**Ito Integral on a Non-Random Simple Process**

Let us take an example. Let’s consider a time period [0, T ] partitioned into n intervals of equal length $\Delta t = T/n$ with endpoints $ t_{k}= k\Delta t,$ k = 0, …, n. These are the times at which the market is open and trade in shares can take place. An investor buys $q(t_{0})$ shares at time 0 at a price of $S(t_{0})$ each. At time $t_{1}$, market trading establishes a new share price $S(t_{1})$. Once this price has been revealed the investor can change the quantity of shares held, from $q(t_{0})$ to $q(t_{1})$. The same at the subsequent times $t_{2}$ through $t_{n−1}$. At time $t_{n}$ the entire portfolio is liquidated as soon as share price $S(t_{n})$ becomes known. A portfolio of $q(t_{k})$ shares is held from just after the share price $S(t_{k})$ has been revealed, to when the share price $S(t_{k+1})$ at time $t_{k+1}$ becomes known. This portfolio is worth $q(t_{k} )S(t_{k})$ at time $t_{k}$ and $q(t_{k})S(t_{k+1})$ at time $t_{k+1}$. So from time $t_{k}$ to time $t_{k+1}$, with the change in share price $S(t_{k+1})$ − $S(t_{k})$ denoted by $S_{k}$, the portfolio value changes by $q(t_{k}) S_{k}$. The gain over all trades can then be expressed as $I_n = \sum_{k=0}^{n-1} q(t_{k}) \Delta S_{k}$. This expression is an example of a discrete stochastic integral.

At each trading time, the quantity of shares to be held until the next trading opportunity arises, is decided by the investor. Some investors will buy shares and others will sell as a trade requires two parties. The factors which an investor takes into account in determining the shareholdings are unknown, so q is a discrete-time random process. Thus the stochastic integral is constructed from the two random processes q and S:

$$\sum_{k=0}^{n-1} q(t_k )[ \mu S(t_k) \Delta t + \sigma_S(t_k ) \Delta B(t_k )]

= I_n$$

Increasing n makes the time step t smaller and discrete-time trading approaches continuous-time trading. The first term is a summation with respect to time step $\Delta t$. The second term is a summation with respect to Brownian motion increment [B($t_{k+1}$) − B($t_{k}$ )].

Let’s take time period [0, T] partitioned into n intervals of equal length t = T/n, with endpoints $t_{k}$ = $k\Delta t$, k = 0, . . ., n. The simplest integrand f is a step-function whose values are non-random, so the only random quantity is the Brownian motion integrator. Step-function f can be written with indicator notation as

$$f = f(t_{0}) 1_{[t_0,t_1)} + · · · + f (t_{k} ) 1_{[t_k ,t_{k+1})} + · · · + f (t_{n−1}) 1_{[t_{n−1},t_n )}$$

where $1_{[tk ,tk+1)}$ denotes the indicator function which has value 1 on the interval shown in the subscript, and 0 elsewhere. Define the discrete stochastic integral $I_{n}( f )$:

$$I_n( f ) = \sum_{k=0}^{n-1}f(t_{k} )[B(t_{k+1}) − B(t_{k} )]$$

Therefore, integral of a non-random process over Brownian motion is the limit of the integrals of non-random simple processes approaching this non-random process.

### Ito Integral on a Simple Adapted Process

Let f be at random level $f(t_{k}, ω)$ for $t_{k}$ < t < $t_{k+1}.$ When the history of the Brownian motion process becomes known progressively at each time $t_{k}$, it must be possible to determine $f (t_{k})$ from this history alone. The current terminology is that $f (t_{k})$ must be adapted to the filtration ($t_{k}$ ).

**Properties:**

**Linearity:**$\int_0^T (\alpha X(t) +\beta Y(t)) d B(t) = \alpha \int_0^T X(t) d B(t) + \beta \int_0^T Y(t) d B(t)$ for constant $\alpha$ and $\beta$ and simple adapted processes $X(t)$ and $Y(t)$

