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October 29th, 2017, 01:27 PM   #1
ZMD
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Expected Value

X and Y are independent positive random variables with the same distribution, and the finite expected value µ:

1) E[(X)/(X+Y)] = 1/2

Any proof?
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October 29th, 2017, 02:18 PM   #2
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Off hand: E[(x)/(X+Y)]=E[(y)/(X+Y)] E[(X+Y)/(X+Y)]=1, so each term =1/2.
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October 29th, 2017, 07:52 PM   #3
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Quote:
Originally Posted by mathman View Post
Off hand: E[(x)/(X+Y)]=E[(y)/(X+Y)] E[(X+Y)/(X+Y)]=1, so each term =1/2.
What wrong with the following argument:
E[(X)/(X+Y)]=E[X]/E[X+Y]=1/2
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October 30th, 2017, 07:12 AM   #4
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1. E[(X)/(X+Y)] = E[(Y)/(X+Y)]
2. E[(X)/(X+Y)] + E[(Y)/(X+Y)] = E[(X+Y)/(X+Y)] =1
Therefore:
E[(X)/(X+Y)] = 1/2
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October 30th, 2017, 04:50 PM   #5
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Quote:
Originally Posted by ZMD View Post
What wrong with the following argument:
E[(X)/(X+Y)]=E[X]/E[X+Y]=1/2
You haven't proved anything. It is like saying a=a, therefore a=1/2
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