Brownian motion(Wiener process)과 quadratic variation

1. Definition of Brownian motion

어떤 실수 값으로 된 랜덤 프로세스(Stochastic process)가 다음의 성질(property)을 따를 때 브라운 운동(Brownian motion)이라 칭한다.

1. $W(0)=0$

2. $W(t)-W(s) \sim N(0, t-s)$, for t > s

3. $W(t_2)-W(t_1), W(t_3)-W(t_2), \cdots, W(t_n)-W(t_{n-1}) $들은 모두 독립이다. 

위 2번째 성질에서 브라운 운동의 증분은 그 정의상 평균은 0, 분산은 time index의 거리가 된다.
즉 다음과 같이 일반화할 수 있다.

$$
E[\Delta W] = 0 \\
$$
$$
\begin{align}
Var[\Delta W] &= E[(\Delta W)^2] - (E[\Delta W])^2\\
&= E[(\Delta W)^2]\\
&= \Delta t
\end{align}
$$

$$
\therefore E[(\Delta W)^2]= \Delta t
$$

여기서 브라운 운동의 증분을 제곱하면 랜덤성이 사라지고 '평균적으로' 증분만큼의 시간으로 deterministic하게 나온다는 중요한 성질을 엿볼 수 있다. 저기서 기댓값만 없애면 "브라운 운동 증분의 제곱은 그 시간 차이만큼의 제곱이다"라고 명징하게 주장할 수 있을텐데 그 부분이 아쉽다. 저 기댓값을 처리할 수는 없을까?
이를 확인하기 위해선 먼저 브라운 운동 증분의 모멘트들을 살펴봐야한다.

 

2. Moment of Brownian Motion

먼저 모멘트 생성 함수(Moment Generating function)를 리뷰해보자. 확률 변수 $z$에 대한 모멘트 생성 함수는 다음과 같은 꼴을 갖는다.

$$
M_Z(\theta) = E[e^{\theta z}]
$$

이를 모멘트 생성 함수라 부르는 이유는, $M_Z(\theta)$를 미분할수록 테일러 근사에 의해 아래의 꼴을 갖는데,
$$
\frac{d}{d\theta}M_Z(\theta) = E[z] + \theta E[z^2] + \frac{\theta^2}{2!} E[z^3] + \frac{\theta^3}{3!} E[z^4] + \cdots
$$


$$
\frac{d^2}{d\theta^2}M_Z(\theta) = E[z^2] + \frac{\theta}{2!} E[z^3] + \frac{\theta^2}{3!} E[z^4] + \cdots
$$

$$
\vdots
$$

여기서 $\theta$를 0으로 두면 $z$에 대한 모멘트들이 생성되기 때문이다.
$$
\frac{d}{d\theta}M_Z(\theta) \bigg |_{\theta=0}=E[z]
$$

$$
\frac{d^2}{d\theta^2}M_Z(\theta) \bigg |_{\theta=0}=E[z^2]
$$

$$
\frac{d^3}{d\theta^3}M_Z(\theta) \bigg |_{\theta=0}=E[z^3]
$$

$$
\vdots
$$

이제 브라운 운동의 모멘트 생성 함수를 알아보자. "Geometric Brownian Motion의 해, 평균, 분산" 포스팅의 식 (20)에서 지수로 정규분포가 포함된 꼴에 대해서 $E[e^Y]=e^{E[Y]+\frac{1}{2}Var[Y]}$를 갖게 되는 걸 살펴봤다.
이에 따라 브라운 운동의 모멘트 생성 함수는 다음과 같다.

$$
E[e^{\theta W_t}] = e^{0+\frac{1}{2}\theta^2 t} = e^{\frac{1}{2}\theta^2 t}
$$

이제 여기서 1,2,3,4차 모멘텀을 구해보자.

