Probability/Random variables question

Click For Summary

Homework Help Overview

The discussion revolves around a probability question involving joint probability density functions (PDFs) and area calculations related to a uniform distribution within a unit square and a quarter circle.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of areas related to the joint PDF and question the correctness of the original poster's approach. There is a discussion about using polar coordinates as an alternative method.

Discussion Status

Participants are engaging in a back-and-forth about the validity of the original calculations, with some affirming the approach while others suggest alternative methods. There is no explicit consensus on the correctness of the answer, but guidance has been provided regarding the nature of the calculations.

Contextual Notes

Some participants question the assumptions made regarding the uniform distribution and the interpretation of the area calculations in the context of the problem.

ashah99
Messages
55
Reaction score
2
Homework Statement
Finding the probability of randomly choosing a point within the unit square constrained within the quarter circle
Relevant Equations
P(( X, Y ) ∈ A) = ∫∫ fXY ( x, y )dxdy
Hello all, I am wondering if my approach is coreect for the following probability question? I believe the joint PDF would be 1 given that the point is chosen from the unit square. To me, this question can be reduced down to finding the area of 1/4 of a circle with radius 1. Any help is appreciated!
1662300114874.png
 
Physics news on Phys.org
It's clear from your calculations that you are indeed simply calculating the area. Which is what you would expect for a uniform distribution.
 
PeroK said:
It's clear from your calculations that you are indeed simply calculating the area. Which is what you would expect for a uniform distribution.
Ok, makes sense. Would you agree that my answer is correct? Just want to make sure I understand.
 
ashah99 said:
Ok, makes sense. Would you agree that my answer is correct? Just want to make sure I understand.
Yes, it's just a ratio of areas, as you've calculated.
 
  • Like
Likes   Reactions: ashah99
Or use polar coordinates: <br /> \int_0^1 \int_0^{\sqrt{1-x^2}} 1\,dy\,dx = \int_0^{\pi/2}\int_0^1 r\,dr\,d\theta.
 
  • Like
Likes   Reactions: ashah99

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K