Stochastic process is not just the Brownian motion, it contains many more things. Today what we are going to is- find the Ito-Doeblin formula for much more generalized version of stochastic process, which is called the Ito Process.
Ito Processes: Let \(\mathbb{W(t), \ t \geq 0}\) be a Brownian motion, and let \(\mathbb{F(t), \ t \geq 0}\) be an associated filtration. An Ito process is a stochastic process of the form-
\[\mathbb{X(t)} = \mathbb{X(0)} + \ \mathbb{\int_0^{t} \Delta(u).dW(u) + \ \int_0^{t} \Theta(u).du}\]
where \(\mathbb{X(0)}\) is non-random and \(\mathbb{\Delta(u)}\) and \(\mathbb{\Theta(u)}\) are adapted stochastic processes.
We also assume few things about these functions. We assume- for all \(\mathbb{t > 0}\),
\(\mathbb{E\int_0^{t} \Delta^2(u).du < \infty}\)
\(\mathbb{\int_0^{t} |\Theta(u)|.du < \infty}\)
This conditions are needed so that the integrals on the right hand side are defined and the Ito integral is a martingale.
Now, similar to the Brownian motion are going to see various properties of the Ito Process. For example- what is the quadratic variation of Ito process, is the Ito process a martingale etc.
First, let us see the quadratic variation of an Ito Process. For that we need to calculate accumulated quadratic variation of Ito process.
To simplify the calculation, we are going do define a few notations as follows-
\[\begin{aligned} &\mathbb{I(t)} = \mathbb{\int_{0}^{t}\Delta(u).dW(u)} \\ &\mathbb{R(t)} = \mathbb{\int_{0}^{t}\Theta(u).du} \\ \end{aligned}\]
Both these processes are continuous in their upper limit of integration \(\mathbb{t}\).
To determine the quadratic variation of \(\mathbb{X}\) on the interval \(\mathbb{[0,t]}\), we choose a partition \(\mathbb{\Pi = \{0=t_0, t_1, \cdots,t_n = T \}}\).
For that particular partition, we can define the sample quadratic variation as-
\[\begin{aligned} \mathbb{\sum_{j=0}^{n-1}[X(t_{j+1})-X(t_{j})]^2} &= \mathbb{\sum_{j=0}^{n-1}[I(t_{j+1})-I(t_{j})]^2} + \mathbb{\sum_{j=0}^{n-1}[R(t_{j+1})-R(t_{j})]^2} \\ &+ 2.\mathbb{\sum_{j=0}^{n-1}[I(t_{j+1})-I(t_{j})].[R(t_{j+1})-R(t_{j})]} \end{aligned}\]
Here we have three parts. Let’s see what happens to each parts when the- \(\mathbb{||\Pi|| \rightarrow 0}\).
First part,
It is nothing but the quadratic variation of the generalized Ito integral. \[\mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}[I(t_{j+1})-I(t_{j})]^2} = \mathbb{[I,I](t)} = \mathbb{\int_0^{t}\Delta^2(u).du}\] Second part, \[\begin{aligned} \mathbb{\sum_{j=0}^{n-1}[R(t_{j+1})-R(t_{j})]^2} \ &\leq \ \mathbb{\underset{0 \leq k \leq n-1}{\max} |R(t_{k+1})-R(t_{k})|} . \ \mathbb{\sum_{j=0}^{n-1}|R(t_{j+1})-R(t_{j})|} \\ &= \ \mathbb{\underset{0 \leq k \leq n-1}{\max} |R(t_{k+1})-R(t_{k})|} . \ \mathbb{\sum_{j=0}^{n-1}|\int_{t_j}^{t_{j+1}}\Theta(u).du|} \\ &\leq \ \mathbb{\underset{0 \leq k \leq n-1}{\max} |R(t_{k+1})-R(t_{k})|} . \ \mathbb{\sum_{j=0}^{n-1}\int_{t_j}^{t_{j+1}}|\Theta(u)|.du} \\ &\leq \ \mathbb{\underset{0 \leq k \leq n-1}{\max} |R(t_{k+1})-R(t_{k})|} . \ \mathbb{\int_{0}^{t}|\Theta(u)|.du} \\ \end{aligned}\]
Now, from the definition of Ito process we know that- \(\mathbb{\int_0^{t} |\Theta(u)|.du < \infty}\) for all \(\mathbb{t > 0}\).