$$
\begin{align}
E[W_t]
&=\frac{d}{d\theta} e^{\frac{1}{2}\theta^2 t} \bigg |_{\theta=0} \\
&= e^{\frac{1}{2}\theta^2 t} \frac{d}{d\theta} \left ( \frac{1}{2}\theta^2 t \right ) \bigg |_{\theta=0}\\
&= e^{\frac{1}{2}\theta^2 t} \theta t \bigg |_{\theta=0} \\
&= 0 \\
E[W_t^2]
&=\frac{d^2}{d\theta^2} e^{\frac{1}{2}\theta^2 t} \bigg |_{\theta=0} \\
&=\frac{d}{d\theta} e^{\frac{1}{2}\theta^2 t} \theta t \bigg |_{\theta=0} \\
&=\left ( \frac{d}{d\theta} \left ( e^{\frac{1}{2}\theta^2 t} \right )\theta t + e^{\frac{1}{2}\theta^2 t} \frac{d}{d\theta} \left ( \theta t \right ) \right ) \bigg |_{\theta=0}\\
&=\left ( e^{\frac{1}{2}\theta^2 t} \theta^2 t^2 + e^{\frac{1}{2}\theta^2 t} t \right ) \bigg |_{\theta=0} \\
&= t \\
E[W_t^3]
&=\frac{d^3}{d\theta^3} e^{\frac{1}{2}\theta^2 t} \bigg |_{\theta=0} \\
&=\frac{d}{d\theta} \left ( e^{\frac{1}{2}\theta^2 t} \theta^2 t^2 + e^{\frac{1}{2}\theta^2 t} t \right ) \bigg |_{\theta=0} \\
&=\left ( \frac{d}{d\theta} \left ( e^{\frac{1}{2}\theta^2 t} \right ) \theta^2 t^2 + e^{\frac{1}{2}\theta^2 t} \frac{d}{d\theta} \theta^2 t^2 + \frac{d}{d\theta} \left ( e^{\frac{1}{2}\theta^2 t} \right ) t + e^{\frac{1}{2}\theta^2 t} \frac{d}{d\theta} t\right ) \bigg |_{\theta=0} \\
&= \left (  e^{\frac{1}{2}\theta^2 t} \theta^3 t^3 + e^{\frac{1}{2}\theta^2 t}  2\theta t^2 +  e^{\frac{1}{2}\theta^2 t} \theta t^2 \right ) \bigg |_{\theta=0} \\
&= \left ( e^{\frac{1}{2}\theta^2 t} \left ( \theta^3 t^3 + 3\theta t^2 \right ) \right ) \bigg |_{\theta=0} \\
&= 0 \\
E[W_t^4]
&=\frac{d^4}{d\theta^4} e^{\frac{1}{2}\theta^2 t} \bigg |_{\theta=0} \\
&=\frac{d}{d\theta} \left ( e^{\frac{1}{2}\theta^2 t} \left ( \theta^3 t^3 + 3\theta t^2 \right ) \right ) \bigg |_{\theta=0} \\
&=\left ( \frac{d}{d\theta} \left (  e^{\frac{1}{2}\theta^2 t} \right ) \left ( \theta^3 t^3 + 3\theta t^2 \right ) + e^{\frac{1}{2}\theta^2 t} \frac{d}{d\theta} \left ( \theta^3 t^3 + 3\theta t^2  \right ) \right ) \bigg |_{\theta=0}\\
&= \left ( e^{\frac{1}{2}\theta^2 t} \theta t  \left ( \theta^3 t^3 + 3\theta t^2 \right ) + e^{\frac{1}{2}\theta^2 t} \left ( 3\theta^2 t^3 + 3 t^2  \right ) \right ) \bigg |_{\theta=0} \\
&= 3t^2
\end{align}
$$

$$
\therefore  E[W_t] = 0, \quad E[W_t^2]=t, \quad E[W_t^3]=0, \quad E[W_t^4]=3t^2
$$

 

3. Quadratic variation of Brownian motion

이제 브라운 운동의 증분에 대해 분석할 툴들이 모두 갖춰졌다. 브라운 운동 $W_t$라고 하는 것은 앞서 살펴보았던 정의 상 시점 $0$부터 $t$까지의 증분을 의미한다. 여기서 증분에 대해 $\Delta t$, 즉 브라운 운동의 증분을 $t$부터 $t+\Delta t$의 시점까지라고 일반화하면 2번의 모멘트들 또한 브라운 운동 증분의 모멘트들로 일반화된다.

$$
\therefore  E[\Delta W] = 0, \quad E[\Delta W^2]=\Delta t, \quad E[\Delta W^3]=0, \quad E[\Delta W^4]=3(\Delta t)^2
$$


그리고 $(\Delta W)^2$의 평균과 분산에 대해 분석해보자. 평균은 앞서 정의한대로 $E[(\Delta W)^2]= \Delta t$가 된다. 분산은 아래와 같다.

$$
\begin{align}
Var[(\Delta W)^2] &= E[(\Delta W)^4]-(E[(\Delta W)^2])^2 \\
&= 3(\Delta t)^2 - (\Delta t)^2\\
&= 2 (\Delta t)^2
\end{align}
$$

이때 $\Delta t$를 0으로 극한을 보내어 무한소($dt$)만큼 작은 시간으로 정의하면 $(dW)^2$의 분산은 0으로 수렴한다. 즉 $(dW)^2$의 평균은 $dt$가 되고 분산은 0이 되므로 $(dW)^2$는 무한소 단위의 증분에서 기댓값이 필요 없이 확정적인 $dt$가 되는 것이다.

그러므로 우리는 본 포스팅에서 다음과 같은 결론을 얻을 수 있다.
$$
\therefore E[dW]=0 , \quad (dW)^2=dt
$$