Stochastic integral and Stieltjes integral

Question

My question is on the convergence of the Riemann sum, when the value spaces are square-integrable random variables. The convergence does depend on the evaluation point we choose, why is the case. Here is some background to make this clearer. Suppose $f\colon \Re \mapsto \Re $ is some continuous function on $[a,b]$, the Stieltjes integral of $f$ with respect to itself $f$ is $\int^{b}_{a} f(t)df(t)$ if we take a partition $ \Delta_n = \{t_0, t_1, \cdots, t_n \}$ of $[a,b]$ the Riemmans sums is $$ L_{n} = \sum^{n}_{i=1} f(t_{i-1})(f(t_{i})-f(t_{i-1})) $$
Now if the limit exists say $\lim \limits_{n\to\infty} L_{n}= A$, then if we choose the evaluation point $t_{i}$ then the sum
$$ R_{n} = \sum^{n}_{i=1} f(t_{i})(f(t_{i})-f(t_{i-1})) $$
will also converge to $A$ so $$\lim_{n\to\infty}L_{n} = \lim_{n\to\infty}R_{n} .$$ Now we apply same idea for a stochastic integral. Here $W(t)$ is a wiener process and we wish to find $$\int^{b}_{a}W(t)dW(t) $$ $$ L_{n} = \sum^{n}_{i=1} W(t_{i-1})(W(t_{i})-W(t_{i-1})) $$
$$ R_{n} = \sum^{n}_{i=1} W(t_{i})(W(t_{i})-W(t_{i-1})) $$ in $L^2$ norm the limits of $L_{n}$ and $R_{n}$ exist but are different $$\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = b-a $$ can someone explain why the limits are different ? If the limit exists which in this case it does. I would have expected $\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = 0 $ in $L^2$ norm.

Answer 1

Limits of $R_n$ and $L_n$ coincide when Stieltjes integral exists. Existence of the Stieltjes integral does not follow from the existence of these limits. In general existence and definition of Stieltjes integral can be messy business as the Figure 2.1 in page 6 (page 10 of the ps file) of this document can attest.

Answer 2

Hint 1:

The basis of the explanation is in the different behaviour of the increments $f(t_i)-f(t_{i-1})$ and $W(t_i)-W(t_{i-1})$. The increments of the first are $O(t_i-t_{i-1})$ those of the second are $O(\sqrt{t_i-t_{i-1}})$.

Hint 2: $$R_n - L_n =\sum_{i=1}^n \left[ f(t_i)-f(t_{i-1}) \right]^2$$ and likewise for $W(t)$.

Answer 3

First write $$ R_n - L_n = \sum\limits_{i = 1}^n {[W(t_i ) - W(t_{i - 1} )]^2 }, $$ then consider Quadratic variation of Brownian motion.