Itô Rules in 1D #
Itô processes #
Suppose $(\Omega, \F, (\F _ t), [0, T], P)$ is a complete filtered probability space. Let $(B _ t)$ be a standard Brownian motion.
Suppose for non-anticipating $(F _ t) \in \bA _ 1(0, T)$ and $(G _ t) \in \bA _ 2(0, T)$, $(X _ t)$ is a real-valued stochastic process satisfying $$ \forall 0 \le s \le r \le T: P\p{X _ r = X _ s + \int _ s^r F _ t \dd t + \int _ s^r G _ t \dd B _ t} = 1, $$ we say that $(X _ t)$ is an 1-dimensional Itô process and write $$ \dd X _ t = F _ t \dd t + G _ t \dd B _ t, \quad t \in [0, T] $$
- Notice that $(X _ t)$ has continuous paths almost surely.
- $(G _ t) \in \bA^2(0, T)$ guarantees that $\p{\int _ 0^t G _ s \dd B _ s} \in \bL _ {c, [2], m}^2(0, T)$.
- $(F _ t) \in \bA^1(0, T)$ guarantees that $\p{\int _ 0^t F _ s \dd s} \in \bL _ {c, [1]}^1(0, T)$.
- When $P(X _ 0 = 0) = 1$ the process is said to have a zero start.
- $(G _ t)$ is called the local volatility, $(F _ t)$ is called the local drift.
Basic transforms of the Brownian motion #
Suppose $(\Omega, \F, (\F _ t), [0, T], P)$ is a complete filtered probability space. Let $(B _ t)$ be a standard Brownian motion.
We can demonstrate that $$ \dd B _ T^2 = \dd t + 2B _ t \dd B _ t, \quad t \in [0, T] $$
- Given partition $(t _ {k}) _ {k = 0}^{m _ n} \in P[s, t]$ we have $$ \sum _ {k = 0}^{m _ n - 1} B _ {t _ k}(B _ {t _ {k + 1}} - B _ {t _ k}) = \frac{B _ t^2 - B _ s^2}{2} - \frac{1}{2} \sum _ {k = 0}^{m _ n - 1} (B _ {t _ {k + 1}} - B _ {t _ k})^2 $$
- Now take any sequence of shrinking partition $\p{(t _ {n, k}) _ {k = 0}^{m _ n}} _ {n = 1}^\infty \subseteq P[s, t]$. We have that with probability one: $$ \int _ s^t B _ \tau \dd B _ \tau = \frac{B _ t^2 - B _ s^2}{2} + \frac{t - s}{2} $$
Also we have $$ \dd (t B _ t) = B _ t \dd t + t \dd B _ t,\quad t \in [0, T] $$
- Given partition $(t _ {k}) _ {k = 0}^{m _ n} \in P[s, t]$ we have $$ \sum _ {k = 0}^{m _ n - 1} t _ {k} (B _ {t _ {k + 1}} - B _ {t _ k}) + \sum _ {k = 0}^{m _ n - 1} B _ {t _ {k + 1}}(t _ {k + 1} - t _ k) = t B _ t - sB _ s $$
- Now take any sequence of shrinking partition $\p{(t _ {n, k}) _ {k = 0}^{m _ n}} _ {n = 1}^\infty \subseteq P[s, t]$. We have that with probability one: $$ t B _ t - sB _ s = \int _ s^t B _ \tau \dd \tau + \int _ s^t \tau \dd B _ \tau $$
Itô chain rule TODO #
Suppose $(\Omega, \F, (\F _ t), [0, T], P)$ is a complete filtered probability space. Let $(B _ t)$ be a standard Brownian motion. Suppose $$ \dd X _ t = F _ t \dd t + G _ t \dd B _ t,\quad t \in [0, T] $$ Suppose $u(x, t) \in C^{2, 1}(\R \times [0, T] \to \R)$, that is $D _ 1 u, D _ {11} u, D _ 2 u$ all exist and are continuous.
- For convenience, we denote $u:= u(X _ t, t)$, $u _ x := D _ 1 u(X _ t, t)$, $u _ t := D _ 2 u(X _ t, t)$, $u _ {xx} := D _ {11}u(X _ t, t)$ in the following discussion.