Again since \(\mathbb{R(t)}\) is a continuous function of \(\mathbb{t}\), which implies- \[\mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \quad \mathbb{\underset{0 \leq k \leq n-1}{\max} |R(t_{k+1})-R(t_{k})|} = 0\] which lead us to-
\[\begin{aligned} \mathbb{\sum_{j=0}^{n-1}[R(t_{j+1})-R(t_{j})]^2} \ &\leq 0.\mathbb{\int_0^{t} |\Theta(u)|.du = 0} \end{aligned}\]
Third part,
\[\begin{aligned} 2.\mathbb{\sum_{j=0}^{n-1}[I(t_{j+1})-I(t_{j})].[R(t_{j+1})-R(t_{j})]} & \leq 2 \ \mathbb{\underset{0 \leq k \leq n-1}{\max}|I(t_{k+1})-I(t_{k})|} \ \mathbb{\sum_{j=0}^{n-1}|R(t_{j+1})-R(t_{j})|} \\ & \leq 2 \ \mathbb{\underset{0 \leq k \leq n-1}{\max}|I(t_{k+1})-I(t_{k})|} \ \mathbb{\sum_{j=0}^{n-1}|\int_{t_{j}}^{t_{j+1}}\Theta(u).du|} \\ & \leq 2 \ \mathbb{\underset{0 \leq k \leq n-1}{\max}|I(t_{k+1})-I(t_{k})|} \ \mathbb{\sum_{j=0}^{n-1}\int_{t_{j}}^{t_{j+1}}|\Theta(u)|.du} \\ & \leq 2 \ \mathbb{\underset{0 \leq k \leq n-1}{\max}|I(t_{k+1})-I(t_{k})|} \ \mathbb{\int_{0}^{t}|\Theta(u)|.du} \\ \end{aligned}\]
Similar to the second case, as \(\mathbb{I(t)}\) is a continuous function of \(\mathbb{t}\), which implies- \[\mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \quad \mathbb{\underset{0 \leq k \leq n-1}{\max} |I(t_{k+1})-I(t_{k})|} = 0\]
which lead us to-
\[2.\mathbb{\sum_{j=0}^{n-1}[I(t_{j+1})-I(t_{j})].[R(t_{j+1})-R(t_{j})]} \leq 0.\mathbb{\int_0^{t} |\Theta(u)|.du = 0}\]
Combining the three parts we can write- \[\mathbb{[X,X](t) = [I,I](t) +0 + 0 = \int_0^{t}\Delta^2(u).du} \qquad \text{Q.E.D.}\]
But you can remember the whole thing with a very simple way. Do you remember the following table-
\[\mathbb{dW(t).dW(t) = dt} \qquad \mathbb{dW(t).dt = 0} \qquad \mathbb{dt.dt = 0}\] We ar gonna use this.
The differential form of an Ito process can be written as- \[\mathbb{dX(t) = \Delta(t)\ dW(t)\ + \Theta(t)\ dt}\]
Then are going to write the quadratic variation is differential form as- \[\begin{aligned} \mathbb{dX(t)\ dX(t)} &= \mathbb{\Delta^2(t)\ dW(t)\ dW(t)\ + 2\Delta(t)\ \Theta(t)\ dW(t)\ dt \ + \Theta^2(t)\ dt\ dt } \\ &= \mathbb{\Delta^2(t)\ dt} \\ \end{aligned}\]
This says that, at each time \(\mathbb{t}\), the process \(\mathbb{X(t)}\) is accumulating quadratic at a rate \(\mathbb{\Delta^2(t)/}\text{unit time}\) variation at rate.