Ito’s Integral: Why Riemann-Stieltjes approach does not work, and how does Ito’s approach work?
Vložit
- čas přidán 21. 12. 2018
- Explains visually the Riemann-Stieltjes approach, and why it does not work when the integrator is a Brownian motion. It then explains how a new approach, based on approximation of the integrand by simple functions works, and proves that the Ito’s integral of the step function converges, and is uniquely defined.
I thought "I wish 3Blue1Brown had a stochastic calculus video teaching the intuition behind these weird mathematics." Well, in fact, 3Blue1Brown does not have these videos, but the work you have done is very impressive and effective in teaching these things. Thank you a lot!
Thanks @Matheus Popst!
his work is amazing but the Wikipedia recording is so annoying 😭😭😭
@@concoursmaths8270 same here.
this is without a doubt the damn best information on the interwebz. Forget reddit, forget 4chan, forget even Porhub ... Quantpie is the place to be. Please make a complete course on Real Analysis !!!
First, thank you so much for making this series!!! Please make a complete online course on real analysis in the same way if possible!!! Actually, the contents are profound but concise enough for non-math major students especially for those who haven't learned real analysis and functional analysis. I am now collecting materials from online videos and rigorous textbooks to learn stochastic calculus and your work helps me to get an insight on a deeper level! Respect!!!
We will! Thanks!
Oh my God !!!
What a good quality content is!!
Thanks mam.
God bless you!!!
Thanks a lot 😊
Incredible! Thank you for the excellent explanation!
Glad you enjoyed it! thanks!
Thank you for this great video!! one point was not clear to me. During the moment (9:55 - 10:07), that answers the question of why Riemann-Stieltjes does not work, the video says: "... but how do we show that it converges. and you can see the infinite variation of Brownian motion which manifests itself through the zigzaggy path here, makes the Riemann-Stieltjes approach irrelevant..." I am still not convinced with the stated conclusion given the explanation.
From the graph, even with the zigzaggy path, I can imagine that we would approach the area that we seek to compute as the number of partitions approaches infinity just like how we did with the function g(t) before mentioning the Brownian motion. From this visualization and following the Riemann-Stieltjes approach, I still cannot imagine why it does not converge?! What is the thing in the zigzaggy path, that g(t) does not have, which prevents the convergence? If I have to guess, the answer is the non-differentiable points that prevent that notion of convergence from existing? it would be great if a visualization was provided to help convince a viewer naive in math like me.
I am very thankful to you
I can now imagine (and understand!) the sentence: "5 years behind the bars for not being adapted to filtration!".
Choose the 'filtration' carefully, wrong filtration can take us to infinity!
These videos are absolutely great! Are there any accompanying notes/book?
Many thanks! Sorry don't have a book or notes, but adding the notes on the website over time - though been incredible slow!
I wish you did your own voice over or one that is a bit better, because despite the robotic voice I sticked around as the information was so easy to understand!
You are the reason I'm going to pass my final exam
thanks @Mytherrus, and good luck with the exams! Glad to hear the material complemented your course material, and to many more happy years of learning!
Amazing channel
Many thanks!! Very kind of you!
at 20:31 the I(S_1) summation how can there be a B_2 when the summation only goes to 1. Sadly I don't see how it can be B_2 - B_0 which is crucial to split it up into two terms. Thanks for your help :)
Ah I see! It is the indexing. The interval is from 0 to 2 in both cases (left and right hand sides):
On the LHS, we use one interval, so the partition is just [t_0, t_1) where t_0=0 and t_1=2. Hence B_{t_0}=B_0, and B_{t_1}=B_2.
On the RHS, we use two intervals, so the partition is { [t_0, t_1), [t_1, t_2)} where t_0=0, t_1=1, and t_2=2. Hence B_{t_0}=B_0, and B_{t_1}=B_1, and B_{t_2}=B_2.
@@quantpie ahh clear thank you very much :)
Is steljes integral like a line integral like in multivariable calculus
When plotting the Brownian Motion integral arent some increments negative? I don't get how it is correct to look at it like that.
thanks Luca! Yes indeed the increments can be negative. In the graph on the cover page you can see the green line in the grey plane has ups and downs, so these represent the negatives and positives of Brownian path increments.
@@quantpie Yes but should that not be visible in the integral then somehow? I am probably looking at it the wrong way.
@@lucaalthaus7466 mean square hits a note?
16:08 i found that truely enlightening, integral of product of two functions is an continuous analogue to the the discrete inner product, and that standard deviation is length of the vector and that correlation is the cosine of the angle..... Wow.... Very intuitive, now i know why they call two functions orthogonal if there product of integration equals zero
thank you! glad you found it helpful!
