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Unit circle derived distribution

  1. May 2, 2015 #1
    Hi

    Assume an x-coordinate from the unit circle is picked from a uniform distribution.
    This is the outcome of the random variable X with probability density function Xden(x) = 0.5 (-1 < x < 1).
    The random variable Y is related to the random variable X by Y = f(X) = √(1-X2) and X = g(Y) = ±√(1-Y2).
    What is the probability density function Yden(y)?

    Since Yden(y) = Xden(g(y)) / |f'(g(y))|, i first calculate f'(x) = -x/√(1-x2).
    Then Yden(y) = Xden( √(1-Y2) ) / f'( √(1-Y2) ) + Xden( -√(1-Y2) ) / f'( -√(1-Y2) )

    I get Yden(y) = 0

    Where do i go wrong?

    Rgds
    Rabbed
     
  2. jcsd
  3. May 2, 2015 #2

    wabbit

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    Your last formula looks (almost) correct but the result is not zero. What are the values of the Xden(..) in that formula and where did the absolute values go ? Also, do compute f'(g(y)), it will simplify.

    As an aside, it's a personnal quirk perhaps but I never use the density formulas, it's easy to mishandle them - using cumulative distribution functions instead, you cannot go wrong.
     
  4. May 2, 2015 #3
    Hey

    Thanks for your answer!
    I checked my calculation again (in particular the absolute values) and came up with this:
    Yden(y) = √y2 / √(1-y2)
    Don't know if that's correct.. i'll try do it the other way around, starting with Yden(y) to get Xden(x) = 0.5

    How do you do this using CDF instead?

    Rgds
    Rabbed
     
  5. May 2, 2015 #4

    wabbit

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    That looks ok to me, you might replace ## \sqrt{y^2} ## with just ## y ## though : )

    With the CDF ## P(Y\leq y)=P(\sqrt{1-X^2}\leq y)=P(|X|\geq\sqrt{1-y^2})=1-\sqrt{1-y^2}##, differentiate and you get your result.
    Of more general applicability, differentiate ##P(|X|\geq\sqrt{1-y^2})##, this gives you the Y-PDF from the X-PDF without needing the actual X-CDF.

    Same thing really - for me it's harder to miss a step this way, I just find it more concrete but it's really a matter of taste, and to be perfectly homest, I'm just too lazy to memorize the PDF formula.
     
    Last edited: May 2, 2015
  6. May 4, 2015 #5
    Ok, i'll check out that technique, thanks.

    So what does the answer tell me, that if i pick an x-coordinate at random between -1 and 1 (with all having equal probability), it's very probable that it's corresponding y-coordinate (on the upper half of the unit circle) is close to 1, and very improbable that the y-coordinate is close to 0?

    Rgds
    Rabbed
     
  7. May 4, 2015 #6

    wabbit

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    Something like that, yes, depending how improbable is very improbable :)
    This relates to the fact that the half-circle has a horizontal tangent at x=0 and a vertical one at x=1.
     
  8. May 29, 2015 #7
    Hi again...

    Is this a correct usage of the CDF method?

    Y_den(y) =
    Y_cdf'(y) =
    ( P( Y < y ) )' =
    ( P( f(X) < y ) )' =
    ( P( sqrt(1-X^2) < y ) )' =
    ( P( 1-X^2 < y^2 ) )' =
    ( P( -1+X^2 > -y^2 ) )' =
    ( P( X^2 > 1-y^2 ) )' =
    ( P( X > sqrt(1-y^2) ) )' =
    ( 1 - X_cdf(sqrt(1-y^2) ) )' =
    (-X_cdf'(sqrt(1-y^2))) * (-y/sqrt(1-y^2)) =
    (-X_pdf(sqrt(1-y^2))) * (-y/sqrt(1-y^2)) =
    (-0.5) * (-y/sqrt(1-y^2)) =
    0.5*y/sqrt(1-y^2)

    Rgds
    Rabbed
     
  9. May 29, 2015 #8

    wabbit

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    Almost. You are off a factor of 2 because of your step ## ( P( X^2 > 1-y^2 ) )' =
    ( P( X > \sqrt{1-y^2} ) )' ##. Looks like you forgot something here : )

    Oh and by the way, do try the tex formatting, it's not difficult (there's a guide in the help/howto section too), and it makes equations much easier to read...
     
