Brownian Motion

Quadratic Variation of Brownian motion- Part-2

In the previous blog I have showed you- how to calculate the First Order Variation. Similarly we are going to calculate the Second Order Variation/ Quadratic Variation.

Quadratic Variation: Let \(f(t)\) be a function defined in \(t \in [0,T]\). The Quadratic Variation of \(f\) up to time T is defined as- \[QV_f(T) = \lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}[f(t_{j+1})-f(t_{j})]^2\] where, \(\Pi = \{0=t_1,t_2, \cdots ,t_n = T \}\) and \(t_0 < t_1 < t_2 < \cdots < t_n = T\). Quadratic Variation of a function \(f\) up to time T is also denoted by- \([f,f](T)\).

Theorem: The Quadratic Variation of a function \(f\), which has a continuous derivative is 0; i.e. \([f,f](T) = 0\).

Proof: Let us start with a function- \(f\), which have a continuous derivative. Then we can write- \[\begin{aligned} \sum_{j=0}^{n-1}[f(t_{j+1})-f(t_{j})]^2 &= \sum_{j=0}^{n-1}|f(t_j^{*})|^2.(t_{j+1}-t_{j})^2 \qquad \text{[Using the Mean Value Theorem]} \\ &\leq \max_{j = 1(1)(n-1)}(t_{j+1}-t_{j}).\sum_{j=0}^{n-1}|f(t_j^{*})|^2.(t_{j+1}-t_{j}) \\ &= ||\Pi||.\sum_{j=0}^{n-1}|f(t_j^{*})|^2.(t_{j+1}-t_{j}) \\ \end{aligned}\]

Now- taking the \(\lim_{||\Pi|| \rightarrow 0}\) for both sides in the above inequality we get- \[\begin{aligned} \lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}[f(t_{j+1})-f(t_{j})]^2 &\leq (\lim_{||\Pi|| \rightarrow 0}||\Pi||).\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1}|f(t_j^{*})|^2.(t_{j+1}-t_{j}) \\ &= (\lim_{||\Pi|| \rightarrow 0}||\Pi||).\int_{0}^{T} |f'(t)|^2 \, dt \\ \end{aligned}\]

Now since we have assumed that \(f\) has a continuous derivative, we can always write- \[\int_{0}^{T} |f'(t)|^2 \, dt < \infty\]

So we have- \[\lim_{||\Pi|| \rightarrow 0}||\Pi||.(\int_{0}^{T} |f'(t)|^2 \, dt) = 0\].

So we can write- \[\begin{aligned} 0 \leq \lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}[f(t_{j+1})-f(t_{j})]^2 &\leq (\lim_{||\Pi|| \rightarrow 0}||\Pi||).\int_{0}^{T} |f'(t)|^2 \, dt = 0\\ \end{aligned}\]

Which implies- \([f,f](T) = 0\), Hence proved.

But, at the end of the proof we used the fact that \(f'(t)\) is continuous to ensure that \(\int_{0}^{T} |f'(t)|^2 \, dt < \infty\). If \(\int_{0}^{T} |f'(t)|^2 \, dt\) is infinite, then- \[\lim_{||\Pi|| \rightarrow 0}\{||\Pi||.\sum_{j=0}^{n-1}|f(t_j^{*})|^2.(t_{j+1}-t_{j})\}\] leads to a \(0.\infty\) situation which is undefined.

Now most functions we deal with (like \(ax^2+bx+c, \sin(x)\) etc.) have continuous derivatives, i.e. why the Mean Value Theorem holds for these function. That’s why the Quadratic variation of these function are zero.

For this reason, we never bothered to calculate Quadratic Variation in ordinary calculus.

But the path of the Brownian motion can not be differentiated with respect to the time variable. So, the Mean Value Theorem does not exist; hence the previous theorem is not applicable for Brownian motion. The best example is to take- \(f(t) = |t|\), whose derivative is undefined at t = 0. In the following picture the chord connecting- \((t_1,f(t_1))\) and \((t_2,f(t_2))\) has slope = \(\frac{1}{5}\). But no where between \(t_1\) and \(t_2\) the function \(|t|\) has derivative = \(\frac{1}{5}\). As- \[\begin{aligned} \dfrac{\mathrm{d}|t|}{\mathrm{d}t} &= \qquad +1 \qquad \qquad \quad \text{if t > 0} \\ &= \qquad -1 \qquad \qquad \quad \text{if t < 0} \\ &= \qquad \text{undeined} \ \ \qquad \text{if t = 0} \\ \end{aligned}\]

For a Brownian motion path W(t), there is no value for which \(\dfrac{\mathrm{d}W(t)}{\mathrm{d}t}\) is defined. So W(t) has a non-zero quadratic variation. In the next blog I am going to calculate the quadratic variation of the Brownian motion, i.e. \([W,W](T)\). Happy reading!!