Random variable and probability

In summary, the conversation discusses two problems involving uniformly distributed random variables in 2D and 3D. The first problem involves finding the probability of ζ<1, where ζ represents the distance between the origin and a point (ξ,η) in a unit square. The solution is to calculate the area below the arc ADB and divide it by the total area of the square, which results in π/4. The second problem involves finding the probability of 1/3<ζ≤2/3, where ζ represents the distance between the origin and a point (ξ1,ξ2,ξ3) on a sphere with a radius of 1. The solution is to calculate
  • #1
Imagin_e
60
0
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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  • #2
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
 
  • #3
(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:
  • #4
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
 
  • #5
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?
 
  • #6
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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Likes Aakash Gupta

1. What is a random variable?

A random variable is a numerical quantity that takes on different values based on the outcome of a random event. It is typically denoted by a capital letter, such as X or Y, and can be discrete or continuous.

2. How is probability related to random variables?

Probability is the likelihood that a particular outcome will occur in a random event. Random variables are used to model probability distributions, which show the possible outcomes and their corresponding probabilities.

3. What is the difference between a discrete and continuous random variable?

A discrete random variable can only take on a finite or countably infinite number of values, while a continuous random variable can take on any value within a certain range. For example, the number of heads in three coin tosses is a discrete random variable, while the height of a person is a continuous random variable.

4. How do you calculate the expected value of a random variable?

The expected value of a random variable is the sum of all possible values multiplied by their corresponding probabilities. For a discrete random variable, it is calculated as E(X) = ∑xP(X=x), where x represents each possible value and P(X=x) is the probability of that value occurring. For a continuous random variable, it is calculated as E(X) = ∫x f(x)dx, where f(x) is the probability density function.

5. What is the central limit theorem and how does it relate to random variables?

The central limit theorem states that when independent random variables are added, their sum tends to follow a normal distribution, regardless of the original distribution of the individual variables. This is important in statistics because many natural phenomena can be modeled as the sum of many random variables, and the normal distribution is a common and well-understood probability distribution.

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