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Uniform Probability Density Function

  • Thread starter cse63146
  • Start date
1. Homework Statement

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. Homework Equations



3. The Attempt at a Solution

Not sure how to set this up. Any suggestions/hints?
 

lanedance

Homework Helper
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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....
 
Would it be a series of integrals or something?

Sorry, still not getting it.
 

HallsofIvy

Science Advisor
Homework Helper
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864
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.
 
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:
32,575
4,304
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?
 
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?
 
32,575
4,304
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.
 
Sorry again, but I'm still stuck. Would it be possible to get another hint?
 
32,575
4,304
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.
 
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:

lanedance

Homework Helper
3,304
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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...
 
since [tex]r^2 = 1/4 \rightarrow r = 1 / \sqrt{4} \rightarrow r = 1/2[/tex]
 

lanedance

Homework Helper
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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
 
Might seem like a stupid question at this point, how would I find the volume of the sphere?
 
32,575
4,304
For a sphere, V = 4/3 * pi * r^3
 
For a sphere, V = 4/3 * pi * r^3
*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]
 
Like I said...
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
 
32,575
4,304
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.
 
in that case, I'm not sure where I made the mistake in:

[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]
 

lanedance

Homework Helper
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why did you change to dividing the sphere in half? It is fully contained in the cube (try drawing it)
 
I was thinking of something else.

I thought I made some sort of mistake whem mark said "Like I said" but I think he's reffering to the part of me banging my head (which I actually did)
 

lanedance

Homework Helper
3,304
2
i know the feeling....
 
Just to make sure, the answer of 0.5236 is the correct one, right?
 

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