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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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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

  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    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

  4. anonymous
    • one year ago
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    sorry, i dont completely understand.

  5. ganeshie8
    • one year ago
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    Hey!

  6. anonymous
    • one year ago
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    |dw:1439457387740:dw|

  7. anonymous
    • one year ago
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    yes, but how is that range an expected value?

  8. ganeshie8
    • one year ago
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    |dw:1439558175814:dw|

  9. ganeshie8
    • one year ago
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    you can pick the ordered pair (x,y) from that shaded region, yes ?

  10. anonymous
    • one year ago
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    I would also draw a line from point zero up to the top of that @ganeshie8

  11. anonymous
    • one year ago
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    yes

  12. ganeshie8
    • one year ago
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    familiar with double integrals ?

  13. anonymous
    • one year ago
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    err.. no sir

  14. anonymous
    • one year ago
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    is that like calculus?

  15. anonymous
    • one year ago
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    you only need a single integral

  16. ganeshie8
    • one year ago
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    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

  17. anonymous
    • one year ago
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    eh, closed interval @ganeshie8 ?

  18. anonymous
    • one year ago
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    so its like just the average?

  19. anonymous
    • one year ago
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    ohh i understant

  20. anonymous
    • one year ago
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    thanks so much sir!

  21. ganeshie8
    • one year ago
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    expected value of "sum of two random variables" is same as the sum of expected values of each of them

  22. ganeshie8
    • one year ago
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    http://www.milefoot.com/math/stat/rv-sums.htm

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