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Uniform Probability Density Function

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that a point [tex](X_1 , X_2 , X_3)[/tex] is chosen at random, that is, in accordance with the uniform probability function over the following set S:

    [tex]S = {(x_1, x_2, x_3) : 0 \leq x_i \leq 1 for i =1,2,3}[/tex]

    Determine [tex]P[(X_1 - 1/2)^2 + (X_2 - 1/2)^2 + (X_3 - 1/2)^2) \leq 1/4] [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Not sure how to set this up. Any suggestions/hints?
     
  2. jcsd
  3. Mar 17, 2009 #2

    lanedance

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    Hi cse63149

    the allowable variable space defines a cube.... I'd try thinking about volume and how you could relate volume it to probabilty....
     
  4. Mar 17, 2009 #3
    Would it be a series of integrals or something?

    Sorry, still not getting it.
     
  5. Mar 17, 2009 #4

    HallsofIvy

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    Not a series of integrals- a single integral. The volume of the given cube is 1 so the probability that [itex](x- 1/2)^2+ (y- 1/2)^2+ (z- 1/2)^2\le 1/4[/itex] is is the volume of that figure.
     
  6. Mar 17, 2009 #5
    so:

    [tex]\int^{1/4}_0 (X_1 - 1/2)^2 dX_1 + \int^{1/4}_0 (X_2 - 1/2)^2 dX_2 + \int^{1/4}_0 (X_3 - 1/2)^2 dX_3[/tex]

    Is that correct?
     
    Last edited: Mar 17, 2009
  7. Mar 17, 2009 #6

    Mark44

    Staff: Mentor

    What geometric figure does this inequality describe?
    [itex](x- 1/2)^2+ (y- 1/2)^2+ (z- 1/2)^2\le 1/4[/itex]
    And what is the volume of this figure?
     
  8. Mar 17, 2009 #7
    The shape is a sphere, but wouldn't the triple I set up in my earlier post be correct, or did I make a mistake?
     
  9. Mar 17, 2009 #8

    Mark44

    Staff: Mentor

    No, they're not correct. All three produce the same value, the area under a parabola, and you're adding these identical areas to get, what?

    Your probability represents a sphere of a certain size, within a cube-shaped region with volume 1. You don't need any calculus for this problem, just a bit of garden-variety geometry.
     
  10. Mar 17, 2009 #9
    Sorry again, but I'm still stuck. Would it be possible to get another hint?
     
  11. Mar 17, 2009 #10

    Mark44

    Staff: Mentor

    When you get it, you're going to slap yourself in the forehead and say "Doh!"

    The probability you're trying to find in this uniform distribution is nothing more than the ratio of the volume of the ball to the volume of your 1 x 1 x 1 cube. (A ball is a solid object, all of whose surface points are equidistant from its center. A sphere is the skin of a ball.)

    The ball is the set [itex](x- 1/2)^2+ (y- 1/2)^2+ (z- 1/2)^2\le 1/4[/itex].
    From this inequality, I can see by inspection the center of the ball and its radius, so I pretty much know everything there is to know about the ball.
     
  12. Mar 17, 2009 #11
    I got the ratio part, but what I'm trying to find is the volume of the ball, unless it's in fact 1/4 (which I doubut it is)

    would the center of the ball be (1/2, 1/2, 1/2) and its radius is 1/8?
     
    Last edited: Mar 17, 2009
  13. Mar 17, 2009 #12

    lanedance

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    getting close radius looks a bit off... the equation of a sphere at center (a,b,c) and radius r will look like

    [tex](x-a)^2 + (y-b)^2 + (y-c)^2 = r^2 [/tex]

    to understand the consider the points defined setting y=b, z=c
    then
    (x-a)^2 = r^2

    ie the distance for x to a is the radius as expected...
     
  14. Mar 17, 2009 #13
    since [tex]r^2 = 1/4 \rightarrow r = 1 / \sqrt{4} \rightarrow r = 1/2[/tex]
     
  15. Mar 17, 2009 #14

    lanedance

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    looks good

    so think about how the sphere of centre (1/2,1/2,1/2) and radius 1/2 sits in the cube

    first you should ask is it fully contained in the cube?

    then what are the ratio of volumes?

    as the variable are uniformly distributed, this ratio should give you yoru probability
     
  16. Mar 17, 2009 #15
    Might seem like a stupid question at this point, how would I find the volume of the sphere?
     
  17. Mar 17, 2009 #16

    Mark44

    Staff: Mentor

    For a sphere, V = 4/3 * pi * r^3
     
  18. Mar 17, 2009 #17
    *bangs head repeteadly*

    [tex]\frac{volume of sphere}{volume of cube} = \frac{ (4/3) \pi (1/2)^3}{1} ~ \frac{0.5236}{1} = 0.5236[/tex]
     
  19. Mar 18, 2009 #18

    Mark44

    Staff: Mentor

    Like I said...
     
  20. Mar 18, 2009 #19
    yeah, sorry. I only realised it during class the sphere isnt fully contained inside the cube, and I didn't have access to the internet to fix my mistake

    since the sphere is centred at (1/2,1/2,1/2), only half of it would be inside the cube it should be:

    [tex]\frac{volume of sphere}{volume of cube} =\frac{\frac{volume of sphere}{2}}{volume of cube} = \frac{ ((4/3) \pi (1/2)^3)/2}{1} ~ \frac{0.2618}{1} = 0.2618[/tex]

    hopefully I didnt make another stupid mistake somewhere
     
  21. Mar 18, 2009 #20

    Mark44

    Staff: Mentor

    The ball is of radius 1/2 and its center is at (1/2, 1/2, 1/2). It is entirely inside (or just touching) the cube.
     
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