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.

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

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

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


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

$$⋮$$

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

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

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

$$⋮$$

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.

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

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

 

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 $\mathrm{\Delta }t$, 즉 브라운 운동의 증분을 $t$부터 $t+\mathrm{\Delta }t$의 시점까지라고 일반화하면 2번의 모멘트들 또한 브라운 운동 증분의 모멘트들로 일반화된다.

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


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

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

이때 $\mathrm{\Delta }t$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.
$$\therefore E[dW]=0,(dW{)}^2=dt$$