In the previous blog, we saw how Stochastic calculus is different from ordinary calculus. That’s why now we have to define all the rules for Stochastic calculus. It will not be completely different, but a kind of an extension.
In ordinary calculus we first learn about differentiation; then we move towards the integration part. Let us start with the Ito- Doeblin formula. It is the rule of differentiation in the world of Stochastic calculus.
Ito- Doeblin formula, named after famous Japanese mathematician Kiyoshi Ito and French-German mathematician Wolfgang Doeblin. Wolfgang Doeblin was a Jewish-German mathematician. His family escaped from Nazi Germany to France where he became a French citizen.
He wrote down his latest work on the Chapman–Kolmogorov equation, and sent this as a “pli cacheté” (sealed envelope) to the French Academy of Sciences. He became an untimely victim of World War-II at the age of 25.
The sealed envelope was opened in 2000, revealing that Doeblin had obtained major results on the Chapman-Kolmogorov equation and the theory of diffusion processes. [Source: Wikipedia]
We want a rule to differentiate expressions of the form \(\mathbb{f(W(t))}\), where \(\mathbb{f(x)}\) is a differentiable function and \(\mathbb{W(t), \ t > 0}\) is a Brownian motion.
If \(\mathbb{f(W(t))}\) was also differentiable, the chain rule of ordinary calculus will give us- \[\mathbb{\frac{d}{dt}f(W(t)) = f'(W(t)).W'(t)}\] In differential notation, this can also be written as- \[\mathbb{df(W(t))= f'(W(t)).W'(t).dt = f'(W(t)).dW(t)}\]
Since Brownian motion \(\mathbb{(W(t))}\) is continuous but not differentiable. \(\mathbb{W(t)}\) has non-zero quadratic variation, i.e. why differential notation has to be corrected a little bit. It becomes-
\[\mathbb{df(W(t)) = f'(W(t)).dW(t)+ \frac{1}{2}f''(W(t)).dt}\] It’s like we are considering up to the second term of the Taylor series expansion due to the non-zero quadratic variation. This is the Ito-Doeblin formula in the differential form. Integrating this, we obtain the Ito-Doeblin formula in the integral form which is-
\[\begin{aligned} &\mathbb{\int_{0}^{t} df(W(u))} = \mathbb{\int_{0}^{t}f'(W(u)).dW(u)+ \frac{1}{2}\int_{0}^{t}f''(W(u)).du} \\ \implies &\mathbb{f(W(t))-f(W(0))} = \mathbb{\int_{0}^{t}f'(W(u)).dW(u)+ \frac{1}{2}\int_{0}^{t}f''(W(u)).du} \\ \end{aligned}\]
The integral-form is more meaningful form of the Ito-Doeblin formula. Because the right hand side now have precise definition.
\(\mathbb{\int_{0}^{t}f'(W(u)).dW(u)}\)- is an Ito integral.
\(\mathbb{\int_{0}^{t}f''(W(u)).du}\)- is a orinary integral or Lebesgue integral with respect to time.
Theorem: Ito-Doeblin formula for Brownian Motion
Let \(\mathbb{f(t,x)}\) be a function for which the partial derivatives-
\[\begin{aligned} \mathbb{\frac{\delta}{\delta x}f(t,x)} = \mathbb{f_x(t,x)}; \quad \mathbb{\frac{\delta}{\delta t}f(t,x)} = \mathbb{f_t(t,x)}; \quad \mathbb{\frac{\delta^2}{\delta^2 x}f(t,x)} = \mathbb{f_{xx}(t,x)} \\ \end{aligned}\] are defined and continuous, and let \(\mathbb{W(t), \ t> 0}\) be a Brownian motion. Then for every \(\mathbb{T \geq 0}\)-
\[\begin{aligned} \mathbb{f(T,W(T))} =& \ \mathbb{f(0,W(0)) + \int_{0}^{T}f_t(t,W(t)).dt} \\ & + \mathbb{\int_{0}^{T}f_x(t,W(t)).dW(t)} + \frac{1}{2}\int_{0}^{T}f_{xx}(t,W(t)).dt \end{aligned}\]
Proof: First we are going to show that the above equation holds for- \(\mathbb{f(x) = \frac{1}{2}x^2}\). In this case- \[\mathbb{f'(x) = x}\quad \text{;} \quad \mathbb{f''(x) = 1} \quad \text{&} \quad \mathbb{f^{(j)}(x) = 0} \quad \text{for} \ \mathbb{j \geq 3}\]
Let \(\mathbb{x_{j+1}}\) and \(\mathbb{x_{j}}\) be numbers. Then the Taylor’s rule implies-
\[\mathbb{f(x_{j+1})-f(x_{j})} = \mathbb{f'(x_{j}).(x_{j+1}-x_{j}) + \frac{1}{2!}f''(x_{j}).(x_{j+1}-x_{j})^2}\]
In this case, we have expanded the Taylor series expansion up to second order because- \(\mathbb{f^{(3)}(.)}\) and higher order terms are zero.
