Random variable and probability

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Discussion Overview

The discussion revolves around calculating probabilities related to two-dimensional and three-dimensional random variables that are uniformly distributed over specified geometric shapes. The first part involves a uniform distribution over a square, while the second part concerns a uniform distribution over a sphere.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Post 1 presents two problems involving uniform distributions: one in a square and the other in a sphere, expressing uncertainty about how to start solving them.
  • Post 2 emphasizes that the uniform distribution implies the probability can be calculated as the area (for the square) or volume (for the sphere) of specific regions divided by the total area or volume.
  • Post 3 asserts that the probability for part (a) can be calculated as the area below a specific arc in the square, suggesting that P(ζ<1) = π/4 based on the area of the square being 1.
  • Post 4 challenges the calculation in Post 3, questioning the area of the circle and its relation to the square, suggesting a different value of P(ζ<1) = π/2.
  • Post 5 reiterates the challenge to Post 3's calculation, prompting a reevaluation of the area of the circle within the square.
  • Post 6 acknowledges the correction to π/4 and seeks confirmation on the accuracy of the revised answer.

Areas of Agreement / Disagreement

There is disagreement regarding the correct calculation of the probability P(ζ<1) for the first problem, with some participants proposing π/4 and others suggesting π/2. The discussion remains unresolved as participants have not reached a consensus on the correct approach or answer.

Contextual Notes

Participants express uncertainty about the geometric relationships involved in the problems, particularly regarding the areas and volumes relevant to the uniform distributions. There are also unresolved assumptions about the integration methods and the definitions of the regions involved.

Imagin_e
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Hi!
I'm searching for guidance and help since I don't know how to solve this problem. Here it is:

a) The two-dimensional random variable (ξ,η) is uniformly distributed over the square
K={(x,y): 0≤x≤1 , 0≤y≤1} . Let ζ=√ξ22 me the distance between the origo and the point (ξ,η) . Calculate the probability P(ζ<1)

b) The three-dimensional random variable ξ=(ξ1,ξ2,ξ3) is uniformly distributed on a sphere, which has origo as it's center and a radius of 1. Calculate the probability that ξ ends in the area A which is given by:
A={(x1,x2,x3): 1/3 <√x12+x22+x32≤2/3} . You should, in other words, calculate :
P(1/3<ζ≤2/3) , where ζ=√ξ12+ξ22+ξ32

Note: √ means square root of.. I literarily have no idea where to start.
 
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Hi Imagin_e:

A key point about these problems is "uniformly distributed".

For (a) this means the probability P is the area of a part of the unit square ζ<1. This square has total area = 1.

For (b) this means P will be a volume of a part of a sphere (1/3<ζ≤2/3) divided by the volume of the whole sphere.

Hope this helps.

Regards,
Buzz
 
(a) Since the probability distribution is uniform in the square K(take it as ABCD here), there P(ξ,η) = 1 for all points inside the square(why? : because ∫∫P(ξ,η)dξdη with 0<ξ<1 & 0<η<1 should be 1 and marginal distribution for ξ and η are uniform)
Now we need to find probability of all those points lying below the arc ADB(to the opposite of e, poly1 side), since the distribution is uniform,
therefore, we will have P(ζ<1) = Area below Arc / Area of square = π/4 (area of square is 1)
phyProb.PNG
(b)Now I hope that you can do this as its similar to (a), but if you face problem, feel free to reply
 
Last edited:
Aakash Gupta said:
P(ζ<1) = Area below Arc / Area of square = π/2
Hi Aakash:

Your answer is flawed. What is the total area of the circle of which a part is inside the square? What is the fraction of the whole circle that is inside the square?

Regards,
Buzz
 
Buzz Bloom said:
Hi Aakash:

Your answer is flawed. What is the total area of the circle of which a part is inside the square? What is the fraction of the whole circle that is inside the square?

Regards,
Buzz
Oh sorry that must be pi/4, is there anything wrong apart from that?
 
Aakash Gupta said:
Oh sorry that must be pi/4, is there anything wrong apart from that?
Hi Aakash:

That's OK now.

Regards,
Buzz
 
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