Then $Y _ t = u(X _ t, t)$ has stochastic differential $$ \dd Y _ t = \dd u(X _ t, t) = u _ t \dd t + u _ x \dd X _ t + \frac{1}{2} u _ {xx} G _ t^2 \dd t = \p{u _ t + u _ x F _ t + \frac{1}{2} u _ {xx} G _ t^2} \dd t + u _ x G _ t \dd B _ t, \quad t \in [0, T] $$
- We always have $\p{u _ t + u _ x F _ t + \frac{1}{2} u _ {xx} G _ t^2} \in \bA _ {1}(0, T)$ and $(u _ x G _ t) \in \bA _ {2}(0, T)$, they are both non-anticipating.
Generalized Itô chain rule TODO #
Suppose $(\Omega, \F, (\F _ t), [0, T], P)$ is a complete filtered probability space. Let $(B _ t)$ be a standard Brownian motion. Suppose $$ \forall i \in \c{1, \ldots, n}:\dd X _ t^{(i)} = F _ t^{(i)} \dd t + G _ t^{(i)} \dd B _ t, \quad t \in [0, T] $$ Suppose $u \in C^{1, 2}(\R^n \times [0, T] \to \R)$. Then let $u _ t := D _ t u(\symbf X _ t, t)$, $u _ i:= D _ {x _ i} u(\symbf X _ t, t)$ and so on. $$ \dd u(\symbf X _ t, t) = u _ t \dd t + \sum _ {i = 1}^{n} u _ {i} \dd X^{(i)} _ t + \frac{1}{2} \sum _ {i, j = 1}^n u _ {ij} G _ t^{(i)} G _ t^{(j)} \dd t,\quad t \in [0, T] $$
Itô's product rule #
Suppose $(\Omega, \F, (\F _ t), [0, T], P)$ is a complete filtered probability space. Let $(B _ t)$ be a standard Brownian motion.
And suppose $$ \dd X _ t = G _ t \dd t + U _ t \dd B _ t, \quad \dd Y _ t = H _ t \dd t + V _ t \dd B _ t, \quad t \in [0, T] $$ Then immediately $$ \begin{aligned} \dd (X _ t + Y _ t) & = (G _ t + H _ t) \dd t + (V _ t + U _ t) \dd B _ t,& t \in [0, T]\\ \dd (X _ t Y _ t) & = \frac{1}{2} \p{\dd (X _ t + Y _ t)^2 - \dd X _ t^2 - \dd Y _ t^2}, & t \in [0, T] \end{aligned} $$ Consider $u(x, t) = x^2$, we have $$ \begin{aligned} &\dd X _ t^2 = 2X _ t \dd X _ t + U _ t^2\dd t = 2X _ t (G _ t \dd t + U _ t \dd B _ t) + U _ t^2\dd t = (2X _ t G _ t + U _ t^2) \dd t + 2X _ t U _ t \dd B _ t\\ &\dd Y _ t^2 = 2Y _ t \dd Y _ t + V _ t^2 \dd t = 2Y _ t (H _ t \dd t + V _ t \dd B _ t) + V _ t^2\dd t = (2Y _ t H _ t + V _ t^2) \dd t + 2Y _ t V _ t \dd B _ t\\ &\dd(X _ t + Y _ t)^2 = 2(X _ t+ Y _ t) \dd (X _ t + Y _ t) + (U _ t + V _ t)^2 \dd t\\ &\dd (X _ t + Y _ t)^2 = \p{2(X _ t + Y _ t)(H _ t + G _ t) + (U _ t + V _ t)^2} \dd t + 2(X _ t + Y _ t)(U _ t+ V _ t) \dd B _ t \end{aligned} $$ Therefore by subtraction, $$ \begin{aligned} \dd (X _ t Y _ t) &= X _ t \dd Y _ t + Y _ t \dd X _ t + U _ t V _ t \dd t\\ &= X _ t(H _ t \dd t + V _ t \dd B _ t) + Y _ t (G _ t \dd t + U _ t \dd B _ t) + U _ t V _ t \dd t\\ & = (X _ t H _ t + Y _ t G _ t + U _ t V _ t) \dd t + (X _ t V _ t + Y _ t U _ t) \dd B _ t \end{aligned} $$
Notice that the same result also follows from the generalized Itô's chain rule.