Random walk with drift

Поділитися
Вставка
  • Опубліковано 16 січ 2025

КОМЕНТАРІ • 26

  • @Alumiss682
    @Alumiss682 7 років тому +42

    4:44 - 5:03 He explained in 19 seconds what the textbook took to explain in 30 pages.

  • @matthiastz1378
    @matthiastz1378 7 років тому +7

    With you videos more people were saved from being killed in university than from a doctors hand.. thanks a lot statistical life saver!!

  • @mahmoudsamy9407
    @mahmoudsamy9407 5 років тому +2

    you're the most brilliant economics teacher on the internet! :D

  • @theinstigatorr
    @theinstigatorr 10 років тому +9

    Life saver!

  • @jongcheulkim7284
    @jongcheulkim7284 2 роки тому

    Thank you. This is very helpful.

  • @awesomefootypro
    @awesomefootypro Рік тому

    I think Expectation of constant term is constant so E(Xt) should be at+Xo

  • @mah2mah
    @mah2mah 4 роки тому

    I would like to watch the videos of first hitting models or first passage model also.

  • @verdaderoamor980
    @verdaderoamor980 4 роки тому

    Thank you Sir for this instructive video !

  • @vedanandramiad9661
    @vedanandramiad9661 3 роки тому

    How does the summation of the alphas lead to alpha t?

  • @double_j3867
    @double_j3867 9 років тому

    Hi Ben -- just curious what a random walk with drift would look like in state-space form? I just want to confirm my understanding. Ideally something where the drift term was dynamic....Thanks!

  • @lizi9019
    @lizi9019 10 років тому +1

    why alpha*t doesn't enter the variance please? I think you are saying it is uncorrelated with the error terms (e0, e1, e2...) thank you!

    • @casasmanejsle9776
      @casasmanejsle9776 9 років тому +4

      +Lizi zhu Because alpha*t is equal to (alpha+alpha+alpha...) which is the same as adding a constant t times. Now, given that the variance of a constant is zero; therefore, the variance of alpha*t is equal to zero.

    • @ianpan0102
      @ianpan0102 4 роки тому

      alpha * t is a constant, and a variance of a constant is 0 -- doesn't contribute to the total variance.

  • @marxattac
    @marxattac 9 років тому

    Really nice!

  • @Exolofy
    @Exolofy 6 років тому +1

    nice

  • @tanushreedutta960
    @tanushreedutta960 7 років тому

    Why Xo is assumed to be0? What happens if it isn't 0? It will only be added by a constant term and that will still be independent of time. So why are we assuming it to be zero?

    • @c0t556
      @c0t556 6 років тому

      This is an unimportant assumption.

    • @ianpan0102
      @ianpan0102 4 роки тому

      Normally one assumes to be starting at the "origin" on some coordinate system, which can be represented with zero

    • @ianpan0102
      @ianpan0102 4 роки тому

      @@johnnyjonas564 I believe it's an "irrelevant" assumption that we start at the origin to make every move relative and normalized.

    • @alexy7634
      @alexy7634 3 роки тому

      @@johnnyjonas564 To my mind it is just made to make the sums clean and clear. It doesn't matter it it doesn't equal zero, if it equalled 5 then the expectation would be 5 and the variance for it would be zero as it is constant over time, like like the alpha coefficient is constant and therefore has a variance of zero.

  • @shunlilu2247
    @shunlilu2247 3 роки тому

    game changer in 5 mins

  • @Shafique96
    @Shafique96 7 років тому

    @1:10 shouldn't this equation exclude the Xt-2 as you are substituting it out - similarly as you did in the equation before, but with of Xt-1.

    • @ianpan0102
      @ianpan0102 4 роки тому

      He did exclude Xt-2 when substituting it out. You must have mistaken Epsilon t-2 with Xt-2.

  • @yiacinecleakameni8185
    @yiacinecleakameni8185 7 років тому

    sorry for my question, but i didn´t understand why expectation of Xo is zero.

    • @mauritsoever9016
      @mauritsoever9016 4 роки тому

      It started in the infinite past (we assume) so no innovations (epsilons) or drifts (alpha's) happened yet.