Fix \(\mathbb{T > 0}\). Let \(\mathbb{\Pi = \{t_0,t_1, \cdots, t_n\}}\) be a partition of \(\mathbb{[0,T]}\). We are interested in the difference between \(\mathbb{f(W(T))}\) and \(\mathbb{f(W(0))}\). We can write this as-
\[\begin{aligned} & \ \mathbb{f(W(T))-f(W(0))} \\ = & \ \mathbb{\sum_{j=0}^{n-1}f(W(t_{j+1}))-f(W(t_j))} \\ = & \ \mathbb{\sum_{j=0}^{n-1}f'(W(t_j)).(W(t_{j+1})-W(t_j))} \\ & \ +\mathbb{\sum_{j=0}^{n-1}\frac{f''(W(t_j))}{2!}.(W(t_{j+1})-W(t_j))^2} \end{aligned}\]
For the function \(\mathbb{f(x) = \frac{1}{2}x^2}\), the above expression will be-
\[\mathbb{\sum_{j=0}^{n-1}W(t_j).(W(t_{j+1})-W(t_j))} \ +\mathbb{\sum_{j=0}^{n-1}\frac{1}{2}.(W(t_{j+1})-W(t_j))^2}\]
If we let \(\mathbb{||\Pi|| \rightarrow 0}\), we get-
\[\begin{aligned} \mathbb{f(W(T))-f(W(0))} &= \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}W(t_j).(W(t_{j+1})-W(t_j))} \\ &+ \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\frac{1}{2}.\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))^2} \\ &= \mathbb{\int_{0}^{T}W(t).dW(t) + \ \frac{1}{2}T} \\ &= \mathbb{\int_{0}^{T}f'(W(t)).dW(t) + \ \frac{1}{2} \int_{0}^{T}f''(W(t)).dt} \\ \end{aligned}\]
You can see- \[\text{RHS} = \text{Ito integral} + \frac{1}{2}\ \text{quadratic variation of Brownian motion}\]
This is the Ito- Doeblin formula in integral form, for the function \(\mathbb{f(x) = \frac{1}{2}x^2}\).
If we have much general form of function, What will you do?
Let’s go back and see where things will change from the previous case.
First, start with the Taylor series expansion for two numbers \(\mathbb{x_{j+1}}\) and \(\mathbb{x_{j}}\). Previously we omitted the terms 3-rd and higher order terms because they were zero. Now we have to consider them.
\[\begin{aligned} &\mathbb{f(W(t_{j+1}))-f(W(t_{j}))} \\ = \ &\mathbb{f'(W(t_{j})).(W(t_{j+1})-W(t_{j})) + \frac{1}{2!}f''(W(t_{j})).(W(t_{j+1})-W(t_{j}))^2} \\ & \ + \mathbb{\frac{1}{3!}f^{(3)}(W(t_{j})).(W(t_{j+1})-W(t_{j}))^3} + \cdots \end{aligned}\]
But consider the following sum-
\[\begin{aligned} \mathbb{\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_{j})|^3} &\leq \ \mathbb{\underset{0 \leq k \leq n-1}{\max}|W(t_{k+1})-W(t_{k})|}.\mathbb{\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_{j})|^2} \\ \implies \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}}\mathbb{\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_{j})|^3} &\leq \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \{ \mathbb{\underset{0 \leq k \leq n-1}{\max}|W(t_{k+1})-W(t_{k})|}\}.\mathbb{\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_{j})|^2} \\ \implies \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}|W(t_{j+1})-W(t_{j})|^3} &\rightarrow \ 0 \\ \end{aligned}\]
Similar for higher order terms.
