Our First Ito Integral

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  • เผยแพร่เมื่อ 30 ม.ค. 2025

ความคิดเห็น • 16

  • @junaidjawaid547
    @junaidjawaid547 2 ปีที่แล้ว +1

    The explanation is really good. Thank you for putting so much thought and effort.

  • @steez54321
    @steez54321 2 ปีที่แล้ว +3

    Thank you so much! This makes so much more sense! The Ito Integral was such a mystery to me even after reading 3 separate textbooks on this topic all of which did a bunch of hand waving for even this exact same integral problem. Now I now how to actually to compute an Ito integral which is was I was was looking to learn.

    • @johnthequant
      @johnthequant  2 ปีที่แล้ว +2

      Me too, man. Textbook explanations of Ito Integrals are needlessly complicated. Glad I could help!

  • @MichaelScoleri-v7h
    @MichaelScoleri-v7h 9 วันที่ผ่านมา

    Great video, thank you so much!

  • @ghostwhowalks5623
    @ghostwhowalks5623 2 ปีที่แล้ว +1

    this is AWESOME!! Thanks for explaining this so clearly!

  • @aali4957
    @aali4957 3 หลายเดือนก่อน +1

    The explanation is really good

  • @stochasticxalid9853
    @stochasticxalid9853 2 ปีที่แล้ว +3

    In history, there is John the Baptist and John the Quant. John the Baptist made the path easy to people to get to know the messiah. John the quant is showing the very shortcut to getting acquainted with quant finance. Thank you so much, John :)

    • @johnthequant
      @johnthequant  2 ปีที่แล้ว

      This is an incredible comment; the most flattering compliment I've ever received. Thank you!

  • @remarljosephdamalerio8921
    @remarljosephdamalerio8921 2 ปีที่แล้ว +2

    Thank you! You really helped me a lot!

  • @xXDamolaXx
    @xXDamolaXx 2 ปีที่แล้ว +1

    Hello, the video helped a lot. Thank you
    How did you use the property of Brownian Motion at 13:15? Thanks again

    • @johnthequant
      @johnthequant  2 ปีที่แล้ว +3

      Thanks for watching! Brownian motion is a Wiener process, and one of the defining characteristics of a Wiener process is that each time-step is an independent increment. In the case that increments are Gaussian, like in Brownian motion, each group of timesteps is a sum of independent Gaussian increments. The sum of independent Gaussian increments is also normally distributed; the mean and the variance of the sum is the sum of the means/variances of the increments.
      So,
      B(s)~N(0, s) and B(t)~N(0, t) => B(s) - B(t) ~ N(0, s-t),
      and B(s-t)~N(0, s-t).
      Since B(s) - B(t) and B(s-t) have exactly the same distribution, we can substitute one for the other inside the expected value.
      I hope that helps!

    • @xXDamolaXx
      @xXDamolaXx 2 ปีที่แล้ว

      @@johnthequant That's amazing! Thank you very much, all the best.

  • @pierrebesouhof9987
    @pierrebesouhof9987 2 ปีที่แล้ว +1

    beautiful explanation! thank you

  • @atmonotes
    @atmonotes 3 ปีที่แล้ว +1

    noice. this reminds me of number theory class, the professor would always use an ipad to write long page proofs exactly like that..

  • @tteej
    @tteej 10 หลายเดือนก่อน

    Brilliant, thank you!

  • @steez54321
    @steez54321 2 ปีที่แล้ว

    Why do you omit the 1/2 from the variance computation on the second summation?