  10. May 30, 2015 #9
    Do you mean this?
    ##
    ( P( X^2 > 1-y^2 ) )' =
    ( P( X > \pm\sqrt{1-y^2} ) )' =
    ( P( (X > -\sqrt{1-y^2}) OR (X > \sqrt{1-y^2}) ) )' =
    ##
    And then i can use the OR/+ formula to continue?

    Shouldn't ## P( (X > -\sqrt{1-y^2}) OR (X > \sqrt{1-y^2}) ) = P( X > -\sqrt{1-y^2} ) ## ?
     
  11. May 30, 2015 #10

    wabbit

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    Yes the OR was sthe missing part - but it's easy to handle here : from the symmetry of the distribution of X, ## P(|X|>x)=2P(X>x)##

    Note : it is true of course that in general ## P(|X|>x)=P(X>x)+P(X<-x) ## but in this case the two terms are equal, hence the simplification.
     
  12. May 30, 2015 #11

    HallsofIvy

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    This seems to me to be ambiguous. Yes, the x coordinates of points on the circle all lie between -1 and 1. But does this mean that x is uniformly distributed on -1 to 1, as you are assuming here, or does it mean that x is x coordinate of a point on the unit circle and the points are uniformly distributed?
    Those are not at all the same. In the latter case, X is [itex]cos(\theta)[/itex] with [itex]\theta[/itex] uniformly distributed between [itex]0[/itex] and [itex]2\pi[/itex].

     
  13. Jun 13, 2015 #12
    Thanks, i've got the same answer with the CDF method now.

    HallsofIvy:
    Yes, i mean the x-coordinate should be uniformly distributed, not the angle. But I do want to check that also later, to see the difference.

    My next concern is how i get from:
    Y_pdf(y) = y/√(1-y2)
    X = g(Y) = ± √(1-Y2)
    to:
    X_pdf(x) = 0.5

    I think i go wrong here:
    ( P( ± √(1-Y2) < x ) )' =
    ( P( -√(1-Y2) < x OR √(1-Y2) < x ) )' =
    ( P( √(1-Y2) > -x OR √(1-Y2) < x ) )' =
    ( P( 1-Y2 > x2 OR 1-Y2 < x2 ) )'

    Any tips? :)

    Rgds
    Rabbed
     
    Last edited: Jun 13, 2015
  14. Jun 13, 2015 #13

    ChrisVer

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    Your inequalities are wrong... [itex]x^2>f(y) \Rightarrow x < -\sqrt{f(y)} ~~\text{or}~~ x> + \sqrt{f(y)}[/itex]
     
  15. Jun 13, 2015 #14

    ChrisVer

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    Also when you have something like: [itex] \pm \sqrt{1+Y^2} \le x[/itex] it doesn't really make sense...

    edit \*what happened with the square roots and latex?*\
     
  16. Jun 19, 2015 #15
    Thanks,

    But how do you go from [tex]X = g(Y) = ± \sqrt{1-Y^2}[/tex]
    to insert into the CDF definition [tex]P(X < x)[/tex]

    Should i start by squaring them both, so that i can insert
    [tex]X^2 = g(Y)^2 = (± \sqrt{1-Y^2})^2 = |1-Y^2|[/tex]
    into
    [tex]P(X^2 < x^2)[/tex]
    and get
    [tex]P(x^2 > |1-Y^2|) = P(x < -\sqrt{|1-Y^2|}\ OR\ x > \sqrt{|1-Y^2|})[/tex]
    ?

    Rgds
    Rabbed
     
  17. Jul 24, 2015 #16
    Any pointers on this, please?

    Rgds
    Rabbed
     
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