Convergence and non-convergence of left-point and mid-point Riemann sums - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:30:05Z http://mathoverflow.net/feeds/question/16664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16664/convergence-and-non-convergence-of-left-point-and-mid-point-riemann-sums Convergence and non-convergence of left-point and mid-point Riemann sums vonjd 2010-02-28T08:41:38Z 2010-02-28T22:06:21Z <p>In standard calculus it is a well known fact that left-point and mid-point <a href="http://en.wikipedia.org/wiki/Riemann_sum" rel="nofollow">Riemann sums</a> do become equal in the limit. When it comes to stochastic integration this is no longer the case.<br> Taking the left-hand sums renders the <a href="http://en.wikipedia.org/wiki/Ito_calculus" rel="nofollow">Ito integral</a> with an extra term, taking the midpoints renders the <a href="http://en.wikipedia.org/wiki/Stratonovich_integral" rel="nofollow">Stratonovich integral</a> (see for example: <a href="http://www.caam.rice.edu/~cox/stoch/dhigham.pdf" rel="nofollow">Higham</a>, p 531).<br> While the Ito integral is the usual choice in applied math, the Stratonovich integral is frequently used in physics. Unlike the Itō calculus, Stratonovich integrals are defined such that e.g. the chain rule of ordinary calculus holds.</p> <p><strong>My question</strong><br> 1.) What is/are the deeper reason(s) that we have a convergence in ordinary calculus but have a non-convergence here?<br> 2.) Are there examples in non-stochastic calculus where we also have a non-convergence of these limiting cases? If yes, how do they look like (and why)? Could you give a toy example of such a function?</p> http://mathoverflow.net/questions/16664/convergence-and-non-convergence-of-left-point-and-mid-point-riemann-sums/16705#16705 Answer by Gerald Edgar for Convergence and non-convergence of left-point and mid-point Riemann sums Gerald Edgar 2010-02-28T17:33:17Z 2010-02-28T17:33:17Z <p>The reason in ordinary calculus is: if the function is Riemann integrable, then <em>any</em> tags can be chosen in <em>any</em> partitions (maximum length going to zero) and it converges to the integral. </p> <p>And of course this answer is irrelevant for stochastic integration.</p> http://mathoverflow.net/questions/16664/convergence-and-non-convergence-of-left-point-and-mid-point-riemann-sums/16706#16706 Answer by George Lowther for Convergence and non-convergence of left-point and mid-point Riemann sums George Lowther 2010-02-28T17:38:14Z 2010-02-28T22:06:21Z <p>The reason that in stochastic calculus the left-hand and right-hand sums give different integrals really all boils down to quadratic variations. Processes such as Brownian motion have non-zero quadratic variation.</p> <p>Suppose that you are integrating a process X with respect to some other process Y, then choosing a partition 0=t<sub>0</sub>&le;...&le;t<sub>n</sub>=t the approximations using left and right hand sums respectively are, $$\int_0^t X\ dY\approx \sum_{k=1}^nX_{t_{k-1}}(Y_{t_k}-Y_{t_{k-1}})$$ $$\int_0^t X\ \overleftarrow{d}Y\approx \sum_{k=1}^nX_{t_{k}}(Y_{t_k}-Y_{t_{k-1}})$$ The difference between these can be bounded as follows $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\max_k\vert X_{t_k}-X_{t_{k-1}}\vert\sum_k\vert Y_{t_k}-Y_{t_{k-1}}\vert$$ The final term on the right hand side converges to the variation of Y as the mesh of the partition goes to zero and, if X is continuous, the first term goes to zero. So, in standard calculus where integration is always with respect to finite variation functions, it makes no difference whether the left hand or right hand sums are used. Alternatively, the Cauchy-Schwarz inequality can be applied to get the following bound. $$\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})(Y_{t_k}-Y_{t_{k-1}}) \le\sqrt{\sum_{k=1}^n(X_{t_k}-X_{t_{k-1}})^2\sum_{k=1}^n(Y_{t_k}-Y_{t_{k-1}})^2}.