In the previous blog I have mentioned- how Brownian motion is connected to economics and finance. The quadratic variation of the Brownian motion is the source of volatility in asset prices driven by Brownian motion.
Let \(\alpha\) and \(\sigma > 0\) be two constants and define the Geometric Brownian motion \[S(t) = S(0)\exp\{\sigma W(t)+ (\alpha-\frac{1}{2}\sigma^2)t\}\] This is the asset price model used in the Black-Scholes-Merton Option Pricing formula.
Here we show, how to use quadratic variation of the Brownian motion to identify the volatility \(\sigma\) from a path of this process.
Let \(0 \leq T_1 \leq T_2\) is given. Suppose you observe the geometric Brownian motion \(S(t)\) for \(T_1 \leq t \leq T_2\). Consider the partition \(\Pi = \{T_1 = t_0 < t_1 < t_2 < \cdots < t_n = T_2 \}\), and observe the log returns- \[\log{\frac{S(t_{j+1})}{S(t_j)}} = \sigma(W(t_{j+1})-W(t_j))+(\alpha-\frac{1}{2}\sigma^2)(t_{j+1}-t_j)\] over each of the sub-interval \([t_j,t_{j+1}]\).
Then, the sum of the square of the log returns, also called the realized volatility is-
\[\begin{aligned} &\sum_{j=0}^{n-1}(\log{\frac{S(t_{j+1})}{S(t_j)}})^2 \\ = &\sigma^2 \sum_{j=0}^{n-1} (W(t_{j+1})-W(t_j))^2 +(\alpha-\frac{1}{2}\sigma^2)^2 \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2+ 2\sigma(\alpha-\frac{1}{2}\sigma^2)\sum_{j=0}^{n-1} (W(t_{j+1})-W(t_j))(t_{j+1}-t_j) \\ \end{aligned}\]
Now, we will decrease the maximum step size of partition \(\Pi\) to zero, i.e. \(\max_{j= 0,1, \cdots \overline{m-1}}(t_{j+1}-t_j) = ||\Pi|| \rightarrow 0\). Then we see that- \[\begin{aligned} &\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(\log{\frac{S(t_{j+1})}{S(t_j)}})^2 \\ = &\sigma^2. \lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1} (W(t_{j+1})-W(t_j))^2 +(\alpha-\frac{1}{2}\sigma^2)^2. \lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(t_{j+1}-t_j)^2+ 2\sigma(\alpha-\frac{1}{2}\sigma^2).\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1} (W(t_{j+1})-W(t_j))(t_{j+1}-t_j) \\ = &\sigma^2. (T_2-T_1) +(\alpha-\frac{1}{2}\sigma^2)^2.0+ 2\sigma.(\alpha-\frac{1}{2}\sigma^2).0 \\ = &\sigma^2(T_2-T_1) \\ \end{aligned}\]
From this we can write- \[\frac{1}{T_2-T_1}\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(\log{\frac{S(t_{j+1})}{S(t_j)}})^2 \approx \sigma^2\]
If the asset price \(S(t)\) really follows geometric Brownian motion with constant volatility \(\sigma\), we can estimate \(\sigma^2\) by calculating \(\hat{\sigma}^2=\frac{1}{T_2-T_1}\lim_{||\Pi|| \rightarrow 0} \sum_{j=0}^{n-1}(\log{\frac{S(t_{j+1})}{S(t_j)}})^2\).
Today we are going to prove that the Brownian motion is a Markov process. For this we need some definitions and some lemmas which we will state below. These are very easy. Let’s start.