@@quantpie im still abit unclear why we need step functions to deal with brownian motion
In 22:04 the c is out of expectation because it is "not anticipating". However the c means approximation of random function. Random means it is represented some moments. Ex) mean, variance.. So I think it could be some expecting behavior. I mean what is the detail reason that c is out of expectation?
Why do we need a 3d function for visual representation of reimann stieltjes integral. When we only have a single independent variable t and two functions which take on their values dependent on this t. Can anybody help me with this?
A 3D plot is made because you got 3 axes: "t", "g(t)", and "f(t)". If you look at the plane g(t) vs "t" you can plot a t² curve. Then you can use this curve to identify the different dg(t) intervals and relate these intervals with a specific f(t) value. If you project the different f(t) values evaluated along the curve into the f(t) vs g(t) plane, you will end up with another curve that has an area equal to the initial integral.
I hope my explanation was clear and helped you. It's kind of hard to explain something in a CZcams comment 😅...
7:01 I don't understand why you say the Stieltjez integral computes the area of the green fence. To me, the area of the green fence would be given by the integral of f(t) * sqrt(dg^2 + dt^2), as we need to integrate the vertical strips, each of which has an area given by the height f(t) times an infinitesimal hypotenus in the horizontal plane given by the red and green axes. I think the Stieltjez integral computes the area of the projection of that fence onto the vertical plane given by the blue and green axes, as the strips being integrated are all parallel to that plane. We're integrating f(t) * dg(t), not f(t) * sqrt(dg^2 + dt^2), and dg is the projection of sqrt(dg^2 + dt^2) onto the green axis.
8:02 I don't understand what you mean by "flatten the line tracing the function g to x-axis". 🤷
Yes, I had the same thoughts process, the area of the green fence is given by a line integral as you described with differential closely related to pythagorean theorem. In my view, is no the same as the projection of that fence onto the f-g plane (x,x^2).
I do have a sense of the meaning to "flatten g to x-axis",
I will use S as for integral sign ---- S f(x) dx (integral f)
S x dx^2 , let t = x^2, then sqrt(t) = x
now we change the limits a' = 0^2 = 0 and b' = 2^2 = 4
then:
S_[0,2] x dx^2 = S_[0,4] sqrt(t) dt
In this sense you have flatten the integration curve now as a line, therefore the function x have to be slowed down to behave like sqrt(t).
In fact this is the projection you mentioned now looking at the f-g plane perspective.
You're right man. She (he?) made a mistake in that part. In reality the area we get from that integral is the projected one onto the f(t) vs g(t) plane. Otherwise we would be using a different differential which is the arc length ds instead of dg(t).
I'm glad I'm not the only one who found this weird.
you are amazing
very kind of you!! thank you!
I was thrown off by the narrator but it turned about to be a fantastic explanation!
thank you!
Maybe I am stupid but I still do not understand why the infinite variation of Brownian motion becomes not a problem any more once you introduce the simple process. It looks to me that, if I want to approximate the simple process to the original process by increasing n, the infinite variation of Brownian motion should become a problem again as it is in Rieman Stieltjes Intergarl.. Can you explain?
Is it because Ito integral is not defined by the convergence of lower sum and upper sum but by the mean square convergence in the probability space?
Is there a particular reason you believe that this might be the case? Even if we don't trust the numerical calculations when the numbers get small, the theory still suggests that it should work. As an aside, even though brownian motion has infinite variation, at the end of the day it is a continuous process, so if we one goes very tiny steps, then the amount by which the process changes will be very small.
mean square is quite a strong convergence - 'mean' might give the impression of 'on average' but that is not really the case. For the simple Brownian you can find proofs in other modes as well.
at 7:29 you said 2 squared is 8 but t^2 should be 4 and then the area of integration should be 8
Thanks for highlighting the confusion it might cause. It is meant to say as you pointed out: We start with one interval, and take the upper sum as the value of the function, f, at the right end of the interval, which we know is 2, and multiply it by the change in the value of the function, g, which is along the, z axis and equals 2 squared, so we get 8. i.e., 2* 2^2.
@@quantpie yes thank you you are awesome
Quite a lot of work that you put into the graphical representation. This video would profit a lot from being explained/read by a human being with a sense for intonation
many thanks for the feedback! much appreciated!
I think this is wrong. Nobody wants to integrate with respect to the total length of the random walk. If the price walks up a bit then down a bit, the effect is zero. On the previous slide the video said that a monotonically increasing or decreasing function was needed. In Σ q(t) . [ p(t+1) - p(t) ] , that's clear.
Bruh, thanks
Awesome content ! Don't shy to show us your real voice
many thanks! much appreciated!