$$ As the mesh of the partition goes to zero, the terms inside the square root converge to the quadratic variations of X and Y respectively, denoted by [X] and [Y]. Again, in standard calculus, we use (continuous) finite variation functions, which have zero quadratic variation. However, in stochastic calculus, processes such as Brownian motion have non-zero quadratic variation. Convergence to the quadratic variation along partitions occurs in the sense of convergence in probability - it does not have to converge in the usual sense with any positive probability. A Brownian motion B has [B]<sub>t</sub>=t, so the left and right hand sums can converge to different numbers.</p> <p>With the Stratonovich integration the correct thing to use is the average of the left and right hand sums, <em>not</em> the mid-point. For Ito processes, which are integrals with respect to Brownian motion and time, it makes no difference. This is because their quadratic variations are absolutely continuous. However, for general continuous semimartingales, the mid-point sums don't actually have to converge to anything. (See <a href="http://www-mathdoc.ujf-grenoble.fr/cgi-bin/spitem?id=1136" rel="nofollow">Sur quelques approximations d'intégrales stochastiques</a> by Marc Yor).</p> <p>So, Stratonovich integration uses the average of the left and right hand sums, and the difference between this and the Ito integral is precisely half of what you get using the right-hand sums.</p> <p>As to the question of whether this difference shows up in standard (non-stochastic) calculus, the answer is, as far as I know, hardly ever. In fact, given any continuous functions then you can choose a sequence of partitions along which the quadratic variation vanishes as the mesh goes to zero (I'll leave this as an exercise!). So, given any continuous function with non-zero quadratic variation with respect to some sequence of partitions, so that the left and right hand sums converge to different numbers then, there will be other partitions along which the quadratic variation vanishes. So the integral isn't really defined at all in this case.</p> <p>However, there is one case I have seen where left and right hand sums for deterministic functions converge to different numbers. For this to happen, you have to fix some sequence of partitions with mesh going to zero, and stick to using these to define the integral. Different partitions could lead to different results. Hans Follmer published a paper using this idea in 1981 (<a href="http://www-mathdoc.ujf-grenoble.fr/cgi-bin/spitem?id=1509" rel="nofollow">Calcul d'Ito Sans Probabilities</a>). There aren't any natural (and useful) cases that I know of where this occurs, but you can construct some examples.</p> <p>Given a Brownian motion, the quadratic variation along a sequence of partitions where each is a refinement of the previous one will converge with probability one. So, selecting a Brownian motion sample path at random, you can then define integrals using these partitions in a pathwise sense, rather than using the machinery of stochastic integration. They will converge to the same thing though, so this is a bit of a cheat.Alternatively, you could construct a continuous function along the lines of the <a href="http://en.wikipedia.org/wiki/Weierstrass_function" rel="nofollow">Weierstrass function</a> while forcing the quadratic variation to converge to a nonzero number along a given sequence of partitions. Then, left and right hand Riemann sums along these partitions will converge to a different answer.</p> <p>For example, let s(t) be the 'sawtooth' function s(t) = 1-|1-2{t/2}| ({t}=fractional part of t). Then, define the following function on the unit interval, $$f(t) = s(t)+\sum_{n=1}^\infty 2^{-(n+1)/2}s(2^nt).$$ This has quadratic variation 1, calculated along <a href="http://en.wikipedia.org/wiki/Dyadic_rational" rel="nofollow">dyadic</a> partitions. The following left-hand, right-hand and `Stratonovich' integrals are easily verified,</p> <p>$$\int_0^1 f df = (f(1)^2-f(0)^2-[f]_1)/2.$$</p> <p>$$\int_0^1 f \overleftarrow{d}f = (f(1)^2-f(0)^2+[f]_1)/2.$$</p> <p>$$\int_0^1 f \partial f = (f(1)^2-f(0)^2)/2.$$</p>