Ito Process
Table of contents
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Ito Process
The Ito-Doeblin theorem from the previous section was specifically for intergrating with respect to a Brownian motion $W(t)$. Now we extend this to a more general case, Ito processes.
An Ito process is defined as the following:
\[X(T) = X(0) + \int_0^T \Delta(u)d W_u + \int_0^T \Theta(u) du\]and in differential form:
\[\partial X(t) = \Delta(t) \partial W(t) + \Theta(t)\partial t\]Here, $W_u$ is a Brownian motion, and $\Delta(u)$ and $\Theta(u)$ are adapted processes. Also $\delta(u)^2$ is integrable and $\Theta(u)$ is absolutely integrable.
The Ito process is defined as a sum of an Ito integral and a Lebesgue intergal. We denote the Ito portion as $I(T) = \int_0^T \Delta(u)dW(u)$ and the Lebesgue portion as $L(T) = \int_0^T \Theta(u)du$. The quadratic variation of such an Ito process is $QV_X(T) = \int_0^T \Delta(u)^2 du$ as demonstrated below.
Quadratic Variation for Ito Process
For a given partition, denote \(\vert \cv{T}_n \vert\) as the length of the largest partition. For a given $n$, let the partition be $0 < t_1 < … < T$. Then the quadratic variation without the limit (via expanding squares) is:
\[\sum_{j=0}^{n-1}(X(t_{j+1}) - X(t_j))^2 = \sum_{j=0}^{n-1}(I(t_{j+1}) - I({t_j}))^2 + \sum_{j=0}^{n-1}(L(t_{j+1}) - L({t_j}))^2 + 2\sum_{j=0}^{n-1}(I(t_{j+1}) - I({t_j}))(L(t_{j+1}) - L({t_j}))\]We shall see that several terms go to 0 in the limit as required by the definition of QV, and the only term remaining will be $\int_0^T \Delta(t)^2dt$.
In the limit, the $\sum_{j=0}^{n-1}(I(t_{j+1}) - I({t_j}))^2$ term becomes $QV_{I}(T) = \int_0^T \Delta^2(u)du$.
We bound the second term by: \(\begin{align*} \sum_{j=0}^{n-1}(L(t_{j+1}) - L({t_j}))^2 &\leq \max_{(t_j, t_{j+1})\in\cv{T}_n}\{R(t_{j+1})-R({t_j})\}\sum_{j=0}^{n-1}\vert R(t_{j+1})-R({t_j})\vert \\ &= \max_{(t_j, t_{j+1})\in\cv{T}_n}\{R(t_{j+1})-R({t_j})\}\left\lvert \int_0^T \Theta(u)du \right\rvert \\ &\leq \max_{(t_j, t_{j+1})\in\cv{T}_n}\{R(t_{j+1})-R({t_j})\} \int_0^T \vert \Theta(u) \vert du \\ &\to 0 \end{align*}\)
We bound the third term with
\[\begin{align*} 2\sum_{j=0}^{n-1}(I(t_{j+1}) - I({t_j}))(L(t_{j+1}) - L({t_j})) &\leq 2\max_{(t_j, t_{j+1})\in\cv{T}_n}\{(I(t_{j+1}) - I({t_j}))\}\sum_{j=0}^{n-1}\vert L(t_{j+1}) - L({t_j})\vert \\ &= 2\max_{(t_j, t_{j+1})\in\cv{T}_n}\{(I(t_{j+1}) - I({t_j}))\}\left\lvert \int_0^T \Theta(u)du \right\rvert \\ &\leq 2\max_{(t_j, t_{j+1})\in\cv{T}_n}\{(I(t_{j+1}) - I({t_j}))\} \int_0^T \vert \Theta(u) \vert du \\ &\to 0 \end{align*}\]So all that remains after taking limits, is $QV_I(T) = \int_0^T \Delta^2(u)du$.
Definition of Integral of Ito Process
For a given adapted processs $\Gamma(t)$, it can be integrated with respect to the Ito process $X(t)$ as $\int_0^T \Gamma(t)dX(t)$. It is defined as follows:
\[\int_0^T \Gamma(t)dX(t) = \int_0^T \Gamma(t)\Delta(t)dW(t) + \int_0^T \Gamma(t) \Theta(t) dt\]That is, we split up the two parts of the Ito process and integrate with respect to the Brownian motion portion, and then again with respect to the Lebesgue portion.
Ito-Doeblin Theorem for Ito Processes
The Ito-Doeblin theorem for an Ito process is analogous to that in the Brownian motion only case. It states that (in both integral and differential form):
\[f(T, X(T)) = f(0, X(0)) + \int_0^T f_t(t,X(t))dt + \int_0^T f_x(t,X(t))dX(t) + \frac{1}{2}\int_0^T f_{xx}(t,X(t))dX(t)dX(t)\] \[df(t,X(t)) = f_t(t,X(t))dt + f_x(t,X(t))dX(t) + \frac{1}{2}f_{xx}(t,X(t))dX(t)dX(t)\]Note the final $dX(t)dX(t)$ is a product of the quadratic variation on $X(t)$ which is $\int_0^T \Delta^2(u)du$ as opposed to the quadratic variation on $W(t)$ which is $T$.