$\int_0^T 1_{(a,b]} d B(t) = B(b)-B(a)$, and

$\int_0^T 1_{(a,b]} X(t) d B(t) = \int_a^b X(t) d B(t)$-
**Zero mean property**

$\mathbf E\int_0^T X(t) d B(t) = 0$ **Isometry property:**

$\mathbf E\left(\int_0^T X(t) d B(t)\right)^2=\int_0^T \mathbf E(X^2(t)) dt$

**Ito Integral of Regular Adapted Processes**

After looking at the specific case of simple adaptive process, we want to find the Ito integral for general integrands i.e. any regular adapted process. So let’s assume $X^n(t)$ to be a sequence of simple processes convergent in probability to the process $X(t)$. $$ X^{(n)}(t)\rightarrow X(t)$$ Then, under some conditions, the sequence of their integrals $\int_0^TX^n(t)dB(t)$ also convergence in probability. Hence, if, $$\int_0^TX^n(t)dB(t)\rightarrow J^n$$

Then $$\int_0^TX(t)dB(t)\rightarrow J$$

**For Example:**

Find $\int_0^T B(t)dB(t)$:

let’s assume $0=t_0^n<t_1^n<t_2^n< … <t_n^n = T$ be a partition of [0, T] and let- $$X^n(t) = \sum_{i=0}^{n-1} B(t_i^n)I_{(t_i^n, t_{i+1}^n)}(t)$$

Also $\delta_n = max_i(t_{i+1}^n – t_i^n)\rightarrow 0$ then $X_n(t)\rightarrow B(t)$ is approximately the continuity of the Brownian paths.

Hence, $$\int_0^TX^n(t)dB(t) = \sum_{i=0}^{n-1} B(t_i^n)(B(t_{i+1}^n)- B(t_{i+1}^n))$$

where, $$B(t_i^n)(B(t_{i+1}^n)- B(t_{i+1}^n))=1/2(B^2(t_{i+1}^n) – B^2(t_{i}^n) – (B(t_{i+1}^n) – B(t_{i}^n))^2)$$

Therefore, $$\int_0^TX^n(t)dB(t)$$

$$ = \sum_{i=0}^{n-1} 1/2(B^2(t_{i+1}^n) – B^2(t_{i}^n) – (B(t_{i+1}^n) – B(t_{i}^n))^2)$$

$$=1/2(B^2(T) – 1/2(B^2(0)) -1/2 \sum_{t=0}^{n-1}(B(t_{i+1}^n) – B(t_i^n))^2$$

Here, the last summation part converges to the quadratic variation T of Brownian motion on [0, T] in probability.

Therefore $\int_0^T B(t)dB(t) = lim \int_0^T X^n(t)dB(t) = 1/2B^2(T) – 1/2T$

**Properties:**

Any regular adaptive process has the same properties as we already discussed for the simple adaptive process.

**Linearity:**

$\int_0^T (\alpha X(t) +\beta Y(t)) d B(t) = \alpha \int_0^T X(t) d B(t) + \beta \int_0^T Y(t) d B(t)$

for constant $\alpha$ and $\beta$- $\int_0^T 1_{(a,b]} X(t) d B(t) = \int_a^b X(t) d B(t)$
**Zero mean property:**$\mathbf E\int_0^T X(t) d B(t) = 0$**Isometry property**: $\mathbf E\left(\int_0^T X(t) d B(t)\right)^2=\int_0^T \mathbf E(X^2(t)) dt$

**Ito Integral Process**

$$Y(t)=\int_0^TX(s)dB(s)$$

where,

$$\int_0^TX^2(s)ds<\infty$$

In simple terms, ito intergal is a way of integrating a stochastic process variable with respect to another stochastic process that result into a new stochastic process. Here, X(s) is a square-integrable process (hence the equation B) that is adapted to the filteration generated by B(s). Standard calculous doesn’t apply to such functions since the function X(S) is not differentiable at any point and has infinite variateion over every time interval.

Ito integral plays crucial part in mathemetical finance to evaluate strategies to make descisions over time variable stochastic processes or Brownian processes such as stoch prices.

**Properties:**

**Martingale Property:**Probabilities of current state doesn’t depend on any knowledge of past events, given that it’s square is inegral is finite. I.e. Expected value of the state Y at time t doesn’t depend on the given $F_s$. And hence it is a stochastic variable Y(s) only.

$$E(Y(t)|F_s) = Y(s)$$**Quadratic Variation:**The quadratic variation of an Ito processes can be given by $\int_0^t X^2 (s)ds$

**Ito’s Formula for Brownian motion**

We know that B(T) is a Brownian motion from 0 to T and since we know the first order derivative and second order derivate of F then we can find out the value of Brownian motion at time T as in the example below.

