Stochastic Integration

What is a stochastic integral?

Let $X_1,X_2,\dots$ be independent random variables with $\mathbb{P}\{X_i=1\}=\mathbb{P}\{X_i=-1\}=1/2$ and let $S_n$ denote the simple random walk $S_n=X_1+\cdots +X_n$

Let ${\mathcal{F}}_n$ denote the information contained in $X_1,\dots ,X_n$ and let $B_n$ be the “bet” on the $n$th game such $B_n$ is measurable with respect to ${\mathcal{F}}_{n-1}$

The winnings up to time $n$ can be written as $Z_n={\sum }_{i=1}^n{B}_i{X}_i={\sum }_{i=1}^n{B}_i[S_i-S_{i-1}]={\sum }_{i=1}^{n}B\Delta S_i$

We call $Z_n$ the integral of $B_n$ with respect to $S_n$

$Z_n$ is a martingale, that is $E(Z_n\mid {\mathcal{F}}_m)=Z_m$ and $\mathbb{E}(Z_n)=0$
In addition, assuming $\mathbb{E}(B_n^2)<\infty$, we can derive $\text{Var}(Z_n)={\sum }_{i=1}^n\mathbb{E}(B_i^2)$

The continuous analogue of the random walk is the a standard one-dimensional Brownian motion $W_t$

If one bets one unit for the period $[s,t]$ then one’s winnings in this time period are $W_t-W_s$

We want to define the continuous analogue $Z_t={\int }_0^t{Y}_{s}d{W}_s$, which should denote the total amount won up to time if the amount bet at $s$ is $Y_s$

We assume $\mathbb{E}(Y_s^2)<\infty$ and ${\int }_0^t\mathbb{E}(Y_s^2)ds<\infty$, implying only bounded strategies, and assume $Y_t$ is ${\mathcal{F}}_t$-measurable, such that bets only use information up through $t$

This is difficult to define because of the roughness of Brownian motion, but we can make it easier by assuming a finite set of bet changes $t_1<t_2<\cdots <t_n$

$Y_t=\{\begin{array}{ll}{Y}_0,& 0\le t<t_1,\\ Y_1,& t_1\le t<t_2,\\ ⋮& \\ Y_n,& t_n\le t<\infty .\end{array}$

We call this a simple strategy, and we can define the stochastic integral for $t_j\le t<t_{j+1}$ with $Z_t={\int }_0^t{Y}_{s}d{W}_s={\sum }_{i=1}^j{Y}_{t-1}[W_{t_i}-W_{t_{i-1}}]+Y_j[W_t-W_{t_j}]$, where the summation adds together the completed intervals and the final term represents the incomplete interval

This integral satisfies linearity and is a martingale with respect to ${\mathcal{F}}_t$, i.e. it’s ${\mathcal{F}}_t$-measurable, $\mathbb{E}(|Z_t|)<\infty$, and most importantly $E(Z_t\mid {\mathcal{F}}_s)=Z_s$

To prove this last point, we consider two cases

If $t_j\le s<t\le t_{j+1}$ for some $j$
$Z_t=Z_s+Y_j[W_t-W_s]$
$E(Z_t\mid {\mathcal{F}}_s)=Z_s+Y_{j}E(W_t-W_s\mid {\mathcal{F}}_s)=Z_s+Y_j\mathbb{E}(W_t-W_s)=Z_s$

If $t_i\le s<t_{i+1}$, $t_j\le t<t_{j+1}$ for some $i<j$, then
$E(Z_t\mid {\mathcal{F}}_s)=E(E(Z_t\mid {\mathcal{F}}_{t_{i+1}})\mid {\mathcal{F}}_s)=E(Z_{t_{i+1}}\mid {\mathcal{F}}_s)=Z_s$

$Z_t$ also satisfies $\mathbb{E}(Z_t^2)={\int }_0^t\mathbb{E}(Y_s^2)ds$ (this is a bit tedious to prove)

To define the stochastic integral for non-simple betting rules, we take the limit over discrete parts

Let $Y_s$ be measurable with respect to ${\mathcal{F}}_s$, satisfying the same second moment conditions
Also assume $Y_s$ is right continuous and has left limits

For each $n\ge 0$, $Y_s^{(n)}={\int }_{(k-1)/n}^{k/n}{Y}_{r}dr,\frac{k}{n}<s\le \frac{k+1}{n}$ is the approximate betting strategy over that interval, and $Y_s^{(n)}=0$ for $s\le 1/n$

