Ito’s Integral: Why Riemann-Stieltjes approach does not work, and how does Ito’s approach work? 

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!

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!!!

I can now imagine (and understand!) the sentence: "5 years behind the bars for not being adapted to filtration!".

Incredible! Thank you for the excellent explanation!

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I am very thankful to you

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.

These videos are absolutely great! Are there any accompanying notes/book?

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 :)

You are the reason I'm going to pass my final exam

Amazing channel

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?

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.

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?

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!

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

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?

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". 🤷

you are amazing

at 7:29 you said 2 squared is 8 but t^2 should be 4 and then the area of integration should be 8

I was thrown off by the narrator but it turned about to be a fantastic explanation!

This is a cool video, however it feels like you have shown the construction of the so-called Stratonovich integral instead of the Itô integral. (These two integral concepts are not the same.)

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.

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

Awesome content ! Don't shy to show us your real voice

Bruh, thanks