Conditional Expectation Decomposition in Regression Analysis
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I am currently working on my understanding of regression fundamentals and I checked this source (one can find the (even exact) same statement in multiple sources).
In Theorem 3.1.1, the author claims to prove the so called Conditional Expectation Function (CEF) Decomposition property, which states:
Theorem 3.1.1:
We have
$$Y_i=E(Y_i|X_i)+epsilon_i$$
with the property that $(a)$ $E(epsilon_i|X_i)=0$ and $(b)$ $E(f(X_i)epsilon_i)=0$ for any function $f$.
The problem is, the proof only checks $(a)$ and $(b)$ but never checks the actual existence of the decomposition of $Y_i=E(Y_i|X_i)+epsilon_i$.
The author then claims that
This theorem says that any random variable,
$Y_i$
, can be decomposed into a piece that's "explained by
$X_i$", i.e., the CEF, and a piece left over which is orthogonal to (i.e., uncorrelated with) any function of $X_i$
.
Do we have to prove this or comes this as some sort of assumption (couldn't find it, but surely one could just build the model like this) or is this a result from somewhere else, like factorization lemma or so?
The whole thing sounds a bit sloppy, clearly things like measurability are dropped, but still, what do I miss?
probability-theory statistics regression linear-regression regression-analysis
asked Jan 10 at 9:54
user190080user190080
3,30621427
$endgroup$
add a comment |
$begingroup$
I am currently working on my understanding of regression fundamentals and I checked this source (one can find the (even exact) same statement in multiple sources).
In Theorem 3.1.1, the author claims to prove the so called Conditional Expectation Function (CEF) Decomposition property, which states:
Theorem 3.1.1:
We have
$$Y_i=E(Y_i|X_i)+epsilon_i$$
with the property that $(a)$ $E(epsilon_i|X_i)=0$ and $(b)$ $E(f(X_i)epsilon_i)=0$ for any function $f$.
The problem is, the proof only checks $(a)$ and $(b)$ but never checks the actual existence of the decomposition of $Y_i=E(Y_i|X_i)+epsilon_i$.
The author then claims that
This theorem says that any random variable,
$Y_i$
, can be decomposed into a piece that's "explained by
$X_i$", i.e., the CEF, and a piece left over which is orthogonal to (i.e., uncorrelated with) any function of $X_i$
.
Do we have to prove this or comes this as some sort of assumption (couldn't find it, but surely one could just build the model like this) or is this a result from somewhere else, like factorization lemma or so?
The whole thing sounds a bit sloppy, clearly things like measurability are dropped, but still, what do I miss?
probability-theory statistics regression linear-regression regression-analysis
asked Jan 10 at 9:54
user190080user190080
3,30621427
$endgroup$
add a comment |
$begingroup$
I am currently working on my understanding of regression fundamentals and I checked this source (one can find the (even exact) same statement in multiple sources).
In Theorem 3.1.1, the author claims to prove the so called Conditional Expectation Function (CEF) Decomposition property, which states:
Theorem 3.1.1:
We have
$$Y_i=E(Y_i|X_i)+epsilon_i$$
with the property that $(a)$ $E(epsilon_i|X_i)=0$ and $(b)$ $E(f(X_i)epsilon_i)=0$ for any function $f$.
The problem is, the proof only checks $(a)$ and $(b)$ but never checks the actual existence of the decomposition of $Y_i=E(Y_i|X_i)+epsilon_i$.
The author then claims that
This theorem says that any random variable,
$Y_i$
, can be decomposed into a piece that's "explained by
$X_i$", i.e., the CEF, and a piece left over which is orthogonal to (i.e., uncorrelated with) any function of $X_i$
.
Do we have to prove this or comes this as some sort of assumption (couldn't find it, but surely one could just build the model like this) or is this a result from somewhere else, like factorization lemma or so?
The whole thing sounds a bit sloppy, clearly things like measurability are dropped, but still, what do I miss?
probability-theory statistics regression linear-regression regression-analysis
asked Jan 10 at 9:54
user190080user190080
3,30621427
$endgroup$
I am currently working on my understanding of regression fundamentals and I checked this source (one can find the (even exact) same statement in multiple sources).