It can be proved that $Y_s^{(n)}\to Y_s$ in mean-square (${\lim }_{n\to \infty }{\int }_0^t\mathbb{E}([Y_s-Y_s^{(n)}{]}^2)ds=0$) which also shows that our three properties still hold

1. ${\int }_0^t[a{X}_s+b{Y}_s]d{W}_s=a{\int }_0^t{X}_{s}d{W}_s+b{\int }_0^t{Y}_{s}d{W}_s$
2. $Z_t$ is a martingale with respect to ${\mathcal{F}}_t$ and $\mathbb{E}(Z_t)=0$
3. $\text{Var}\left(Z_t\right)=\mathbb{E}[Z_t^2]={\int }_0^t\mathbb{E}[Y_s^2]ds$

Thus $Z_t={\int }_0^t{Y}_{s}d{W}_s$ is the mean-square limit of the random variables $Z_t^{(n)}={\int }_0^t{Y}_s^{(n)}d{W}_s$

We often write this in the differential form $d{Z}_t=Y_{t}d{W}_t$

We can think of $Z_t$ as a process at time $t$ that looks like a Brownian motion with variance $Y_t^2$

Likewise, if we have a process $Z_t={\int }_0^t{X}_{t}ds+{\int }_0^t{Y}_{s}d{W}_s$ then we can write the differential form as $d{Z}_t=X_{t}dt+Y_{t}d{W}_t$, which looks like a Brownian motion with variance $Y_t^2$ and drift $X_t$

Itô’s Formula

How do we actually calculate these integrals? Regular calculus rules give us ${\int }_0^t{W}_{s}d{W}_s=\frac{1}{2}{W}_t^2$, but this cannot be true because the left side has expectation $0$ and the right has expectation $t/2$

To start, we review the ordinary fundamental theorem of calculus

In general, we can expand a continuously differential function
$f(t)=f(t_0)+f^{\prime }(t_0)(t-t_0)+o(t-t_0)$
$f\left(\frac{(j+1)t}{n}\right)=f\left(\frac{jt}{n}\right)+f^{\prime }\left(\frac{jt}{n}\right)\frac{t}{n}+to\left(\frac{1}{n}\right)$

Using this, we can write
$f(t)=f(0)+{\sum }_{j=0}^{n-1}[f((j+1)t/n)-f(jt/n)]$
$f(t)-f(0)={\sum }_{j=0}^{n-1}{f}^{\prime }\left(\frac{jt}{n}\right)\frac{t}{n}+{\sum }_{j=0}^{n-1}to\left(\frac{1}{n}\right)$

As $n\to \infty$, this goes to $f(t)-f(0)={\int }_0^t{f}^{\prime }(s)ds$ as wanted

Now let’s try to derive a similar result for stochastic integrals
Let $W_t$ be a Brownian motion, and $f$ a twice continuously differentiable function

$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{1}{2}{f}^{\prime \prime }(x_0)(x-x_0{)}^2+o((x-x_0{)}^2)$
$f(W_{(j+1)t/n})=f(W_{jt/n})+f^{\prime }(W_{jt/n})[W_{(j+1)t/n}-W_{jt/n}]$
$+\frac{1}{2}{f}^{\prime \prime }(W_{jt/n})[W_{(j+1)t/n}-W_{jt/n}{]}^2+to\left(\frac{1}{n}\right)$

The reason we can simplify to $to\left(\frac{1}{n}\right)$ is because that term amounts to the variance of $W_{(j+1)t/n}-W_{jt/n}$ which is of order $(t/n)$

Using this, we can write
$f(W_t)=f(W_0)+{\sum }_{j=0}^{n-1}[f(W_{(j+1)t/n})-f(W_{jt/n})]$
$f(W_t)-f(W_0)={\sum }_{j=0}^{n-1}{f}^{\prime }(W_{jt/n})[W_{(j+1)t/n}-W_{jt/n}]$
$+\frac{1}{2}{\sum }_{j=0}^{n-1}{f}^{\prime \prime }(W_{jt/n})[W_{(j+1)t/n}-W_{jt/n}{]}^2+{\sum }_{j=0}^{n-1}to\left(\frac{1}{n}\right)$

As $n\to \infty$, the third term goes to $0$ and the first term approaches ${\int }_0^t{f}^{\prime }(W_s)d{W}_s$, however we still have the second term which is a bit more confusing

