What is the Expected Value of Two Random Variables with a Joint PDF?

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SUMMARY

The discussion focuses on calculating the expected value of two random variables with a joint probability density function (PDF) defined as f(x,y) = c for x^2 + y^2 ≤ 1, x > 0, y > 0. The constant c is determined to be 4/π based on the area of the quarter circle. The expected value E{xy} is approached using polar coordinates, but an error occurs when the integral yields a negative result, which contradicts the positivity of x and y. The correct transformation to polar coordinates is emphasized as essential for accurate computation.

PREREQUISITES
  • Understanding of joint probability density functions (PDFs)
  • Knowledge of polar coordinates and transformations
  • Familiarity with double integrals in calculus
  • Basic concepts of expected value in probability theory
NEXT STEPS
  • Study the derivation of joint PDFs and their properties
  • Learn about polar coordinate transformations in double integrals
  • Explore the calculation of expected values for continuous random variables
  • Review common pitfalls in integral calculus, particularly with sign errors
USEFUL FOR

Students in statistics or mathematics, particularly those studying probability theory and calculus, as well as educators looking for examples of joint PDFs and expected value calculations.

rzn972
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Homework Statement



two random variables have a join pdf
f(x,y)= c , x^2+y^2<= 1, x >0, y>0
0, otherwise
Find c.
Find E{xy}.

Homework Equations





The Attempt at a Solution


For the first part, since area= pi/4,
c(pi/4) = 1
c=(4/pi)
b)
x=cos θ
y= sin θ
∫∫-(sin θ)^2(cos θ)^2 dθ dθ = -308. (integral from 0 to pi/2 for both bounds)
This doesn't make sense to me. x,y are always positive how is the expected value negative? Am I doing something wrong?
 
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rzn972 said:

Homework Statement



two random variables have a join pdf
f(x,y)= c , x^2+y^2<= 1, x >0, y>0
0, otherwise
Find c.
Find E{xy}.

Homework Equations





The Attempt at a Solution


For the first part, since area= pi/4,
c(pi/4) = 1
c=(4/pi)
b)
x=cos θ
y= sin θ
∫∫-(sin θ)^2(cos θ)^2 dθ dθ = -308. (integral from 0 to pi/2 for both bounds)
This doesn't make sense to me. x,y are always positive how is the expected value negative? Am I doing something wrong?

You need ##x=r\cos\theta,~y=r\sin\theta,~dydx=rdrd\theta##. And where did that minus sign come from anyway?
 

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