So in general case also, the third and higher order terms don’t make any contribution to the final answer.
Now we take bi-variate function \(\mathbb{f(t,x)}\), which takes both time variable \(\mathbb{t}\) and other variable \(\mathbb{x}\). Then the Taylor series expansion says-
\[\begin{aligned} \mathbb{f(t_{j+1},x_{j+1})-f(t_{j},x_{j})} &= \mathbb{f_t(t_{j},x_{j}).(t_{j+1}-t_{j})} + \mathbb{f_x(t_{j},x_{j}).(x_{j+1}-x_{j})} \\ & \ + \frac{1}{2}\mathbb{f_{tt}(t_{j},x_{j}).(t_{j+1}-t_{j})^2} + \frac{1}{2}\mathbb{f_{xx}(t_{j},x_{j}).(x_{j+1}-x_{j})^2} \\ & \ + \mathbb{f_{tx}(t_{j},x_{j}).(t_{j+1}-t_{j}).(x_{j+1}-x_{j})} + \text{higher order terms} \end{aligned}\]
Here we replace \(\mathbb{x_j}\) by \(\mathbb{W(t_j)}\) and \(\mathbb{x_{j+1}}\) by \(\mathbb{W(t_{j+1})}\) and sum over \(\mathbb{j}\), we get-
\[\begin{aligned} & \mathbb{f(T,W(T))-f(0,W(0))} \\ = & \ \mathbb{\sum_{j=0}^{n-1}[f(t_{j+1},x_{j+1})-f(t_{j},x_{j})]} \\ = & \ \mathbb{\sum_{j=0}^{n-1}f_t(t_{j},W(t_{j})).(t_{j+1}-t_{j})} + \mathbb{\sum_{j=0}^{n-1}f_x(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))} \\ & + \ \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{tt}(t_{j},W(t_{j})).(t_{j+1}-t_{j})^2} + \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{xx}(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))^2} \\ & + \mathbb{\sum_{j=0}^{n-1} f_{tx}(t_{j},W(t_{j})).(t_{j+1}-t_{j}).(W(t_{j+1})-W(t_{j}))} + \text{higher order terms} \end{aligned}\]
Now if \(\mathbb{||\Pi|| \rightarrow 0}\), the first part \(\mathbb{f(T,W(T))-f(0,W(0))}\) remains un-changed. RHS-
First part, \[\begin{aligned} \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}f_t(t_{j},W(t_{j})).(t_{j+1}-t_{j})} = \mathbb{\int_{0}^{T} f_t(t,W(t)).dt} \end{aligned}\]
Second part, \[\begin{aligned} \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1}f_x(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))} = \mathbb{\int_{0}^{T}f_x(t,W(t)).dW(t)} \end{aligned}\]
Third part, \[\begin{aligned} & \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{xx}(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))^2} \\ = & \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{xx}(t_{j},W(t_{j})).(t_{j+1}-t_{j})} \\ = & \ \mathbb{\frac{1}{2}\int_{0}^{T}f_{xx}(t,W(t)).dt} \end{aligned}\]
Fourth part, \[\begin{aligned} & \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1} f_{tx}(t_{j},W(t_{j})).(t_{j+1}-t_{j}).(W(t_{j+1})-W(t_{j}))} \\ \leq & \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{|\sum_{j=0}^{n-1} f_{tx}(t_{j},W(t_{j})).(t_{j+1}-t_{j}).(W(t_{j+1})-W(t_{j}))|} \\ \leq & \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1} |f_{tx}(t_{j},W(t_{j}))|.|(t_{j+1}-t_{j}).|W(t_{j+1})-W(t_{j})|} \\ \leq & \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ (\mathbb{\underset{0 \leq k \leq n-1}{\max}|W(t_{k+1})-W(t_{k})|}). \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \mathbb{\sum_{j=0}^{n-1} |f_{tx}(t_{j},W(t_{j}))|.(t_{j+1}-t_{j})} \\ = & \ 0. \mathbb{\int_{0}^{T}|f_{tx}(t,W(t)|.dt} = 0 \end{aligned}\]
Fifth part, \[\begin{aligned} & \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{tt}(t_{j},W(t_{j})).(t_{j+1}-t_{j})^2} \\ \leq & \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \frac{1}{2} \mathbb{|\sum_{j=0}^{n-1} f_{tt}(t_{j},W(t_{j})).(t_{j+1}-t_{j})^2|} \\ \leq & \ \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ (\mathbb{\underset{0 \leq k \leq n-1}{\max}|t_{k+1}-t_k|}). \mathbb{\underset{||\Pi|| \rightarrow 0}{\lim}} \ \frac{1}{2} \mathbb{\sum_{j=0}^{n-1} |f_{tt}(t_{j},W(t_{j}))|.(t_{j+1}-t_{j})} \\ = & \mathbb{\frac{1}{2}}.0. \mathbb{\int_{0}^{T}|f_{tt}(t,W(t))|.dt} = 0 \end{aligned}\]
And the higher order terms contributes zero.
