Setting Up Stochastic Integration

in prog

Variations

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 here).

Why is the infinite total variation $TV(B) = \infty$ a problem?

Suppose we’d like to compute

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

the so-called Wiener integral. $B_s(w)$ is almost surely continuous, but we can’t treat this as a Riemann-Stieltjes integral.

Consider a nonrandom function $f$ continuous 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 $|\prod_n| \rightarrow 0$. The limit should exist independently of our choice of partition and choice of $s_i$’s.

When does the limit exist?

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

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

with $\tilde s^*_i$ 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 and does not depend on choice of partition.

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. $\blacksquare$

What if $TV(g; [0,1]) = \infty$? Then there exists simple functions $||f_n - f||_\infty \rightarrow 0$ such that $I(f_n) \uparrow \infty.$

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. $\blacksquare$

Integrating w.r.t. Brownian Motion

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

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

Ultimately, if we’d want to integrate Brownian motion, we’d consider 2 cases:

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

We can think of these as choices to evaluate $f$ within the interval (e.g., choosing sides of the rectangle for left/right/midpoint Riemann integral).

Note that

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

is either $0$ (scenario a) or $1$ (scenario b). Both seem like good approximations to $f(t,w) = B_t(w),$ but have different integrals!

Generally:

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