Setting Up Stochastic Integration

The Riemann-Stieltjes integral generalizes the Riemann integral, allowing us to integrate functions (an integrand) with respect to other functions (the integrator). In stochastic integration, we’d like to consider a random integrator. We can imagine this as integrating a function weighted randomly, rather than by interval width (Riemann) or a deterministic function (Riemann-Stieltjes). Is this possible? There are a couple of bumps in the road…

in progress.

Variations

We are motivated by Riemann-Stieltjes integration, which usually assumes an integrator with finite or bounded variation. We want to model randomness, so how can we understand the variation of Brownian motion?

For $B_t$ Brownian motion,

$$TV(\{B_t\}_{[a,b]} ; [a,b]) = \infty \text{ a.s. for any } a<b$$

and

$$QV(\{B_t\}_{[a,b]} ; [a,b]) = b-a \text{ in probability and } \mathcal{L}^2.$$

Pf.

Let $\prod_k[a,b]$ be the set of all partitions on the interval $[a,b]$ with $k+1$ pieces: $a = t_0 < t_1 < ... < t_k < t_{k+1} = b.$

Let $\prod[a,b] = \bigcup_k \prod_k[a,b]$ be the set of all partitions.

We need to find a sequence of partitions $\pi_n \in \prod_{[a,b]}$ such that

$$\sum_{t_i \in \pi_n} |B_{t_{i+1} - B_{t_i}}| \rightarrow \infty \text{ a.s.}$$

Consider for any $\pi \in \prod_{[a,b]}$:

$$\E \left[ \sum_{t_i \in \pi} (B_{t_{i+1}} - B_{t_i})^2 \right] = \sum_{t_i \in \pi} (t_{i+1}- t_i) = b-a.$$

Let $Z_i = (B_{t_{i+1}} - B_{t_i})^2 - (t_{i+1}- t_i).$ The $Z_i$ are independent with law $(\zeta^2 - 1)(t_{i+1} - t_i),$ with $\zeta \sim N(0, 1).$

This implies that

$$\begin{align*} &\E \left[ \left( \sum_{t_i} (B_{t_{i+1}} - B_{t_i})^2 - (b-a)\right)^2 \right] \\ &= \E \left[ \sum_{t_i} Z_i^2 \right] \\ &= \E \left[(\zeta^2 - 1)^2 \right] \cdot \sum(t_{i+1} - t_i)^2. \end{align*}$$

Choose $\{ \pi_n\} \subset \prod_{[a,b]}$ such that $\sup_{t_i \in \pi_n} |t_{i+1} - t_i| \rightarrow 0.$ Then the above is

$$\leq \E((\zeta^2 -1)^2) \cdot \sup_{t_i \in \pi} |t_{i+1}-t_i| \cdot \sum(t_{i+1}-t_i).$$

The first term is $< \infty,$ the second $\rightarrow 0$ as $n \rightarrow \infty,$ and the final term equals $b-a$. So

$$\sum(B_{t_{i+1}} - B_{t_i})^2 \rightarrow (b-a) \text { in } \mathcal{L}^2.$$

and therefore $\rightarrow b-a$ in probability (by Chebyshev). Then there must exist a subsequence $m_n \uparrow \infty$ such that

$$\sum_{t_i \in m_n} |B_{t_{i+1}} - B_{t_i}|^2 \rightarrow b-a \text{ a.s.},$$

so almost surely,

$$\begin{align*} b-a &= \lim_n \sum (B_{t_{i+1}} - B_{t_i})^2 \\ &\leq \lim_n \left[ \max_{t_i \in \pi_n} |B_{t_{i+1}} - B_{t_i}| \cdot \sum |B_{t_{i+1}} - B_{t_i}| \right] \end{align*}$$

But the first term in the limit is $0$ by continuity of sample paths, so the second term must grow to infinity. Then $TV(\{B_t\}; [a,b]) = \infty$ almost surely. $\blacksquare$

This isn’t great–this means that we can’t just stick $B_t$ into a Riemann-Stieltjes integral and expect a result. Let’s think about the problems that infinite total variation $TV(B) = \infty$ might pose.

Riemann Stieltjes

Suppose we’d like to compute

$$\int_0^t f(s) dB_s,$$

the Wiener integral (which we don’t yet know exists…). $B_s(w)$ is almost surely continuous, but we of course cannot apply the Riemann-Stieltjes treatment.

