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Statistics expectation problem involving circle.

  • Thread starter peripatein
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  • #1
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Hi,

Homework Statement


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].


Homework Equations





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.
 

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Answers and Replies

  • #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:
 
  • #3
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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?
 
  • #4
tiny-tim
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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?
yees, but it's f(θ) that you know, not f(x) …

so what is E(X) expressed as an integral over θ ? :smile:
 
  • #5
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E(X) = ∫(2rsin(θ/2)/2π)(rcos(θ/2))dθ from 0 to 2π?
 
  • #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) ?
 
  • #7
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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?
 
  • #8
tiny-tim
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IHere'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?
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:
 
  • #9
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4r/pi?
How may I now determine whether X obeys an even distribution U?
 
  • #10
tiny-tim
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show how you got it :confused:
 
  • #11
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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?
 
  • #12
tiny-tim
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Well, shouldn't it be E(X) = ∫ 2rsin(θ/2) f(θ) dθ = ∫ (r/pi)sin(θ/2) dθ between 0 and 2pi?
correct :smile:
How may I now determine whether X obeys an even distribution U?
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:)
 
  • #13
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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?
 
  • #14
tiny-tim
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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?
(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​
 
  • #15
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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?
 
  • #16
tiny-tim
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Well, integration by parts, twice.
why not just once? :confused:
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?
i] what is xne-x when x = 0 ?

ii] what is (the limit of) xne-x when x = ∞ ?
 
  • #17
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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?
 
  • #18
tiny-tim
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Now, suppose I define f(x) = (xne-x)/n!. I have found E(X) = n + 1
yes that looks correct :smile:
and V(X) = 0.
what is V(X) ? :confused:
 
  • #19
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Variance (=E(x^2) - (E(x))^2)
Is it not zero in this case?
 
  • #20
tiny-tim
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no :confused:

how did you get E(x2) ?
 
  • #21
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My bad, I made a mistake along the way. How's the following:
E(X2) = ∫ (from zero to infinity) (x2xne-x)/n!dx, right? And that is In+2/n!, which is in turn (n+2)(n+1), is it not? Now, (n+2)(n+1) - (n+1)2 = n+1 = Var(X). Is it correct now?
 
  • #22
tiny-tim
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yes!! :rofl:

you see now how important it is to write everything out carefully? :wink:
 
  • #23
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Yes, tiny-tim, thank you very much! :-)
 

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