Simplifying Expectations of Dependent Variables

In summary, when the variables x1 and x2 are dependent, and y1 and y2 are dependent, but all the x variables are independent of all the y variables, the expression E(x1y1x2y2) can be simplified to E(x1x2)E(y1y2) according to the textbook. This rule states that you cannot separate two independent variables that are multiplied with a third dependent variable, but you can separate a group of variables where all the elements in one group are independent of all the elements in the other group. This is because the product of two independent variables is also independent. However, the result used in this simplification may require proof.
  • #1
MaxManus
277
1

Homework Statement



if x1 and x2 are dependent, and y1 and y2 are dependent, but all the x are independent of all the y.

Then how can one simplify

E(x1y1x2y2)?

the textbook says
E(x1x2)E(y1y2)

So is the rule that you can not just separate two independent variables which they are multiplied with a third variable which are dependent on the first variable? But you can separate a group of variables which all the elements in the first group is independent of all the elements in the other group?
 
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  • #2
MaxManus said:

Homework Statement



if x1 and x2 are dependent, and y1 and y2 are dependent, but all the x are independent of all the y.

Then how can one simplify

E(x1y1x2y2)?

the textbook says
E(x1x2)E(y1y2)

So is the rule that you can not just separate two independent variables which they are multiplied with a third variable which are dependent on the first variable? But you can separate a group of variables which all the elements in the first group is independent of all the elements in the other group?

Just look at what things are independent. Since the X_i are independent of the Y_j
the product U = X1X2 is independent of the product V = Y1Y2, so E(X1Y1X2Y2) = E(UV) = E(U)E(V).

Note: in principle, the (true) result I used--that the Xi being independent of the Yj implies that X1X2 is independent of Y1Y2--needs proof, although it is more-or-less "obvious".

RGV
 
  • #3
Thanks for the help.
 

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