In Theorem 3.1.1, the author claims to prove the so called Conditional Expectation Function (CEF) Decomposition property, which states:
Theorem 3.1.1:
We have
$$Y_i=E(Y_i|X_i)+epsilon_i$$
with the property that $(a)$ $E(epsilon_i|X_i)=0$ and $(b)$ $E(f(X_i)epsilon_i)=0$ for any function $f$.
The problem is, the proof only checks $(a)$ and $(b)$ but never checks the actual existence of the decomposition of $Y_i=E(Y_i|X_i)+epsilon_i$.
The author then claims that
This theorem says that any random variable,
$Y_i$
, can be decomposed into a piece that's "explained by
$X_i$", i.e., the CEF, and a piece left over which is orthogonal to (i.e., uncorrelated with) any function of $X_i$
.
Do we have to prove this or comes this as some sort of assumption (couldn't find it, but surely one could just build the model like this) or is this a result from somewhere else, like factorization lemma or so?
The whole thing sounds a bit sloppy, clearly things like measurability are dropped, but still, what do I miss?
probability-theory statistics regression linear-regression regression-analysis
probability-theory statistics regression linear-regression regression-analysis
asked Jan 10 at 9:54
user190080user190080
3,30621427
asked Jan 10 at 9:54
user190080user190080
3,30621427
asked Jan 10 at 9:54
user190080user190080
3,30621427
asked Jan 10 at 9:54
user190080user190080
3,30621427
asked Jan 10 at 9:54
user190080user190080
3,30621427
3,30621427
add a comment |
add a comment |
1 Answer
1
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oldest
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This follows from the equality
$$
Y_i=mathbb Eleft[Y_imid X_iright]+Y_i-mathbb Eleft[Y_imid X_iright];
$$
defining $varepsilon_i:=Y_i-mathbb Eleft[Y_imid X_iright]$ gives that
$mathbb Eleft[varepsilon_imid X_iright]=0$, from which $mathbb Eleft[fleft(X_iright)varepsilon_iright]=0$ follows.
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This follows from the equality
$$
Y_i=mathbb Eleft[Y_imid X_iright]+Y_i-mathbb Eleft[Y_imid X_iright];
$$
defining $varepsilon_i:=Y_i-mathbb Eleft[Y_imid X_iright]$ gives that
$mathbb Eleft[varepsilon_imid X_iright]=0$, from which $mathbb Eleft[fleft(X_iright)varepsilon_iright]=0$ follows.
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
add a comment |
$begingroup$
This follows from the equality
$$
Y_i=mathbb Eleft[Y_imid X_iright]+Y_i-mathbb Eleft[Y_imid X_iright];
$$
defining $varepsilon_i:=Y_i-mathbb Eleft[Y_imid X_iright]$ gives that
$mathbb Eleft[varepsilon_imid X_iright]=0$, from which $mathbb Eleft[fleft(X_iright)varepsilon_iright]=0$ follows.
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
add a comment |
$begingroup$
This follows from the equality
$$
Y_i=mathbb Eleft[Y_imid X_iright]+Y_i-mathbb Eleft[Y_imid X_iright];
$$
defining $varepsilon_i:=Y_i-mathbb Eleft[Y_imid X_iright]$ gives that
$mathbb Eleft[varepsilon_imid X_iright]=0$, from which $mathbb Eleft[fleft(X_iright)varepsilon_iright]=0$ follows.
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
$endgroup$
This follows from the equality
$$
Y_i=mathbb Eleft[Y_imid X_iright]+Y_i-mathbb Eleft[Y_imid X_iright];
$$
defining $varepsilon_i:=Y_i-mathbb Eleft[Y_imid X_iright]$ gives that
$mathbb Eleft[varepsilon_imid X_iright]=0$, from which $mathbb Eleft[fleft(X_iright)varepsilon_iright]=0$ follows.
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
answered Jan 10 at 10:04
Davide GiraudoDavide Giraudo
127k16152266
127k16152266
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
add a comment |
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
$begingroup$
Ah, that makes sense. I was able to follow the proof for $(a/b)$ (even without knowing this decomposition exists) but somehow missed this idea, thanks!
$endgroup$
– user190080
Jan 10 at 10:17
add a comment |
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