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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.

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Maybe it has to do with the $L^2$ norm. After all, $||\textbf{x}|| = \sqrt{x_{1}^2+ \cdots + x_{n}^2}$. –  PEV Dec 28 '10 at 16:59
    
I answered a similar question on MO a while ago. Maybe my answer there will help: mathoverflow.net/questions/16664/… –  George Lowther Dec 28 '10 at 21:52
    
@George Lowther thank your answer is indeed very helpful –  almost_sure Dec 28 '10 at 23:36

3 Answers 3

up vote 1 down vote accepted

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.

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thank you. I made the naive assumption if the limits $R_n$ and $L_n$ coincide then the stieltjes integral exists. thanks of the document –  almost_sure Dec 28 '10 at 20:41
    
@almost_sure, you are welcome. I did not know that Stieltjes integrals can be complicated until I took a course where the professor was one of the authors of that document. –  mpiktas Dec 28 '10 at 20:47

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 $W(t)$.

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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.

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Thanks i understand why $R_n - L_n \to {\rm E}(Z_1^2)$ my question is, why does the evaluation point make such a difference when the sums converge in $L^2$ norm? but not in the real valued function case. like other posters said is it just have to do with the Norm. If the limit existed to me it seemed intuitive for $R_n - L_n \to 0 $ in $L^2$ I am still not sure why this is not the case, the hint below have got me thinking it just the norms are different –  almost_sure Dec 28 '10 at 20:20

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