Conditional expectation of $X$ given $X+Y$
$X$ and $Y$ are two random independent variables, they have the same distribution and have finite expected value. Find $ mathbb{E} [X|X+Y] $
My attempt:
$$ mathbb{E}[X|X+Y] + mathbb{E}[Y|X+Y] = X + Y $$
Whats more $$ mathbb{E}[X|X+Y] = mathbb{E}[Y|X+Y] $$
because $X$ and $Y$ have the same distribution.
Hence:
$$ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $$
Edit:
From comments below I conclude that this is a sufficient solution.
probability
asked Nov 29 '18 at 22:19
WywanaWywana
585
add a comment |
$X$ and $Y$ are two random independent variables, they have the same distribution and have finite expected value. Find $ mathbb{E} [X|X+Y] $
My attempt:
$$ mathbb{E}[X|X+Y] + mathbb{E}[Y|X+Y] = X + Y $$
Whats more $$ mathbb{E}[X|X+Y] = mathbb{E}[Y|X+Y] $$
because $X$ and $Y$ have the same distribution.
Hence:
$$ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $$
Edit:
From comments below I conclude that this is a sufficient solution.
probability
asked Nov 29 '18 at 22:19
WywanaWywana
585
3
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
4
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28
add a comment |
$X$ and $Y$ are two random independent variables, they have the same distribution and have finite expected value. Find $ mathbb{E} [X|X+Y] $
My attempt:
$$ mathbb{E}[X|X+Y] + mathbb{E}[Y|X+Y] = X + Y $$
Whats more $$ mathbb{E}[X|X+Y] = mathbb{E}[Y|X+Y] $$
because $X$ and $Y$ have the same distribution.
Hence:
$$ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $$
Edit:
From comments below I conclude that this is a sufficient solution.
probability
asked Nov 29 '18 at 22:19
WywanaWywana
585
$X$ and $Y$ are two random independent variables, they have the same distribution and have finite expected value. Find $ mathbb{E} [X|X+Y] $
My attempt:
$$ mathbb{E}[X|X+Y] + mathbb{E}[Y|X+Y] = X + Y $$
Whats more $$ mathbb{E}[X|X+Y] = mathbb{E}[Y|X+Y] $$
because $X$ and $Y$ have the same distribution.
Hence:
$$ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $$
Edit:
From comments below I conclude that this is a sufficient solution.
probability
probability
asked Nov 29 '18 at 22:19
WywanaWywana
585
asked Nov 29 '18 at 22:19
WywanaWywana
585
edited Dec 2 '18 at 14:06
Wywana
asked Nov 29 '18 at 22:19
WywanaWywana
585
asked Nov 29 '18 at 22:19
WywanaWywana
585
asked Nov 29 '18 at 22:19
WywanaWywana
585
585
3
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
4
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28
add a comment |
3
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
4
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28
3
3
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
4
4
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28
add a comment |
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3
If $X$ and $Y$ have the same distribution, this should hold by definition; they are the same function over $A$.
– platty
Nov 29 '18 at 22:21
4
Don't you have your answer at the point you say "Hence: $ mathbb{E}[X|X+Y] = frac{1}{2}(X+Y) $" ?
– Henry
Nov 29 '18 at 22:28