The proof is as follows, with Taylor expansion:
\[\begin{align*} f(T,X(T)) - f(0,X(0)) &= \sum_{j=0}^{n-1}f_t(t_j, X(t_j))(t_{j+1}-t_j) + \sum_{j=0}^{n-1}f_x(t_j, X(t_j))(X(t_{j+1})-X(t_j)) + \\ &\qquad \qquad \frac{1}{2}\sum_{j=0}^{n-1}f_{xx}(t_j, X(t_j))(X(t_{j+1})-X(t_j))^2 + \frac{1}{2}\sum_{j=0}^{n-1}f_{tt}(t_j, X(t_j))(t_{j-+1}-t_j)^2 + \\ &\qquad \qquad \sum_{j=0}^{n-1}f_{tx}(t_j, X(t_j))(t_{j+1}-t_j)(X(t_{j+1})-X(t_j)) + R \end{align*}\]The first line are the first order terms, the second line are the quadratic terms, and the third line is the cross term + higher order terms.
$\frac{1}{2}\sum_{j=0}^{n-1}f_{tt}(t_j, X(t_j))(t_{j+1}-t_j)^2$ converges to 0 in the same manner as discussed before. So does the cross term. The first order terms also converge to their integral counter parts. The quadratic $X(t)$ term results in $\int_0^T f_{xx}(t_j, X_j) d(X,X)t$ and since $QV_X(T) = \int_0^T \Delta^2(u)du$ then $d(QV_X(T)) = \Delta^2(u)du$. Thus the second term becomes $\int_0^T f_{xx}(t,X(t))\Delta^2(u)du$.
We can further simplify $\int_0^T f_x(t,X(t))dX(t)$ as $\int_0^T f_x(t,X(t))\Delta(t)dW(t) + \int_0^T f_x(t, X(t))\Theta(t)dt$. This is by definition of the integral of Ito processes.
Usages of Ito-Doeblin
We can use Ito-Doeblin to solve stochastic integrals and also stochastic differential equations.
Stochastic Integration
While we solved $\int_0^T W_t dW_t$ by way of sums, we can also solve it using Ito-Doeblin, and in general, it is easier to solve it this way. Let $g(W_t) = \frac{1}{2}W_t^2$ so that $g’ = W_t$ and $g’’ =1$. Then we have: \(\begin{align*} dg(W_t) &= g'(W_t) dW_t + \frac{1}{2}g''(W_t)dt \\ &= W_t dW_t + \frac{1}{2}dt \\ \int_0^T dg(W_t) &= \int_0^T W_t dW_t + \frac{1}{2}\int_0^T dt \\ \int_0^T W_t dW_t &= \int_0^T dg(W_t) - \frac{1}{2}\int_0^T dt \\ &= \frac{1}{2}W_T^2 - \frac{1}{2}T \end{align*}\) This result agrees with what we have above.
It is important to note that the resulting integral is a stochastic process evaluate at time $T$ and is not a ``numeric’’ solution in the Lebesgue sense, even though this is a definite integral. Say we wish to evaluate $\int_0^T W_t^2 dW_t$. First, let $g(W_t) = \frac{1}{3}W_t^3$ which is what we would expect from ordinary calculus, or the Stratanovich integral. Then $g’(W_t) = W_t^2$ and $g’‘(W_t) = 2W_t$.
\[\begin{align*} d\left(\frac{1}{3}W_t^3\right) &= g'(W_t) dW_t + \frac{1}{2}g''(W_t) dt \\ &= W_t^2 dW_t + 2W_t dt \\ \frac{1}{3}W_t^3 &= \int_0^T W_t^2 dW_t + 2\int_0^T W_t dt \\ \int_0^T W_t^2 dW_t &= \frac{1}{3}W_T^3 - 2\int_0^T W_t dt \end{align*}\]Here, $\int_0^T W_t dt$ is a stochastic process evaluated at time $T$.
Stochastic Differential Equations
A standard Ito process can be an asset price obeying:
\[d S_t = \mu S_t dt + \sigma S_t dW_t\]That is, the stochastic process tracking the stock price at time $t$ given by $S_t$ changes by some multiplicative drift term and some random noise term supplied by $W_t$, Brownian motion. Then to find $S_t$, we can use a candidate function. It turns out $g(S_t) = \log S_t$. Then $g’(S_t) = \frac{1}{S_t}$ and $g’‘(S_t) = =\frac{1}{S_t^2}$. By Ito-Doeblin:
\[\begin{align*} d\log S_t &= \frac{1}{S_t} dS_t - \frac{1}{2S_t^2}(dS_t)^2 \\ &= \frac{1}{S_t} dS_t - \frac{1}{2S_t^2}(dS_t)^2 \\ &= \frac{1}{S_t} (\mu S_t dt + \sigma S_t dW_t) - \frac{1}{2S_t^2}(\mu S_t dt + \sigma S_t dW_t)^2 \\ &= \mu dt + \sigma dW_t - \frac{1}{2S_t^2}(\mu^2 S_t^2 dtdt + 2\mu\sigma S_t^2 dt dW_t + \sigma^2 S_t^2 dW_t dW_t) \\ &= \mu dt + \sigma dW_t - \frac{1}{2}(\sigma^2 dW_t dW_t) \\ &= \mu dt + \sigma dW_t - \frac{1}{2} \sigma^2 dt \\ &= \left(\mu - \frac{1}{2}\sigma^2\right)dt + \sigma dW_t \\ \int_0^T d\log S_t &= \left(\mu - \frac{1}{2}\sigma^2\right)\int_0^T dt + \sigma \int_0^T dW_t \\ \log S_T &= \log S_0 + \left(\mu - \frac{1}{2}\sigma^2\right) T + \sigma W_T \\ S_T &= S_0\exp\left(\left(\mu - \frac{1}{2}\sigma^2\right) T + \sigma W_T\right) \end{align*}\]Thus the solution to the differential equation provides the formula for the stochastic process tracking the stock price that obeys the differential equation above.