Why is $ Var(X)= EVar(X|Y)+Var(E(X|Y)) $
$begingroup$
I See this identity but it does not make sense to me.
$ Var(X)= EVar(X|Y)+Var(E(X|Y)) $
is this something that should intuitively make sense? Is it saying the variance of X is equal to the expected variance of X given Y plus the variance of Y? I say the second part because since E(X|Y) is a function with Y as variable Var(E(X|Y)) would capture the variance of Y.
probability
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
$endgroup$
add a comment |
$begingroup$
I See this identity but it does not make sense to me.
$ Var(X)= EVar(X|Y)+Var(E(X|Y)) $
is this something that should intuitively make sense? Is it saying the variance of X is equal to the expected variance of X given Y plus the variance of Y? I say the second part because since E(X|Y) is a function with Y as variable Var(E(X|Y)) would capture the variance of Y.
probability
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
$endgroup$
1
$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02
add a comment |
$begingroup$
I See this identity but it does not make sense to me.
$ Var(X)= EVar(X|Y)+Var(E(X|Y)) $
is this something that should intuitively make sense? Is it saying the variance of X is equal to the expected variance of X given Y plus the variance of Y? I say the second part because since E(X|Y) is a function with Y as variable Var(E(X|Y)) would capture the variance of Y.
probability
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
$endgroup$
I See this identity but it does not make sense to me.
$ Var(X)= EVar(X|Y)+Var(E(X|Y)) $
is this something that should intuitively make sense? Is it saying the variance of X is equal to the expected variance of X given Y plus the variance of Y? I say the second part because since E(X|Y) is a function with Y as variable Var(E(X|Y)) would capture the variance of Y.
probability
probability
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
asked Oct 20 '16 at 14:43
QwertfordQwertford
321212
321212
1
$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02
add a comment |
1
$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02
1
1
$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02
$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02
add a comment |
1 Answer
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$begingroup$
The main thing here is this one: $mathbb{E}(mathbb{E}[f(X) | Y]) = mathbb{E} f(X)$.
Just denote $psi (Y) = mathbb{E}[X^2 | Y]$, $phi(Y) = mathbb{E}[X | Y]$.
Then $$mathbb{E} text{Var}[X|Y] = mathbb{E} psi (Y) - mathbb{E} psi (Y)^2,$$ $$text{Var}mathbb{E}[X | Y] = text{Var} phi(Y) = mathbb{E} psi (Y)^2 - (mathbb{E}X)^2.$$
Sum this expressions and obtain the equality needed.
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
The main thing here is this one: $mathbb{E}(mathbb{E}[f(X) | Y]) = mathbb{E} f(X)$.
Just denote $psi (Y) = mathbb{E}[X^2 | Y]$, $phi(Y) = mathbb{E}[X | Y]$.
Then $$mathbb{E} text{Var}[X|Y] = mathbb{E} psi (Y) - mathbb{E} psi (Y)^2,$$ $$text{Var}mathbb{E}[X | Y] = text{Var} phi(Y) = mathbb{E} psi (Y)^2 - (mathbb{E}X)^2.$$
Sum this expressions and obtain the equality needed.
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
$endgroup$
add a comment |
$begingroup$
The main thing here is this one: $mathbb{E}(mathbb{E}[f(X) | Y]) = mathbb{E} f(X)$.
Just denote $psi (Y) = mathbb{E}[X^2 | Y]$, $phi(Y) = mathbb{E}[X | Y]$.
Then $$mathbb{E} text{Var}[X|Y] = mathbb{E} psi (Y) - mathbb{E} psi (Y)^2,$$ $$text{Var}mathbb{E}[X | Y] = text{Var} phi(Y) = mathbb{E} psi (Y)^2 - (mathbb{E}X)^2.$$
Sum this expressions and obtain the equality needed.
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
$endgroup$
add a comment |
$begingroup$
The main thing here is this one: $mathbb{E}(mathbb{E}[f(X) | Y]) = mathbb{E} f(X)$.
Just denote $psi (Y) = mathbb{E}[X^2 | Y]$, $phi(Y) = mathbb{E}[X | Y]$.
Then $$mathbb{E} text{Var}[X|Y] = mathbb{E} psi (Y) - mathbb{E} psi (Y)^2,$$ $$text{Var}mathbb{E}[X | Y] = text{Var} phi(Y) = mathbb{E} psi (Y)^2 - (mathbb{E}X)^2.$$
Sum this expressions and obtain the equality needed.
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
$endgroup$
The main thing here is this one: $mathbb{E}(mathbb{E}[f(X) | Y]) = mathbb{E} f(X)$.
Just denote $psi (Y) = mathbb{E}[X^2 | Y]$, $phi(Y) = mathbb{E}[X | Y]$.
Then $$mathbb{E} text{Var}[X|Y] = mathbb{E} psi (Y) - mathbb{E} psi (Y)^2,$$ $$text{Var}mathbb{E}[X | Y] = text{Var} phi(Y) = mathbb{E} psi (Y)^2 - (mathbb{E}X)^2.$$
Sum this expressions and obtain the equality needed.
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
answered Oct 20 '16 at 15:14
Denis KorzhenkovDenis Korzhenkov
60329
60329
add a comment |
add a comment |
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$begingroup$
Intuitively to me, the $E[text{Var}(Xmid Y )]$ term is the dispersion of $X$ that is not driven by changes in $Y$, and the $text{Var}(E[Xmid Y] )$ is the dispersion of $X$ that is driven by changes in $Y$. So if $X$ is independent of $Y$ then the second term is zero, while if $X$ is exactly determined by $Y$ (e.g. is a function of $Y$) then the first term is zero. The second term is not necessarily the variance of $Y$, though it is when $E[Xmid Y]=Y$
$endgroup$
– Henry
Oct 20 '16 at 15:02