Characteristic functions introduction

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  • Опубліковано 2 лис 2024

КОМЕНТАРІ • 22

  • @SpartacanUsuals
    @SpartacanUsuals  11 років тому +18

    Hi, many thanks for your comment. Glad to hear you found it useful. I do intend to make more videos on the more pure statistical theory, alhough I fear it may not be in time for your exam unfortunately. Good luck with the exam, and please let me know any further additions you think would be useful.Thanks, Ben

  • @Elekko
    @Elekko 11 років тому +9

    Amazing lectures, these help me alot. I will have exam next tuesday and you make SENSE! Keep working on these videos, maybe some more examples with higher difficulty.

  • @TranquilSeaOfMath
    @TranquilSeaOfMath 10 місяців тому +1

    Nice introduction to the topic.

  • @josvandeneynde5849
    @josvandeneynde5849 2 роки тому

    Very clear explanation! Thanks for posting this video!

  • @Nera313
    @Nera313 10 років тому +6

    In the property part, you are talking about group of independent random variables. Do you mean if we add up these random variables?
    Because this very looks like the CF of the sum of those random variables.

    • @SpartacanUsuals
      @SpartacanUsuals  10 років тому

      Hi, thanks for your message. Yes I am talking (I can see it wasn't very clear here) about the sum of random variables. Best, Ben

    • @daryl0161
      @daryl0161 5 років тому

      @@SpartacanUsuals m

  • @PF-vn4qz
    @PF-vn4qz 2 роки тому

    Thank you for making these videos

  • @edwardfranklin4835
    @edwardfranklin4835 10 років тому +4

    The formula for the characteristic function is E[e^(itx)], while the formula for the moment generating function is E[e^(tx)]. Why is there a difference (the i)?

    • @SpartacanUsuals
      @SpartacanUsuals  10 років тому

      Hi, thanks for your message. Yes - it appears quite a subtle difference. However, it does make a significant difference to the sort of function which results. The reason being that when we include i, it is necessary to evaluate an integrand in the complex plane. It is possible to show that for any random variable a CF always exists whereas this is not necessarily the case for a MGF. Hope that helps. Best, Ben

    • @motayam67
      @motayam67 9 років тому

      +Ben Lambert
      ? does it the same function which was introduced by Harmsen for the detection of steganography

  • @Golkarian
    @Golkarian 9 років тому +1

    Love the videos. Do you prove the one to one mapping in any of your videos?

  • @pallavlearn5348
    @pallavlearn5348 7 років тому +1

    For the second property relating to "ax", shouldn't "ax" be the domain of probability distribution? Like (e^itax)P(ax)dx

    • @gautamsethi3751
      @gautamsethi3751 6 років тому

      You're right but P(ax) is always identical to P(x) for any non-trivial constant (i.e. a not equal to 0).

  • @OsamaComm
    @OsamaComm 11 років тому +2

    Very Nice. Thank you very much.

  • @alecvan7143
    @alecvan7143 4 роки тому

    Great vid !

  • @bouhababrahim2292
    @bouhababrahim2292 3 роки тому +1

    Tq so much prof

  • @vasuyadav2931
    @vasuyadav2931 4 роки тому

    Does the t here stand for time?

  • @shravanthipabba8122
    @shravanthipabba8122 4 роки тому

    Thank you sir

  • @malts100
    @malts100 7 років тому +1

    You sounded like simon

  •  4 роки тому

    SORT OF