Calculating E(X+Y+Z/X) and E(W/X+Y)

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The discussion focuses on calculating the expected values E(W/X) and E(W/(X+Y)) where W is defined as the sum of the results from three ordinary dice rolls (X, Y, Z). Participants clarify the need to determine how many different combinations of (X, Y, Z) yield specific expected values. The total combinations of three dice rolls amount to 216, and the range of W spans from 3 to 18. The conversation emphasizes the importance of correctly interpreting the expected value calculations and the independence of the variables involved.

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we throw three ordinary dice and X,Y,Z their results and W=X+Y+Z how many differents values have the random values E(W/X) and E(W/X+Y)?

can anyone explain me how to beggin because i am confused.. i will use E(X+Y+Z/X)=E(X/X)+E(Y/X)+E(Z/X)=? for the first one?
 
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ParisSpart said:
we throw three ordinary dice and X,Y,Z their results and W=X+Y+Z how many differents values have the random values E(W/X) and E(W/X+Y)?

can anyone explain me how to beggin because i am confused.. i will use E(X+Y+Z/X)=E(X/X)+E(Y/X)+E(Z/X)=? for the first one?

I think you need to go back and look at the problem again. I don't know what you mean by "how many differents values have the random values E(W/X) and E(W/X+Y)?"
How many different values" of what? E(W/X) and E(W/(X+ Y) are specific numbers. Do you mean how many (X, Y, Z) combinations give W that are equal to those expectations. Actually, I would be surprized if those were integer values. Or do you mean simply "find E(W/X) and E(W/(X+Y))"?

It's not all that difficult to determine the [itex]6^3= 216[/itex] combinations of three dice. W ranges in value from 3 to 18.
 
ParisSpart said:
E(X+Y+Z/X)=E(X/X)+E(Y/X)+E(Z/X)=?
Assuming you mean E((X+Y+Z)/X), that's a good start. E(X/X) is obvious. Since X and Y are independent, can you expand E(Y/X) into separate functions of X and Y?
 

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