Does mean centering reduces covariance?
$begingroup$
It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.
correlation covariance random-vector
$endgroup$
add a comment |
$begingroup$
It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.
correlation covariance random-vector
$endgroup$
add a comment |
$begingroup$
It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.
correlation covariance random-vector
$endgroup$
It might be a very naive question, but assuming I have two non-independent random variables and I want to reduce covariance between them as much as possible without loosing too much "signal". Does mean centering help? I read somewhere that mean centering reduces correlation by a significant factor, so thinking it should do the same for covariance.
correlation covariance random-vector
correlation covariance random-vector
asked 47 mins ago
lvdplvdp
63
63
add a comment |
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1 Answer
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$begingroup$
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
$$
begin{aligned}
operatorname{Cov}(X + a, Y + b)
&= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
&= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
&= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
&= E[(X - E[X])(Y - E[Y])] \
&= operatorname{Cov}(X, Y).
end{aligned}
$$
Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.
Also, since correlation is defined as
$$
operatorname{Corr}(X, Y)
= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
$$
we can see that
$$
begin{aligned}
operatorname{Corr}(X + a, Y + b)
&= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
&= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
end{aligned}
$$
so in particular, correlation isn't affected by centering either.
That was the population version of the story. The sample version is the same: If we use
$$
widehat{operatorname{Cov}}(X, Y)
= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
$$
as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
$$
begin{aligned}
widehat{operatorname{Cov}}(X + a, Y + b)
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
&= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
&= widehat{operatorname{Cov}}(X, Y)
end{aligned}
$$
for any $a$ and $b$.
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
$$
begin{aligned}
operatorname{Cov}(X + a, Y + b)
&= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
&= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
&= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
&= E[(X - E[X])(Y - E[Y])] \
&= operatorname{Cov}(X, Y).
end{aligned}
$$
Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.
Also, since correlation is defined as
$$
operatorname{Corr}(X, Y)
= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
$$
we can see that
$$
begin{aligned}
operatorname{Corr}(X + a, Y + b)
&= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
&= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
end{aligned}
$$
so in particular, correlation isn't affected by centering either.
That was the population version of the story. The sample version is the same: If we use
$$
widehat{operatorname{Cov}}(X, Y)
= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
$$
as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
$$
begin{aligned}
widehat{operatorname{Cov}}(X + a, Y + b)
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
&= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
&= widehat{operatorname{Cov}}(X, Y)
end{aligned}
$$
for any $a$ and $b$.
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
$$
begin{aligned}
operatorname{Cov}(X + a, Y + b)
&= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
&= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
&= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
&= E[(X - E[X])(Y - E[Y])] \
&= operatorname{Cov}(X, Y).
end{aligned}
$$
Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.
Also, since correlation is defined as
$$
operatorname{Corr}(X, Y)
= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
$$
we can see that
$$
begin{aligned}
operatorname{Corr}(X + a, Y + b)
&= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
&= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
end{aligned}
$$
so in particular, correlation isn't affected by centering either.
That was the population version of the story. The sample version is the same: If we use
$$
widehat{operatorname{Cov}}(X, Y)
= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
$$
as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
$$
begin{aligned}
widehat{operatorname{Cov}}(X + a, Y + b)
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
&= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
&= widehat{operatorname{Cov}}(X, Y)
end{aligned}
$$
for any $a$ and $b$.
$endgroup$
add a comment |
$begingroup$
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
$$
begin{aligned}
operatorname{Cov}(X + a, Y + b)
&= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
&= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
&= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
&= E[(X - E[X])(Y - E[Y])] \
&= operatorname{Cov}(X, Y).
end{aligned}
$$
Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.
Also, since correlation is defined as
$$
operatorname{Corr}(X, Y)
= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
$$
we can see that
$$
begin{aligned}
operatorname{Corr}(X + a, Y + b)
&= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
&= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
end{aligned}
$$
so in particular, correlation isn't affected by centering either.
That was the population version of the story. The sample version is the same: If we use
$$
widehat{operatorname{Cov}}(X, Y)
= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
$$
as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
$$
begin{aligned}
widehat{operatorname{Cov}}(X + a, Y + b)
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
&= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
&= widehat{operatorname{Cov}}(X, Y)
end{aligned}
$$
for any $a$ and $b$.
$endgroup$
If $X$ and $Y$ are random variables and $a$ and $b$ are constants, then
$$
begin{aligned}
operatorname{Cov}(X + a, Y + b)
&= E[(X + a - E[X + a])(Y + b - E[Y + b])] \
&= E[(X + a - E[X] - E[a])(Y + b - E[Y] - E[b])] \
&= E[(X + a - E[X] - a)(Y + b - E[Y] - b)] \
&= E[(X - E[X])(Y - E[Y])] \
&= operatorname{Cov}(X, Y).
end{aligned}
$$
Centering is the special case $a = -E[X]$ and $b = -E[Y]$, so centering does not affect covariance.
Also, since correlation is defined as
$$
operatorname{Corr}(X, Y)
= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
$$
we can see that
$$
begin{aligned}
operatorname{Corr}(X + a, Y + b)
&= frac{operatorname{Cov}(X + a, Y + b)}{sqrt{operatorname{Var}(X + a) operatorname{Var}(Y + b)}} \
&= frac{operatorname{Cov}(X, Y)}{sqrt{operatorname{Var}(X) operatorname{Var}(Y)}},
end{aligned}
$$
so in particular, correlation isn't affected by centering either.
That was the population version of the story. The sample version is the same: If we use
$$
widehat{operatorname{Cov}}(X, Y)
= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright)
$$
as our estimate of covariance between $X$ and $Y$ from a paired sample $(X_1,Y_1), ldots, (X_n,Y_n)$, then
$$
begin{aligned}
widehat{operatorname{Cov}}(X + a, Y + b)
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n (X_j + a)right)left(Y_i + b - frac{1}{n}sum_{j=1}^n (Y_j + b)right) \
&= frac{1}{n} sum_{i=1}^n left(X_i + a - frac{1}{n}sum_{j=1}^n X_j + frac{n}{n} aright)left(Y_i + b - frac{1}{n}sum_{j=1}^n Y_j + frac{n}{n} bright) \
&= frac{1}{n} sum_{i=1}^n left(X_i - frac{1}{n}sum_{j=1}^n X_jright)left(Y_i - frac{1}{n}sum_{j=1}^n Y_jright) \
&= widehat{operatorname{Cov}}(X, Y)
end{aligned}
$$
for any $a$ and $b$.
edited 10 mins ago
answered 22 mins ago
Artem MavrinArtem Mavrin
33516
33516
add a comment |
add a comment |
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