# Expected value of continuous joint random variables

Why would the expected value of $g(x,y)$ be the integral of $f(x,y) g(x,y)$ as shown in example 6 of here?

How is $f(x,y)$ related to $g(x,y)$ that it needs to be involved?

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$f(x,y)$ is not related to $g(x,y)$. It is related to (precisely speaking, it's the density of) the probability distribution you are computing expectations over. –  Srivatsan Nov 13 '11 at 19:46
Read the previous section "Expected Value of an Arbitrary Function", where the same thing is explained for one variable. –  leonbloy Nov 13 '11 at 19:58