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Statistics expectation problem involving circle.

  1. May 13, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    A circle of radius r is as shown in the attached diagram. I am asked to first express X as a function of θ, then to compute E(X). It is also stated that θ obeys U[0,2π].


    2. Relevant equations



    3. The attempt at a solution
    Through simple trigonometry I have found X = 2rsin(θ/2). As θ distributes evenly, I know that the density of θ should be 1/2π. What is the density of x then? Am I supposed to simply substitute 2arcsin(x/2r)? Or should it be approached by first computing E(θ) and trying to hence infer E(X)?
    I'd appreciate an advice.
     

    Attached Files:

  2. jcsd
  3. May 13, 2013 #2

    tiny-tim

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    hi peripatein! :smile:

    look at the definition

    if X(θ) is a function of θ, and if the distribution of θ is f(θ),

    then what is the definition of E(X) ? :wink:
     
  4. May 13, 2013 #3
    The definition of E(X) is integral [would the boundaries be 1 to 2r?] x*f(x) dx, where f(x) is the probability density of X. Is this correct?
     
  5. May 13, 2013 #4

    tiny-tim

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    yees, but it's f(θ) that you know, not f(x) …

    so what is E(X) expressed as an integral over θ ? :smile:
     
  6. May 13, 2013 #5
    E(X) = ∫(2rsin(θ/2)/2π)(rcos(θ/2))dθ from 0 to 2π?
     
  7. May 13, 2013 #6

    tiny-tim

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    no

    stick to the definition for the moment …

    what is the definition of E(X) if X = X(θ), and the distribution of θ is f(θ) ?
     
  8. May 13, 2013 #7
    I am not sure, tiny-tim :-(.
    Should I look for the inverse function of X(θ)?
    I mean, the definition reads E[g(X)] = integral (from -inf. to +inf.) g(x)f(x)dx. I understand that as E[θ(X)] = θ(X)*f(x)dx. I am probably mistaken, yet I am not sure how to approach it.
    Here's another attempt, which makes not much sense to me but for the sake of trying: might it be E(X) = int X(θ)*f(θ)dθ?
     
  9. May 13, 2013 #8

    tiny-tim

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    yes, that's correct …

    E(X) = ∫ X(θ) f(θ) dθ

    … this is exactly the same as the definition you mentioned earlier, ∫ x f(x) dx, but using the distribution of θ rather than the distribution of x (remember, the two f's are different!) :smile:

    so in this case, E(X) = ∫ 2rsin(θ/2) f(θ) dθ,

    which is … ? :wink:
     
  10. May 13, 2013 #9
    4r/pi?
    How may I now determine whether X obeys an even distribution U?
     
  11. May 13, 2013 #10

    tiny-tim

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    show how you got it :confused:
     
  12. May 13, 2013 #11
    Well, shouldn't it be E(X) = ∫ 2rsin(θ/2) f(θ) dθ = ∫ (r/pi)sin(θ/2) dθ between 0 and 2pi?
    How may I now determine whether X obeys an even distribution U?
     
  13. May 13, 2013 #12

    tiny-tim

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    correct :smile:
    it obviously doesn't

    θ is evenly distributed, so sinθ/2 isn't

    (why are you asking? doesn't the question only ask for E(X) ? :confused:)
     
  14. May 13, 2013 #13
    Now, the second part of that question, which is not related to the first, asks for a proof that
    ∫ (from 0 to ∞) xne-x dx = n! for every n > 0, and the hint is to find a recursive argument for In and an initial condition.
    I have found In = (In + 1 + xne-x + nxn-1e-x) / (n(n-1)), where x changes from 0 to ∞, but am not sure how to proceed. Could you advise?
     
  15. May 13, 2013 #14

    tiny-tim

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    (how did you get that ?? :confused:)

    anyway, you shouldn't have any x's in the equation

    x is a dummy variable, the only "real" variable is n​
     
  16. May 13, 2013 #15
    Well, integration by parts, twice. Setting the original integral to be In, I let u = xn and dv be e-x. In the second integration by parts I let u = xn-1 and dv = e-x.
    Does that make more sense now? Is that how it is to be handled? The boundaries should be between 0 and infinity, but I don't know how to incorporate that into the expression (I mean, what do you get for (xne-x + nxn-1e-x) between 0 and inf.? How do you evaluate that?
     
  17. May 13, 2013 #16

    tiny-tim

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    why not just once? :confused:
    i] what is xne-x when x = 0 ?

    ii] what is (the limit of) xne-x when x = ∞ ?
     
  18. May 13, 2013 #17
    You're right, silly me! A single integration by parts led me to the right answer: In+1 = (n+1)In.
    Now, suppose I define f(x) = (xne-x)/n!. I have found E(X) = n + 1 and V(X) = 0.
    Would you please confirm that?
     
  19. May 13, 2013 #18

    tiny-tim

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    yes that looks correct :smile:
    what is V(X) ? :confused:
     
  20. May 13, 2013 #19
    Variance (=E(x^2) - (E(x))^2)
    Is it not zero in this case?
     
  21. May 13, 2013 #20

    tiny-tim

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    no :confused:

    how did you get E(x2) ?
     
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