Example: \[f(B(t)) = f(0)+\int_0^t f'(B(s)) d B(s) + {1\over 2}\int_0^t f”(B(s)) d s\]

where: \[\exp(B(t)) = 1+\int_0^t \exp(B(s)) d B(s) + {1\over 2}\int_0^t \exp(B(s)) d s\]

Proof of the above example is:

\[f(B(t)) = f(0) + \sum_{i=0}^{n-1} \left(f(B(t^n_{i+1}))-f(B(t^n_{i}))\right)\]

After taking the first order derivative,

\[\hphantom{f(B(t)) } = f(0) + \sum_{i=0}^{n-1} f'(B(t^n_{i}))\left(B(t^n_{i+1})-B(t^n_{i})\right) + {1\over 2}\sum_{i=0}^{n-1} f”(\theta^n_{i})\left(B(t^n_{i+1})-B(t^n_{i})\right)^2\] For \[\theta^n_{i}\in \left(B(t^n_{i}),B(t^n_{i+1})\right)\]

\[\sum_{i=0}^{n-1} f'(B(t^n_{i}))\left(B(t^n_{i+1})-B(t^n_{i})\right) \to \int_0^t f'(B(s)) d B(s)\]

After taking the second order derivative and as $n\to\infty$ we have the below equations

\[\sum_{i=0}^{n-1} f”(\theta^n_{i})\left(B(t^n_{i+1})-B(t^n_{i})\right)^2\to \sum_{i=0}^{n-1} f”(B(t^n_{i}))\left(B(t^n_{i+1})-B(t^n_{i})\right)^2\]

In probability as $n\to\infty$

then

\[\sum_{i=0}^{n-1} f”(B(t^n_{i}))\left(B(t^n_{i+1})-B(t^n_{i})\right)^2 \to \sum_{i=0}^{n-1} f”(B(t^n_{i}))\left(t^n_{i+1}-t^n_{i}\right)\]

Hence

\[\sum_{i=0}^{n-1} f”(B(t^n_{i}))\left(B(t^n_{i+1})-B(t^n_{i})\right)^2\to \int_0^t f”(B(s)) d s\]

**Ito’s Process and Stochastic Differentials**

We know B(t) is a Brownian motion from $0<t<T$. An Ito-process Y(t) from 0≤t≤T is given by

\[Y(t) = Y(0) + \int_0^t \mu(s) d s + \int_0^t \sigma(s) d B(s)\] or

\[d Y(t) = \mu(t) d t + \sigma(t) d B(t)\]

where $0\le t\le T$, $Y(0)$ is $F_0$-measurable, $\mu(t)$ and $\sigma(t)$ are $F_t$-adapted, $\int_0^T|\mu(t)| d t<\infty$ and $\int_0^T\sigma^2(t) d t <\infty$.

Example: Taylor’s formula written at the second order for Brownian motion reads as:

Chain rule: $d(f(B(t)))=f'(B(t)) d B(t) + {1\over 2}f”(B(t)) d t$, for $f$ with continuous 2nd derivative:

$d \exp(B(t)) = \exp(B(t)) d B(t) + {1\over 2} \exp(B(t)) d t$

$d B^2(t) = 2 B(t) d B(t) + dt$

**Ito’s Formular for Ito processes**

Let X be an Ito process with stochastic differential then

$d X(t) = \mu(t) d t + \sigma(t) d B(t)$ for 0 ≤ t ≤ T.

Let ($f\in C^2$) then the stochastic differential of the process f exists and can be written as

$$\begin{eqnarray*}

df(X(t)) &=& f'(X(t))dX(t)+\frac{1}{2}f”(X(t))d[X,X](t)\\

&=& f'(X(t))dX(t)+\frac{1}{2}f”(X(t))\sigma^{2}(t)dt\\

&=& \left(f'(X(t))\mu(t)+\frac{1}{2}f”(X(t))\sigma^{2}(t)\right)dt+f'(X(t))\sigma(t)dB(t)

\end{eqnarray*}$$

### Ito’s Integral and Gaussian process

If x is a non- random process satisfying $\int_0^T X^2(s) d s <\infty$ and Y is a Gaussian process where $Y(t)=\int_0^t X(s) d B(s)$ Then it hass a zero mean and the covariance can be written as below:

$\mbox{Cov}(Y(t), Y(t+u)) = \int_0^t X^2(s) d s$, $t,u\ge 0$