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This is a closed interval, you know by the brackets meaning it doesnt include 1 ot 0 therefore answer is: 0 < (x + y) <1
ok, but i dont see how this helps me find the expected value, there are infinite possibilities for the value of x and y... right?
which is just what I said, the value of (x+y) lies in the range between 0 and 1 but does not include 0 or 1

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sorry, i dont completely understand.
Hey!
|dw:1439457387740:dw|
yes, but how is that range an expected value?
|dw:1439558175814:dw|
you can pick the ordered pair (x,y) from that shaded region, yes ?
I would also draw a line from point zero up to the top of that @ganeshie8
yes
familiar with double integrals ?
err.. no sir
is that like calculus?
you only need a single integral
then you may simply do this : expected value of x = 0.5 expected value of y = 0.5 therefore expected value of x+y = 0.5+0.5 = 1
eh, closed interval @ganeshie8 ?
so its like just the average?
ohh i understant
thanks so much sir!
expected value of "sum of two random variables" is same as the sum of expected values of each of them
http://www.milefoot.com/math/stat/rv-sums.htm

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