Joint probability distribution

Click For Summary
SUMMARY

The discussion centers on calculating the joint probability distribution defined by the function f(x,y) = x² + xy³ for the bounds 0 < x < 1 and 0 < y < 2. The user attempts to find P(X+Y < 1) by transforming the bounds and setting up the double integral ∫[0-1]∫[0-y] (x² + xy³) dx dy. The user initially calculates P(X < 1-Y) as 1/8 but later acknowledges this result as incorrect, indicating a need for further clarification on the integration process.

PREREQUISITES
  • Understanding of joint probability distributions
  • Familiarity with double integrals
  • Knowledge of integration techniques in calculus
  • Experience with probability density functions
NEXT STEPS
  • Review the properties of joint probability distributions
  • Study the method of changing variables in double integrals
  • Learn about the cumulative distribution function (CDF) for two variables
  • Practice solving similar problems involving joint distributions and integration
USEFUL FOR

Students in statistics or mathematics, educators teaching probability theory, and anyone seeking to understand joint probability distributions and integration techniques.

Gauss M.D.
Messages
151
Reaction score
1

Homework Statement



f(x,y) = x2 + xy3 for 0 < x < 1, 0 < y < 2
and 0 otherwise.

Calculate P(X+Y < 1)

Homework Equations





The Attempt at a Solution



P(X+Y < 1) = P(X < 1-Y) which means y is now bounded by 0:1 instead of 0:2 and x is bounded by 0:y.

So we get ∫[0-1][0-y] x2 + xy3 dx dy

Integrating twice I get the answer P(X < 1-Y) = 1/8, which is incorrect.

What am I doing wrong?
 
Physics news on Phys.org
Oops.
 
I.e. nevermind.
 

Similar threads

Replies
7
Views
1K
Replies
4
Views
3K
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K