L06.7 Joint PMFs and the Expected Value Rule

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  • Опубліковано 6 вер 2024
  • MIT RES.6-012 Introduction to Probability, Spring 2018
    View the complete course: ocw.mit.edu/RE...
    Instructor: John Tsitsiklis
    License: Creative Commons BY-NC-SA
    More information at ocw.mit.edu/terms
    More courses at ocw.mit.edu

КОМЕНТАРІ • 20

  • @tarunpahuja3443
    @tarunpahuja3443 2 роки тому +2

    Think of each value of Y as a scenario. Under each scenario X can take value x with some probability. To find marginal probability, apply total law probability.

  • @ZH59
    @ZH59 2 роки тому

    At 2:17, could someone explain to me why both P(Y=2) and P(X=1) are clearly greater than 0? But P(X=1 and Y = 2) is 0? Thank you!

    • @youngjim7987
      @youngjim7987 2 роки тому +2

      why this cannot be true? Y can equal 2 and X can equal 1 but they cannot both equal to that number simultaneously

    • @shardx191
      @shardx191 Рік тому

      in this case they are not independent

  • @tommymerelte4399
    @tommymerelte4399 4 роки тому +2

    excellent lecture mr

  • @VuNguyen-pd1uq
    @VuNguyen-pd1uq 4 роки тому +4

    sadly, I can't understand very much. Why don't you demonstrate some specific example with real figure ?

  • @AkashKumar-lr6hc
    @AkashKumar-lr6hc Рік тому

    awesome

  • @nileshkharat236
    @nileshkharat236 4 роки тому

    at 9:18 shouldn't it be double summation to be more accurate?

    • @DeadPool-jt1ci
      @DeadPool-jt1ci 4 роки тому +1

      No,you're not running the summation twice , you are running it once.Summing over PAIRS (x,y) such that g(x,y) = z. You get a single value z each time , and u sum over those z's. Essentially what i mean to say is that u sum over the OUTPUT of g(x,y) , not over the INPUT of g(x,y) which is X and Y , and then you would indeed need double summation.
      An example , lets say g(x,y)= x+y , then for the pair (1,2) u have z=3.You add the 3 to your sum , not the 1 or the 2 to some double sum

    • @nileshkharat236
      @nileshkharat236 4 роки тому

      @@DeadPool-jt1ci got it. Thanks

    • @tarunpahuja3443
      @tarunpahuja3443 2 роки тому

      @@DeadPool-jt1ci We could do it with double summation too. Like For each value of x consider all value of y such that g(x,y) = z. This wont sum twice. Like its being done in Expectation.

    • @samedy00
      @samedy00 2 роки тому

      The explanation of Dead Pool is not quite correct. You indeed sum over pairs (x,y) that give the same particular z, but do not sum over z's, since z is the variable at the left hand side and can not be the index of summation.

  • @joshuac9142
    @joshuac9142 7 місяців тому +1

    Very helpful!

  • @avishshah2186
    @avishshah2186 2 роки тому

    3:15

  • @brainstormingsharing1309
    @brainstormingsharing1309 3 роки тому +1

    👍👍👍👍👍👍👍👍

  • @anas.said27
    @anas.said27 2 роки тому +2

    the little russian accent made me sure that math and probability were made by russians lol ( foreigner student in russia )

  • @charlesmendeslima1389
    @charlesmendeslima1389 3 роки тому

    2min

  • @akashwaghmare3957
    @akashwaghmare3957 4 роки тому +2

    it is not clear explained .