In general, we are considering the limit of ${\sum }_{j=0}^{n-1}g(W_{jt/n})[W_{(j+1)t/n}-W_{jt/n}{]}^2$ where $g$ is continuous

If $g=1$ then we are looking at $Q_t^{(n)}={\sum }_{j=0}^{n-1}[W_{(j+1)t/n}-W_{jt/n}{]}^2$

$Q_t={\lim }_{n\to \infty }{Q}_t^{(n)}$ is called the quadratic variation of $W_t$
$[W_{(j+1)t/n}-W_{jt/n}{]}^2$ has the same distribution as $(t/n)U^2$ where $U$ is standard normal, and note that $\mathbb{E}(U^2)=1$ and $\text{Var}(U^2)=2$

$\mathbb{E}(Q_t^{(n)})={\sum }_{j=0}^{n-1}\mathbb{E}([W_{(j+1)t/n}-W_{jt/n}{]}^2)=t$
$\text{Var}(Q_t^{(n)})={\sum }_{j=0}^{n-1}\text{Var}([W_{(j+1)t/n}-W_{jt/n}{]}^2)=n\text{Var}((t/n)U^2)=\frac{2t^2}{n}$

As $n\to \infty$, the limiting random variable $Q_t$ is therefore a constant, and the quadratic variation of Brownian motion up to time $t$ is $t$

For any $g$,
$Q_t^{(n)}(g)={\sum }_{j=0}^{n-1}g(t)[W_{(j+1)t/n}-W_{jt/n}{]}^2$
$Q_t(g)={\lim }_{n\to \infty }{Q}_t^{(n)}(g)$

If $g$ is a step function of the form $g(s)=u(W_{jt/m})$ for $jt/m\le s<(j+1)t/m$ (a function of the current value of the process), then
$Q_t(g)={\lim }_{n\to \infty }{Q}_t^{(n)}(g)={\lim }_{k\to \infty }{Q}_t^{(km)}(g)$
$={\lim }_{k\to \infty }{\sum }_{j=0}^{m-1}u(W_{jt/m}){\sum }_{i=0}^{k-1}{\left[W_{\frac{kj+i+1}{km}t}-W_{\frac{kj+i}{km}t}\right]}^2$
$={\sum }_{j=0}^{m-1}u(W_{jt/m})\frac{t}{m}$

If $g$ is continuous, let $g_n$ be the step function $g_n(s)=g(t)$ for $\frac{j}{n}t\le s<\frac{j+1}{n}t$
$|Q_t(g)-Q_t(g_n)|\le \Vert g-g_n\Vert Q_t=t\Vert g-g_n\Vert$, where $\Vert g-g_n\Vert ={\sup }_{0\le s\le t}|g(s)-g_n(s)|$ (we essentially factor out the maximum difference between $g$ and $g_n$)

Because $g$ is continuous, $\Vert g-g_n\Vert \to 0$ as $n\to \infty$, so $|Q_t(g)-Q_t(g_n)|\to 0$ and we have $Q_t(g)={\lim }_{n\to \infty }{Q}_t(g_n)={\lim }_{n\to \infty }{\sum }_{j=0}^{n-1}g\left(\frac{j}{n}t\right)\frac{t}{n}$

This is a Reimann integral $Q_t(g)={\int }_0^{t}g(s)ds$

Since $h(W_t)$ is continuous, we obtain our result

Itô’s Formula: If $f$ is a function with two continuous derivatives, and $W_t$ is a standard Brownian motion, $f(W_t)-f(W_0)={\int }_0^t{f}^{\prime }(W_s)d{W}_s+\frac{1}{2}{\int }_0^t{f}^{\prime \prime }(W_s)ds$

In differential form,
$df(W_t)=f^{\prime }(W_t)d{W}_t+\frac{1}{2}{f}^{\prime \prime }(W_t)dt$

For $f(t)=t^2$, we get $W_t^2={\int }_0^{t}2W_{s}d{W}_s+\frac{1}{2}{\int }_0^{t}2ds$
${\int }_0^t{W}_{s}d{W}_s=\frac{1}{2}{W}_t^2-\frac{1}{2}t$

That was a lot of work to solve our original question!

Example: Consider $X_t=e^{W_t}$
We use Itô’s formula with $f(t)=e^t$,
$X_t-1={\int }_0^t{e}^{W_s}d{W}_s+\frac{1}{2}{\int }_0^t{e}^{W_s}ds$
$d{X}_t=X_{t}d{W}_t+\frac{1}{2}{X}_{t}dt$