- #1

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Find E{x|x+y+z=1}.

- Thread starter purplebird
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- #1

- 18

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Find E{x|x+y+z=1}.

- #2

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- #3

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I am not able to understand what you are refering to. Could you please explain. Thanks.

- #4

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Sure thing... If I understand the problem correctly, it doesn't really matter what the exact distributions of x, y, and z are. It only matters that they are identically distributed. Also, note that the condition x+y+z = 1 is unchanged by permuting the letters x,y,z. This, together with their being identically distributed means that

E{x|x+y+z=1} = E{y|x+y+z=1} = E{z|x+y+z=1}

Also, think about what E{x+y+z|x+y+z=1} would be, and remember the additive property of the expected value.

edit: think simple. no hard integrals needed, which was the first thing that came to my mind when I read the problem.

E{x|x+y+z=1} = E{y|x+y+z=1} = E{z|x+y+z=1}

Also, think about what E{x+y+z|x+y+z=1} would be, and remember the additive property of the expected value.

edit: think simple. no hard integrals needed, which was the first thing that came to my mind when I read the problem.

Last edited:

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