[Math] Why do we unavoidably (or not) use Riemann integral to define Itô integral

brownian motionprobability theorystochastic-analysis


$$\int_0^tH_tdB_t\equiv \lim_{n\rightarrow\infty}\sum_{i=1}^nH_{t_i}(B_{t_i}-B_{t_{i-1}})$$

But I'm wondering why not defining this using Lebesgue Integral?

It looks more consistent, meaning we can somehow 'obsolete' Riemann integral after knowing Lebesgue Integral. Moreover, we can integrate processes that are hard to integrate. Isn't this good?

Quote from wikipedia:
Suppose that $B$ is a Wiener process (Brownian motion) and that $H$ is a right-continuous (cadlag), adapted and locally bounded process. If
$\{π_n\}$ is a sequence of partitions of $[0, t]$ with mesh going to zero, then the Itô integral of H with respect to B up to time t is a random variable. It can be shown that this limit converges in probability…

Or maybe because the above condition that the limit converge is strong enough in most situations?

Best Answer

First of all, note that the limit

$$\int_0^t H(s) \, dB_s = \lim_{n \to \infty} \sum_{i=1}^n H_{t_i} (B_{t_i}-B_{t_{i-1}}) \tag{1}$$

is a limit in $L^2$ (or, alternatively, in probability); the limit does, in general, not exist in a pointwise sense. This means, in particular, that the existence of this limit does not imply that the function $t \mapsto H_t(\omega)$ is Riemann-integrable for each $\omega \in \Omega$. This is really something you should be careful about when working with such limits: in which sense does the limit exist?

Let's turn to the question why we cannot introduce the Itô integral in a pointwise sense. There is the following general statement which is a direct consequence of the Banach-Steinhaus theorem (see below for a detailed proof):

Theorem Let $\alpha:[a,b] \to \mathbb{R}$ be a mapping. If the Riemann-Stieltjes integral $$I(f) := \int_a^b f(t) \, d\alpha(t)$$ exists for all continuous functions $f:[a,b] \to \mathbb{R}$, then $\alpha$ is of bounded variation.

This means that if we define a stochastic integral with respect to a stochastic process $(X_t)_{t \geq 0}$ in a pointwise sense

$$\int_0^t f(s) \, dX_s(\omega), \qquad \omega \in \Omega$$

as a Riemann-Stieltjes integral, then this integral is only well-defined for all continuous functions if $t \mapsto X(t,\omega)$ is of bounded variation (on compacts) for all $\omega \in \Omega$. However, it is well-known that the sample paths of a Brownian motion are almost surely of unbounded variation, and therefore the definition of a stochastic integral in a pointwise sense is not a good idea: the class of functions which we can integrate would not even include the continuous functions.

Proof of the above theorem: The idea for this proof is taken from R. Schilling & L. Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Corollary A.41.

Consider the normed spaces $$(X,\|\cdot\|_X) := (C[a,b],\|\cdot\|_{\infty}) \qquad \text{and} \qquad (Y,\|\cdot\|_Y) := (\mathbb{R},|\cdot|).$$ For a partition $\Pi = \{a=t_0<\ldots<t_n=b\}$ of the interval $[a,b]$ define $I^{\Pi}:X \to Y$ by

$$I^{\Pi}(f) := \sum_{t_j \in \Pi} f(t_{j-1}) (\alpha(t_j)-\alpha(t_{j-1})).$$

Since, by assumption, $I^{\Pi}(f) \to I(f)$ as $|\Pi| \to 0$ for all $f \in C[a,b]$, we have

$$\sup_{\Pi} |I^{\Pi}(f)| \leq c_f < \infty$$

for all $f \in C[a,b]$. Applying the Banach Steinhaus theorem we find

$$\sup_{f \in C[a,b], \|f\|_{\infty} \leq 1} \sup_{\Pi} I^{\Pi}(f) < \infty$$

which is equivalent to

$$\sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty. \tag{2}$$

For a partition $\Pi$ let $f_{\Pi}:[a,b] \to \mathbb{R}$ be a piecewise linear continuous function such that

$$f_{\Pi}(t_{j-1}) = \text{sgn} \, (\alpha(t_j)-\alpha(t_{j-1}))$$

for all $j=1,\ldots,n$. By the very choice of $f_{\Pi}$, we have

$$I^{\Pi}(f_{\Pi}) = \sum_{t_j \in \Pi} |\alpha(t_j)-\alpha(t_{j-1}))| \leq \sup_{f \in X, \|f\|_X \leq 1} |I^{\Pi}(f)|. \tag{3}$$

Combining $(2)$ and $(3)$, we conclude

$$\text{VAR}_1(\alpha,[a,b]) = \sup_{\Pi} I^{\Pi}(f_{\pi}) \leq \sup_{\Pi} \sup_{f \in C[a,b],\|f\|_{\infty} \leq 1} |I^{\Pi}(f)| < \infty.$$

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