Independence Lemma: Let \((\Omega,\mathbb{F},\mathbb{P})\) be a probability space and let \(\mathbb{G}\) be a sub-\(\sigma\) algebra of \(\mathbb{F}\). Suppose the random variables \(X_1,X_2, \cdots,X_K\) are \(\mathbb{G}\)-measurable and the random variables \(Y_1,Y_2,\cdots,Y_L\) are independent of \(\mathbb{G}\). Let \(f(x_1,x_2,\cdots,x_K,y_1,y_2,\cdots,y_L)\) be a function of dummy variables \(x_1,x_2,\cdots,x_K\) and \(y_1,y_2,\cdots,y_L\) and define-
\[g(x_1,x_2,\cdots,x_K) = E[f(x_1,x_2,\cdots,x_K,Y_1,Y_2,\cdots,Y_L)]\] Then we can write- \[E[f(X_1,X_2, \cdots,X_K,Y_1,Y_2,\cdots,Y_L)|\mathbb{G}] = g(X_1,X_2, \cdots,X_K)\]
Markov Process: Let \((\Omega,\mathbb{F},\mathbb{P})\) be a probability space, let \(T\) be a positive number, and let- \(\mathbb{F(t)}, 0 \leq t \leq T\) be a sub-\(\sigma\)-algebras of of \(\mathbb{F}\). Consider an adapted stochastic process \(X(t), \ 0 \leq t \leq T\). Assume that for all \(0 \leq s \leq t \leq T\) and for every non-negative, Borel measurable function \(f\) there is another Borel measurable function \(g\) such that- \[E[f(X(t))| \mathbb{F(s)}] = g(X(s))\] Then we say X is a Markov process.
Theorem: Let \(W(t), \ t \geq 0\) be a Brownian motion and let \(\mathbb{F(t)}, \ t \geq 0\) be a filtration for Brownian motion. Then \(W(t), \ t \geq 0\) is a Markov process.
Proof: So, according to the definition of Markov process we just need to prove- whenever \(0 \leq s\leq t\) and a Borel measurable function \(f\), there is another Borel measurable function \(g\) such that- \[E[f(W(t))| \mathbb{F(s)}] = g(W(s))\] Let us start with the left hand side. \[\begin{aligned} E[f(W(t))| \mathbb{F(s)}] &= E[f(W(t)-W(s)+W(s))| \mathbb{F(s)}] \end{aligned}\] The \(W(t)-W(s)\) is independent of \(\mathbb{F(s)}\), and the random variable \(W(s)\) is \(\mathbb{F(s)}\)-measurable.
So we can apply the Independence lemma stated above. In order to calculate the expectation on the right hand side we need to replace the dummy variable \(W(s)\) by \(x\) to hold it constant and then take unconditional expectation of remaining R.V. i.e. \(g(x) = E[f(W(t)-W(s)+x)]\). But \(W(t)-W(s) \sim N(0,,t-s)\), therefore we can write- \[g(x) = \frac{1}{\sqrt{2\pi(t-s)}} \int_{-\infty}^{\infty} f(w+x).e^{-\frac{w^2}{2(t-s)}} \; dw\]
The Independence Lemma states that if we now take the function \(g(x)\) stated above and replace the dummy variable \(x\) by random variable \(W(s)\), then we may have- \[E[f(W(t))| \mathbb{F(s)}] = g(W(s))\]
We set \(\tau = t-s\) & \(y = w+x\) to get- \[g(x) = \frac{1}{\sqrt{2\pi\tau}} \int_{-\infty}^{\infty} f(y).e^{-\frac{(y-x)^2}{2\tau}} \; dy\] We define the transition density \(p(\tau,x,y)\) for the Brownian motion to be- \[p(\tau,x,y) = \frac{1}{\sqrt{2\pi\tau}}e^{-\frac{(y-x)^2}{2\tau}}\] Then we can write- \[\begin{aligned} g(x) &= \int_{-\infty}^{\infty} f(y).p(\tau,x,y) \; dy \\ \end{aligned}\]
Hence we can write-
\[E[f(W(t))| \mathbb{F(s)}] = \int_{-\infty}^{\infty} f(y).p(\tau,W(s),y) \; dy\] So, for every choice of non-negative Borel measurable function \(f\) we can get non-negative Borel measurable function \(g\) for which the above relation holds. Hence the theorem is proved.
The above equation has following interpretation. Conditioned on the information in \(\mathbb{F(s)}\), i.e. information obtained by observing Brownian motion, \(W(t)\) up to time \(s\), the conditional density of \(W(t)\) is given by- \(p(\tau,W(s),y)\). This is the Normal density with \(\mu = W(s)\) and variance \(\tau = t-s\).
In particular, the value of \(W(s)\) is important from \(\mathbb{F(s)}\) not the previous values in \(\mathbb{F(s)}\). This is the essence of Markov property.