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
asked 55 mins ago
lvdplvdp
63
$endgroup$
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
asked 55 mins ago
lvdplvdp
63
$endgroup$
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
asked 55 mins ago
lvdplvdp
63
$endgroup$
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 55 mins ago
lvdplvdp
63
asked 55 mins ago
lvdplvdp
63
asked 55 mins ago
lvdplvdp
63
asked 55 mins ago
lvdplvdp
63
asked 55 mins ago
lvdplvdp
63
63
add a comment |
add a comment |
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 $b$.
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
$endgroup$
add a comment |
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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 $b$.
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
$endgroup$
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 $b$.
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
$endgroup$
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 $b$.
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
$endgroup$
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 $b$.
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
edited 17 mins ago
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
answered 30 mins ago
Artem MavrinArtem Mavrin
33516
33516
add a comment |
add a comment |
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