Expected value of bivariate pdf

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SUMMARY

The expected value of a bivariate probability density function (pdf) fXY(x,y) is calculated using double integrals. Specifically, for the function f(x,y) = x + y, the expected value E[X] is determined by the double integral E[X] = ∫∫ x * (x + y) dx dy over the limits 0 < x < 1 and 0 < y < 1. The result of this calculation yields E[X] = 7/12. This confirms that a single integration only provides the conditional expectation of one variable with respect to the other.

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  • Understanding of bivariate probability density functions
  • Knowledge of double integrals in calculus
  • Familiarity with the concept of expected value in statistics
  • Basic proficiency in mathematical notation and integration techniques
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  • Study the properties of bivariate probability density functions
  • Learn how to perform double integrals in various coordinate systems
  • Explore conditional expectations and their applications in statistics
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Mathematicians, statisticians, and students studying probability theory, particularly those interested in understanding the computation of expected values in bivariate distributions.

boneill3
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Is the expected value E[x]or E[y] of a bivariate pdf fXY(xy)

\int x f(x,y)dx

or

E[y]

\int y f(x,y)dy ?

example if f(x,y) = x+y

E[X] = \int x *(x+y)dx


regards
Brendan
 
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You need to carry out the double integral. A single integration gives the conditional expection of one variable with rexpect to the other.
 
So just say you had the lmits of 0<x<1 0<y<1
The expected value would be:
<br /> E[X] = \int_{0}^{1}\int_{0}^{1} x *(x+y)dxdy<br /> <br /> = 7/12<br /> [\latex]<br /> Is that right?
 
Sorry,

<br /> E[X] = \int_{0}^{1}\int_{0}^{1} x *(x+y)dxdy<br /> <br /> = 7/12<br />

regards
 

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