Z?21?△W)△W]=0这样找到了独立的样本，且注意到此时根据随机微分的性质可得到:
(
△
Z
?
1
2
△
W
)
～
N
(
0
,
1
12
△
t
3
)
(\triangle Z-\frac{1}{2} \triangle W)～N(0,\frac{1}{12}\triangle t^3)
(△Z?21?△W)～N(0,121?△t3)注意到:
∫
t
0
t
∫
t
0
s
d
s
1
d
W
s
=
△
t
△
W
?
△
Z
\int_{t_0}^t\int_{t_0}^sds_1dWs=\triangle t \triangle W-\triangle Z
∫t0?t?∫t0?s?ds1?dWs=△t△W?△Z那么显然地,我们可以根据如下方法来给出1.5阶格式,若我们有
G
1
G_1
G1?,
G
2
G_2
G2?两个独立的同分布正态分布变量
G
1
～
N
(
0
,
1
)
,
G
2
～
N
(
0
,
1
)
G_1～N(0,1),G_2～N(0,1)
G1?～N(0,1),G2?～N(0,1)那么根据我们的推导
K
=
L
0
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
W
s
+
L
1
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
W
s
1
d
s
+
L
0
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
s
+
(
L
1
)
2
σ
(
x
t
0
)
∫
t
0
t
∫
t
0
s
∫
t
0
s
1
d
W
s
2
d
W
s
1
d
W
s
K=L^0\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^sds_1dWs+L^1b(x_{t_0})\int_{t_0}^t\int_{t_0}^s dWs_1ds+L^0b(x_{t_0})\int_{t_0}^t\int_{t_0}^s ds_1ds+(L^1)^2\sigma(x_{t_0})\int_{t_0}^t\int_{t_0}^s\int _{t_0}^{s_1}d{Ws_2}dWs_1dWs
K=L0σ(xt0??)∫t0?t?∫t0?s?ds1?dWs+L1b(xt0??)∫t0?t?∫t0?s?dWs1?ds+L0b(xt0??)∫t0?t?∫t0?s?ds1?ds+(L1)2σ(xt0??)∫t0?t?∫t0?s?∫t0?s1??dWs2?dWs1?dWs
K
=
L
0
σ
(
x
t
0
)
[
△
t
△
W
?
△
Z
]
+
L
1
b
(
x
t
0
)
△
Z
+
L
0
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
s
+
(
L
1
)
2
σ
(
x
t
0
)
[
1
6
△
W
2
?
1
2
△
t
]
△
W
K=L^0\sigma(x_{t_0})[\triangle t \triangle W-\triangle Z]+L^1b(x_{t_0})\triangle Z+L^0b(x_{t_0})\int_{t_0}^t\int_{t_0}^s ds_1ds+(L^1)^2\sigma(x_{t_0})[\frac{1}{6}\triangle W^2-\frac{1}{2}\triangle t]\triangle W
K=L0σ(xt0??)[△t△W?△Z]+L1b(xt0??)△Z+L0b(xt0??)∫t0?t?∫t0?s?ds1?ds+(L1)2σ(xt0??)[61?△W2?21?△t]△W其中:
△
W
=
△
t
G
1
.
\triangle W= \sqrt{\triangle t} G_1.
△W=△t?G1?.
△
Z
=
1
2
△
t
△
W
+
1
2
3
△
t
3
2
G
2
\triangle Z= \frac{1}{2} \triangle t\triangle W+\frac{1}{2 \sqrt{3}} \triangle t^{\frac{3}{2}} G_2
△Z=21?△t△W+23?1?△t23?G2?此时
K
K
K中所有成分可求，这样的1.5阶格式为:
x
t
=
x
t
0
+
b
(
x
t
0
)
(
t
?
t
0
)
+
σ
(
x
t
0
)
(
W
t
?
W
t
0
)
+
1
2
σ
′
(
x
t
0
)
σ
(
x
t
0
)
[
(
W
t
?
W
t
0
)
2
?
(
t
?
t
0
)
]
+
K
x_t=x_{t_0}+b(x_{t_0})(t-t_0)+\sigma(x_{t_0})(W_t-W_{t_0})+\frac{1}{2}\sigma^{'}(x_{t_0})\sigma(x_{t_0})[(W_t-W_{t_0})^2-(t-t_0)]+K
xt?=xt0??+b(xt0??)(t?t0?)+σ(xt0??)(Wt??Wt0??)+21?σ′(xt0??)σ(xt0??)[(Wt??Wt0??)2?(t?t0?)]+K
K
=
L
0
σ
(
x
t
0
)
[
△
t
△
W
?
△
Z
]
+
L
1
b
(
x
t
0
)
△
Z
+
L
0
b
(
x
t
0
)
∫
t
0
t
∫
t
0
s
d
s
1
d
s
+
(
L
1
)
2
σ
(
x
t
0
)
[
1
6
△
W
2
?
1
2
△
t
]
△
W
K=L^0\sigma(x_{t_0})[\triangle t \triangle W-\triangle Z]+L^1b(x_{t_0})\triangle Z+L^0b(x_{t_0})\int_{t_0}^t\int_{t_0}^s ds_1ds+(L^1)^2\sigma(x_{t_0})[\frac{1}{6}\triangle W^2-\frac{1}{2}\triangle t]\triangle W
K=L0σ(xt0??)[△t△W?△Z]+L1b(xt0??)△Z+L0b(xt0??)∫t0?t?∫t0?s?ds1?ds+(L1)2σ(xt0??)[61?△W2?21?△t]△W当然，读者可以自行推导更高阶的
I
t
o
?
T
a
y
l
o
r
Ito-Taylor
Ito?Taylor展开，笔者在这里不过多介绍。