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Taylor series

  1. Jan 13, 2015 #1
    help with the following taylor series:


  2. jcsd
  3. Jan 13, 2015 #2


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    I'm guessing that y is a constant and dy is your variable? and you are trying to expand around dy=0 ? well, you should try to use the definition of the Taylor series, and see what you get :)
  4. Jan 13, 2015 #3
    The thing is that i have an r.v (random variable) of the form Y=X^2

    and by definition : fy(y)dy= P{y<Y<=y+dy}
    i can substitute Y with (X)^2 and then i take the square root from both sides and get :
    P{ sqrt(y)<X<= sqrt(y+dy) }}

    now i want to show the PDF of X by the definition so i want to develop the right side of the inquality into a taylor series.
    sqrt(y+dy) This is the term i want to develop into a taylor series . y and dy both are numbers. i tried to use Taylor's formula but couldn't get it right,

    Hope you understand my question

  5. Jan 13, 2015 #4


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    yes, I think I understand. you want to show how the pdf of X (from first principles) is related to the pdf of Y. Generally, you also need to take into account that X can be negative. (unless you want to specifically say that the random variable X cannot be negative). After this, I think it is easiest to use the cumulative distribution, and differentiate to get the probability density functions.
  6. Jan 13, 2015 #5
    Yes you right X also should be taken negative.
    i didnt understand the step between line 2 and 3
    https://drive.google.com/file/d/0B4wgc0vIE7CCUmpXLXJ0TEtIeUU/view?usp=sharing [Broken]
    Last edited by a moderator: May 7, 2017
  7. Jan 13, 2015 #6

    Stephen Tashi

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    It looks like an argument using differentials. If [itex] f_X [/itex] is the pdf of the random variable [itex] X [/itex] then the probability of the event [itex] \{x: a < x \le a + h\} \approx f_X(a) h [/itex]. This is applied when [itex] a = \sqrt{y} [/itex] and [itex] h = \frac{\triangle y} {2 \sqrt{y} } [/itex] and again when [itex] a = -\sqrt{y} [/itex].
    Last edited by a moderator: May 7, 2017
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