\[\begin{aligned} & \mathbb{f(T,W(T))-f(0,W(0))} \\ = & \ \mathbb{\sum_{j=0}^{n-1}[f(t_{j+1},x_{j+1})-f(t_{j},x_{j})]} \\ = & \ \mathbb{\sum_{j=0}^{n-1}f_t(t_{j},W(t_{j})).(t_{j+1}-t_{j})} + \mathbb{\sum_{j=0}^{n-1}f_x(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))} + \frac{1}{2}\mathbb{\sum_{j=0}^{n-1} f_{xx}(t_{j},W(t_{j})).(W(t_{j+1})-W(t_{j}))^2} \\ = & \ \mathbb{\int_{0}^{T} f_t(t,W(t)).dt} + \mathbb{\int_{0}^{T}f_x(t,W(t)).dW(t)} + \mathbb{\frac{1}{2}\int_{0}^{T}f_{xx}(t,W(t)).dt} \end{aligned}\]
In differential form it becomes- \[\begin{aligned} &\mathbb{df(t,W(t))} \\ = & \mathbb{f_{t}(t,W(t)).dt}+ \mathbb{f_{x}(t,W(t)).dW(t)} + \frac{1}{2} \mathbb{f_{xx}(t,W(t)).dW(t).dW(t)} + \mathbb{f_{tx}(t,W(t)).dW(t).dt} + \frac{1}{2} \ \mathbb{f_{tt}(t,W(t)).dt.dt} \end{aligned}\]
But, we know that- \[\mathbb{dW(t).dW(t) = dt}; \quad \mathbb{dW(t).dt = 0; \quad dt.dt = 0};\]
The Ito-Doeblin formula simplifies to- \[\begin{aligned} &\mathbb{df(t,W(t))} \\ = & \mathbb{f_{t}(t,W(t)).dt}+ \mathbb{f_{x}(t,W(t)).dW(t)} + \frac{1}{2} \mathbb{f_{xx}(t,W(t)).dW(t).dW(t)} + \mathbb{f_{tx}(t,W(t)).dW(t).dt} + \frac{1}{2} \ \mathbb{f_{tt}(t,W(t)).dt.dt} \\ = & \mathbb{f_{t}(t,W(t)).dt}+ \mathbb{f_{x}(t,W(t)).dW(t)} + \frac{1}{2} \mathbb{f_{xx}(t,W(t)).dt} \\ \end{aligned}\]
For Example: For \(\mathbb{f(x) = \frac{1}{2}x^2}\) this formula states that- \[\begin{aligned} \frac{1}{2}\mathbb{W^2(T)} &= \mathbb{f(W(T))-f(W(0))} \\ &= \mathbb{\int_{0}^{T}f'(W(t)).dW(t)} + \mathbb{\int_{0}^{T}\frac{1}{2}f''(W(t)).dt} \\ &= \mathbb{\int_{0}^{T}W(t).dW(t)} + \mathbb{\frac{T}{2}} \quad \text{Q.E.D.} \\ \end{aligned}\]
In the next blog, we will go deeper into Ito processes: how the quadratic variation changes, what will be the Ito integral etc. Happy reading.