Considere
X
{\displaystyle X}
R
n
{\displaystyle \mathbf {R} ^{n}}
equação diferencial estocástica
d
X
t
=
b
(
X
t
)
d
t
+
σ
(
X
t
)
d
B
t
.
{\displaystyle \mathrm {d} X_{t}=b(X_{t})\,\mathrm {d} t+\sigma (X_{t})\,\mathrm {d} B_{t}.\ }
Para um ponto
x
∈
{\displaystyle x\in }
R
n
{\displaystyle \mathbf {R} ^{n}}
P
x
{\displaystyle \mathbf {P} ^{x}}
X
{\displaystyle X}
X
0
=
x
{\displaystyle X_{0}=x}
E
x
{\displaystyle \mathbf {E} ^{x}}
P
x
{\displaystyle \mathbf {P} ^{x}}
Considere
A
{\displaystyle A}
infinitesimal de
X
{\displaystyle X}
C
2
{\displaystyle C^{2}}
compactamente suportadas (duplamente diferenciáveis com segunda derivada contínua)
f
:
R
n
→
R
{\displaystyle f:\mathbf {R} ^{n}\rightarrow \mathbf {R} }
A
f
(
x
)
=
lim
t
↓
0
E
x
[
f
(
X
t
)
]
−
f
(
x
)
t
{\displaystyle Af(x)=\lim _{t\downarrow 0}{\frac {\mathbf {E} ^{x}[f(X_{t})]-f(x)}{t}}\ }
ou, equivalentemente,
A
f
(
x
)
=
∑
i
b
i
(
x
)
∂
f
∂
x
i
(
x
)
+
1
2
∑
i
,
j
(
σ
σ
⊤
)
i
,
j
(
x
)
∂
2
f
∂
x
i
∂
x
j
(
x
)
.
{\displaystyle Af(x)=\sum _{i}b_{i}(x){\frac {\partial f}{\partial x_{i}}}(x)+{\frac {1}{2}}\sum _{i,j}{\big (}\sigma \sigma ^{\top }{\big )}_{i,j}(x){\frac {\partial ^{2}f}{\partial x_{i}\,\partial x_{j}}}(x).\ }
Considere que
τ
{\displaystyle \tau }
E
x
[
τ
]
<
+
∞
{\displaystyle \mathbf {E} ^{x}[\tau ]<+\infty }
f
{\displaystyle f}
C
2
{\displaystyle C^{2}}
[ 1]
E
x
[
f
(
X
τ
)
]
=
f
(
x
)
+
E
x
[
∫
0
τ
A
f
(
X
s
)
d
s
]
.
{\displaystyle \mathbf {E} ^{x}[f(X_{\tau })]=f(x)+\mathbf {E} ^{x}\left[\int _{0}^{\tau }Af(X_{s})\,\mathrm {d} s\right].\ }
Na verdade, se
τ
{\displaystyle \tau }
conjunto limitado
B
⊂
R
n
{\displaystyle B\subset \mathbf {R} ^{n}}
E
x
[
τ
]
<
+
∞
{\displaystyle \mathbf {E} ^{x}[\tau ]<+\infty }
f
{\displaystyle f}
C
2
{\displaystyle C^{2}}
A fórmula de Dynkin pode ser usada para encontrar o primeiro tempo de saída esperado
t
K
{\displaystyle t_{K}}
movimento browniano
B
{\displaystyle B}
bola fechada
K
=
K
R
=
{
x
∈
R
n
|
|
x
|
≤
R
}
,
{\displaystyle K=K_{R}=\{x\in \mathbf {R} ^{n}|\,|x|\leq R\},}
que, quando
B
{\displaystyle B}
a
{\displaystyle a}
interior de
K
{\displaystyle K}
E
a
[
τ
K
]
=
1
n
(
R
2
−
|
a
|
2
)
.
