thanks Brian! Yes well remembered!! There we proved this using almost sure convergence, which involves slightly more advanced maths. But most textbooks use mean square so we have to cover it from mean square perspective for completeness.

agree!

Hello! Hope you been well! Have you tried this one: quant.stackexchange.com/questions

@@quantpie Thanks, I hope my questions won't look too dumb for them ;)

many thanks!! The method we have outlined works when we know the limit, but it is usually relatively easy to infer the limit, then apply the convergence test, if it does not work, then try another limit. We shall see a heuristic way to infer the limit in another video. Thanks, have collected the videos in the series in this playlist: kzfaq.info/sun/PLS3zAvd8OxeyliTrjCvb8EdWd-5M_mbTc

@@quantpie Can you specify which video? Thank you for all your efforts in this series, it is wonderful.

To make sense of the differential dW, put it through the integral

Thanks Tushar! Yes that’s correct, and hence dw^2=dt.

many thanks!! glad you found it useful!

Many thanks for the question! You mean different in the sense of convergence type-e.g., convergence in probability, convergence in mean square etc. or in terms of steps? Mean squared is more common. We do intend to cover other types of convergence in the future, we just could not because people get bored! In terms of books, we don’t follow any textbook as such, they just don’t seem to cover enough details, and most books seem to adopt the same approach (probably because they share the same editorial board!). Is there any particular book you are reading? If you could provide the reference then we can comment on the approach.

thanks for the great question. The interval is supposed to be small, but it is the same interval over which dw^2 is meant. I can see the confusion is caused by the use of t in the integral, think of it as 0 to dt

Hello, it might be late but it could help others to understand. What he did is to use linearity of expectation and at the righthand side you have E(deltaW^2)=delta t because W is a brownian motion. So the icrements (i.e. delta W) are normally distributed with variance delta t. And because because the increments has expectation zero, the variance is E(delta W^2), hence E(delta W^2)=delta t.

Thanks!! very kind of you!!

Many thanks! Sorry if we give you the impression that it is a derivation- it is not meant to be a derivation or a proof, it is more like an interpretation. I.e., in what sense are these expressions to be interpreted.

Great question! No they are the same, it is like dx^2 in Taylor series, not sure why the differential notation developed this- probably because too many brackets will render it unreadable!