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anonymous
 one year ago
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anonymous
 one year ago
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This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a closed interval, you know by the brackets meaning it doesnt include 1 ot 0 therefore answer is: 0 < (x + y) <1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, but i dont see how this helps me find the expected value, there are infinite possibilities for the value of x and y... right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i dont completely understand.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439457387740:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, but how is that range an expected value?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439558175814:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you can pick the ordered pair (x,y) from that shaded region, yes ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would also draw a line from point zero up to the top of that @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1familiar with double integrals ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that like calculus?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you only need a single integral

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1then 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0eh, closed interval @ganeshie8 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its like just the average?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1expected value of "sum of two random variables" is same as the sum of expected values of each of them
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