{\displaystyle \mathbf {E} ^{a}[\tau _{K}]={\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )}.}
Escolha um número inteiro
j
{\displaystyle j}
X
=
B
{\displaystyle X=B}
τ
=
σ
j
=
min
(
j
,
τ
K
)
{\displaystyle \tau =\sigma j=\min(j,\tau _{K})}
f
{\displaystyle f}
C
2
{\displaystyle C^{2}}
f
(
x
)
=
|
x
|
2
{\displaystyle f(x)=\left\vert x\right\vert ^{2}}
K
{\displaystyle K}
Δ
2
{\displaystyle {\frac {\Delta }{2}}}
Δ
{\displaystyle \Delta }
operador de Laplace . Por isso, pela fórmula de Dynkin,
E
a
[
f
(
B
σ
j
)
]
{\displaystyle \mathbf {E} ^{a}\left[f{\big (}B_{\sigma _{j}}{\big )}\right]}
=
f
(
a
)
+
E
a
[
∫
0
σ
j
1
2
Δ
f
(
B
s
)
d
s
]
{\displaystyle =f(a)+\mathbf {E} ^{a}\left[\int _{0}^{\sigma _{j}}{\frac {1}{2}}\Delta f(B_{s})\,\mathrm {d} s\right]}
=
|
a
|
2
+
E
a
[
∫
0
σ
j
n
d
s
]
{\displaystyle =|a|^{2}+\mathbf {E} ^{a}\left[\int _{0}^{\sigma _{j}}n\,\mathrm {d} s\right]}
=
|
a
|
2
+
n
E
a
[
σ
j
]
.
{\displaystyle =|a|^{2}+n\mathbf {E} ^{a}[\sigma _{j}].}
Assim, para qualquer
j
{\displaystyle j}
E
a
[
σ
j
]
≤
1
n
(
R
2
−
|
a
|
2
)
.
{\displaystyle \mathbf {E} ^{a}[\sigma _{j}]\leq {\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )}.}
Agora, considere
j
→
+
∞
{\displaystyle j\rightarrow +\infty }
τ
K
=
lim
j
→
+
∞
σ
j
<
+
∞
{\displaystyle \tau _{K}=\lim _{j\rightarrow +\infty }\sigma j<+\infty }
quase certamente e
E
a
[
τ
K
]
=
1
n
(
R
2
−
|
a
|
2
)
,
{\displaystyle \mathbf {E} ^{a}[\tau _{K}]={\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )},}
como afirmado.[ 2]
![{\displaystyle Af(x)=\lim _{t\downarrow 0}{\frac {\mathbf {E} ^{x}[f(X_{t})]-f(x)}{t}}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/af7b0b23105674a671237235d5ffdaf180866dfe)
![{\displaystyle \mathbf {E} ^{x}[\tau ]<+\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/002f7f648acefd34f6d41dd4e6bb59307a722c4f)
![{\displaystyle \mathbf {E} ^{x}[f(X_{\tau })]=f(x)+\mathbf {E} ^{x}\left[\int _{0}^{\tau }Af(X_{s})\,\mathrm {d} s\right].\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e47693741dc5193bcd1f3064b1215935a2e75017)
![{\displaystyle \mathbf {E} ^{a}[\tau _{K}]={\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eec6e015fb2713fe90af499c6c8525c9b378a430)
![{\displaystyle \mathbf {E} ^{a}\left[f{\big (}B_{\sigma _{j}}{\big )}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ce1999dc5da865b75eebbe2b5958fc8a312d9e1)
![{\displaystyle =f(a)+\mathbf {E} ^{a}\left[\int _{0}^{\sigma _{j}}{\frac {1}{2}}\Delta f(B_{s})\,\mathrm {d} s\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75238efd08ad750982b3ed74042a7b4bc19f9cb8)
![{\displaystyle =|a|^{2}+\mathbf {E} ^{a}\left[\int _{0}^{\sigma _{j}}n\,\mathrm {d} s\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cee70a05be0358cd3beefaebcd90aa1b7176247)
![{\displaystyle =|a|^{2}+n\mathbf {E} ^{a}[\sigma _{j}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6547df5c719935e414cc415cdfa0018d6ca9822c)
![{\displaystyle \mathbf {E} ^{a}[\sigma _{j}]\leq {\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6aef52c85273a84cb14c212f5bac1374f960163)
![{\displaystyle \mathbf {E} ^{a}[\tau _{K}]={\frac {1}{n}}{\big (}R^{2}-|a|^{2}{\big )},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a79fc8f288ff4e6d9f9c44795d58d5c7f6a21f5a)