Consider a nonrandom continuous function $f$ and $g$ to be characterized as follows:

$$\int_0^t f(s) dg(s) = \lim_{\prod_n} \sum_{t_i \in \prod_n} f(s_i) (g(t_{i+1}) - g(t_i))$$

for each $s_i \in [t_i, t_{i+1}]$ and taking the limit of finer meshes $|\prod_n| \rightarrow 0$. Intuitively, the limit should exist independently of our choice of partition and choice of $s_i$, where $f$ is evaluated. Let’s confirm this:

Let $\prod_m$ be a coarser partition than $\prod_n$. Then

$$\begin{align} \sum_{t_i \in \prod_n} f(s_i) (g(t_{i+1}) - g(t_i)) - \sum_{\tilde t_j \in \prod_m} f(\tilde s_j)(g(\tilde t_{j+1}) - g(\tilde t_j)) \\ = \sum_{t_i \in \prod_n} (f(s_i) - f(\tilde s_i^*))(g(t_{i+1}) - g(t_i)), \end{align}$$

where $\tilde s^*_i$ is chosen the obvious way.

$$\begin{align*} \implies (2) &\leq \sum |f(s_i) - f(\tilde s^*_i) | |g(t_{i+1}) - g(t_i) | \\ &\leq TV(g) * \max_i |f(s_i) - f(\tilde s^*_i)|. \end{align*}$$

If $f$ is continuous and $TV(g) < \infty,$ then the sequence of sums is Cauchy.

Then the limit would exist independent of our choice of partition. $\blacksquare$

We know we encounter problems with the Riemann Stieltjes integral with infinite variation. The contrapositive gives:

Pf:…

Pf:

Denote $\lim_{n} \int_0^t f_n dg$ as $I(f_n)$.

$$I(f_n) - I(f_m) = \sum_i (f_n(t_i) - f_m(t_i))(g(t_{i+1}) - g(t_i))$$

for some common refinement of $t$’s. Hence $|I(f_n) - I(f_m)| \leq ||f_n - f_m|| TV(g).$ Since

$$||f_n - f_m|| \leq ||f_n - f|| + ||f - f_m|| \rightarrow 0,$$

we have $I(f_n)$ Cauchy sequence and the limit exists.

Let $(h_n)$ be another sequence of simple functions such that $||h_n -f|| \rightarrow 0$. Note

$$||h_n - f_m|| \leq ||h_n - f|| + ||f - f_m|| \rightarrow 0$$

as both $n,m$ approach $\infty.$ For a common refinement,

$$I(h_n) - I(f_m) = \sum_i (h_n(t_i) - f_m(t_i))(g(t_{i+1}) - g(t_i)),$$

which is again Cauchy. Therefore, the integrals must approach the same limit, regardless of approximating sequence. $\blacksquare$

Pf:

WLOG assume $f=0$.

Let $TV(g) = \infty$. Then there exists a sequence of partitions $(\prod_n)$ such that

$$\sum_{t_i \in \prod_n} |g(t_{i+1}) - g(t_i)| \rightarrow \infty.$$

Define $h_n(t_i) = \text{sgn}(g(t_{i+1} - g(t_i))).$ Then $I(h_n) \uparrow \infty$ but $h_n$ does not converge to $0$ uniformly.

Let $f_n = \frac{h_n}{\sqrt{I(h_n)}}.$ Then $I(f_n) = \sqrt{I(h_n)} \uparrow \infty$ and $\sup_t |f_n(t) - 0 | = \frac{1}{\sqrt{I(h_n)}} \rightarrow 0$.

When $TV(g) = \infty$ the Riemann-Stieltjes limit depends on choice of approximating sequence. $\blacksquare$

Brownian Motion as Integrator

Let’s try to make sense of the integral

$$\int_0^1 f(t,w) dB_t(w)$$

where $(B_t)$ is Brownian motion and $f$ is simple:

$$f_n(t,w) = \sum_j c_j(w) \mathbb{1}_{[j/2^n, (j+1)/2^n]}^{(t)}.$$

Note this means $f_n$ is random piecewise constant. Then we have

$$\int_0^1 f_n(t,w) dB_t(w) = \sum c_j(w) (B_{((j+1)/2^n)}(w) - B_{j / 2^n}(w)).$$

As a toy example, suppose we’re trying to integrate Brownian motion itself. Then for $c_j$, we might consider two natural choices for possible values:

• $c_j(w) = B_{j / 2^n}(w)$, scenario a
• $c_j(w) = B_{(j+1)/2^n}(w)$, scenario b

What happens when we pick either of these?

$$\E[\int_0^1 f_n(t,w) dB_t(w)]$$

is either $0$ (scenario a) or $1$ (scenario b). Both seem like they should be good approximations for $f(t,w) = B_t(w),$ but evaluate to different results!

As we construct the stochastic integral, we’ll see that generally we take one of two choices:

$$\sum f(t_i^*, w) \mathbb{1}_{[t_0, t_{i+1}]}$$

for $t^*_i = t_i$, the Ito choice, or $t^*_i = \frac{t_i + t_{i+1}}{2}$, the Stratonovich choice.

What else should we expect? Certainly $f(t,w)$ needs to be $\mathcal{F}_t$ measurable, with $\mathcal{F}_t$ the natural filtration for $B$, i.e., $\mathcal{F_t} = \sigma(B_s : s\leq t).$

For example, $h_1(t,w) = B_{t/2}(w)$ is adapted, but $h_2(t,w) = B_{2t}(w)$ is not adapted.

Also, as per the variation discussion above, we’ll want $QV(B_t) < \infty,$ so the Wiener integral will be defined in the $\mathcal{L}^2$ limit instead of $\mathcal{L}^1$.