In the previous blog I have proved that- sample quadratic variation of a Brownian motion \((W)\) over a partition \(\Pi\) of \([0,T]\) converges to \(T\) as \(||\Pi|| \rightarrow 0\). For that we used two things- \[\begin{aligned} E(W(t_{j+1})-W(t_{j}))^2 &= t_{j+1} - t_{j} \\ Var[(W(t_{j+1})-W(t_{j}))^2] &= 2(t_{j+1} - t_{j})^2 \\ \end{aligned}\]
Now, \(||\Pi|| \rightarrow 0\) implies for any \(j \in 0(1)\overline{n-1},\)
\[\overline{t_{j+1}-t_j} \quad \text{is very small;} \implies (t_{j+1}-t_j)^2 \quad \text{is very small}.\]
Therefore, although \((W(t_{j+1})-W(t_{j}))^2\) is random, it is near it’s mean \((t_{j+1}-t_j)\) with very high probability. So we are tempted to write- \[\begin{aligned} &(W(t_{j+1})-W(t_{j}))^2 \approx t_{j+1}-t_j \\ &\frac{(W(t_{j+1})-W(t_{j}))^2}{t_{j+1}-t_j} \approx 1 \end{aligned}\]
However, I would be very wrong to write that. \(\frac{(W(t_{j+1})-W(t_{j}))^2}{t_{j+1}-t_j}\)is not near 1, how small we make- \(\overline{t_{j+1}-t_j}\).
But we would be very very wrong to do that. Previously we have seen that for \(t_j < t_{j+1}\) we have- \[\begin{aligned} W(t_{j+1})-W(t_{j}) &\sim N(0,t_{j+1}-t_j) \\ \implies \frac{W(t_{j+1})-W(t_{j})}{\sqrt{t_{j+1}-t_j}} &\sim N(0,1) \\ \end{aligned}\]
Let us denote \(\frac{W(t_{j+1})-W(t_{j})}{\sqrt{t_{j+1}-t_j}}\) by \(Y_{j+1}\) which follows \(N(0,1)\) and its distribution remains the same, however small \(t_{j+1}-t_j\) be.
To make it simpler let us choose large value of n and define- \(t_j = \frac{jT}{n}\) for j = 0,1,… ,n. Then \(t_{j+1}-t_j = \frac{T}{n}\) for \(j = 0(1)\overline{n-1}\)
Now we can write- \[(W(t_{j+1})-W(t_{j}))^2 = \frac{T}{n}Y_{j+1}^2\]
Since \(Y_1,Y_2,\cdots,Y_n\) are independent and identically distributed i.e. i.i.d, applying the Law of Large number we can write- \[\lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} \frac{Y_{j+1}^2}{n} = E(Y_1^2) = 1\]
Hence- \[\begin{aligned} &\lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} (W(t_{j+1})-W(t_{j}))^2 \\ = &\lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} T\frac{Y_{j+1}^2}{n} = T \\ \end{aligned}\]
We formally write this as- \[dW(t)dW(t) = dt\] The above statement states that- on an interval \([0,T]\) the Brownian motion \(W\) accumulates \(T\) units of quadratic variation.
Consider an interval \([0,T_1]\). In this interval the quadratic variation of Brownian Motion is- \([W,W](T_1) = T_1\).
If we compute the quadratic variation over the interval \([0,T_2]\), where \(0 < T_1 < T_2\), we get- \([W,W](T_2) = T_2\).
Therefore, if we consider the interval \([T_1,T_2]\); take its partition \(\Pi\); square the increments of Brownian motion for each of the sub-intervals in \(\Pi\) and take \(||\Pi|| \rightarrow 0\), then we will get- \[[W,W]([T_1,T_2]) = [W,W](T_2)-[W,W](T_1) = T_2-T_1\]
So the Brownian motion accumulates \(T_2-T_1\) units of quadratic variation over the interval \([T_1,T_2]\). Since This is true for any arbitrary interval \([T_1,T_2]\) with \(0 < T_1 < T_2\) we can write-
\[\text{Brownian motion accumulates quadratic variation at rate one per unit time}\]
We are seeing all these things in Brownian motion… It started with physics; but how does it connects with economics and finance?
Here is the important part. I hope you are familiar with the terms like interest rates, assets etc.
\[\text{The quadratic variation of the Brownian motion is the source of volatility in asset prices driven by Brownian motion}\]
Wow!! Right??
In upcoming blogs you will see- we shall eventually scale Brownian motion, sometimes in time dependent ways, sometimes in time dependent ways in order to vary the rate- at which volatility enters these asset prices.
Now, for a partition \(\Pi = \{0=t_0,t_1,\cdots,t_n= T\}\) of \([0,T]\) we already have quadratic variation of \(W(t)\)- \[\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_{j}))^2 = T\]
Similarly we can calculate cross variation of \(W(t)\) with t and quadratic variation of \(t\) with itself. These are denoted as follows-
\[\begin{aligned} &\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j) \quad \text{and} \\ &\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (t_{j+1}-t_j)^2 \end{aligned}\]
Since expressed \(\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_{j}))^2\) by \(dW(t)dW(t)\), we can express the following- \[\begin{aligned} &\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j) = dW(t).dt \quad \text{and} \\ &\lim_{||\Pi|| \rightarrow 0}\sum_{j=0}^{n-1} (t_{j+1}-t_j)^2 = dt.dt \end{aligned}\]
For a partition \(\Pi = \{0=t_0,t_1,\cdots,t_n= T\}\) of \([0,T]\) we can write- \[\begin{aligned} |(W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j)| &\leq \max_{0 \leq k\leq n-1}|W(t_{k+1})-W(t_{k})|.(t_{j+1}-t_j) \\ \implies |\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j)| &\leq \max_{0 \leq k\leq n-1}|W(t_{k+1})-W(t_{k})|.\sum_{j=0}^{n-1}(t_{j+1}-t_j) \\ &\leq \max_{0 \leq k\leq n-1}|W(t_{k+1})-W(t_{k})|.T \\ \end{aligned}\]
Since \(W\) is continuous- \[\lim_{||\Pi|| \rightarrow 0} \max_{0 \leq k\leq n-1}|W(t_{k+1})-W(t_{k})| = 0\] So we have- \[\begin{aligned} 0 \leq &|\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j)| \leq \max_{0 \leq k\leq n-1}|W(t_{k+1})-W(t_{k})|.T = 0 \\ \implies &|\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j)| = 0 \\ \implies &\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_{j}))(t_{j+1}-t_j) = 0 \\ \implies &dW(t).dt = 0 \\ \end{aligned}\]
For a partition \(\Pi = \{0=t_0,t_1,\cdots,t_n= T\}\) of \([0,T]\) we can write- \[\begin{aligned} (t_{j+1}-t_j)^2 &\leq \{\max_{0 \leq k\leq n-1}(t_{j+1}-t_j)\}.(t_{j+1}-t_j) \\ \implies \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2 &\leq ||\Pi||.\sum_{j=0}^{n-1}(t_{j+1}-t_j) \\ \implies \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2 &\leq ||\Pi||.T \\ \end{aligned}\]
Hence we have- \[\begin{aligned} 0 \leq &\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2 \leq \lim_{||\Pi|| \rightarrow 0}||\Pi||.T = 0 \\ \implies &\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2 = dt.dt = 0 \end{aligned}\]
These are very important results. Rather than computing every time you may remember the following for faster calculation-
\[\begin{aligned} &dW(t).dW(t) = dt \\ &dW(t).dt = 0 \\ &dt.dt = 0 \end{aligned}\]