Is covariance transitive? The 2019 Stack Overflow Developer Survey Results Are InCovariance Questioncovariance and correlation of x and yFinding Variance from CovarianceCovariance QuestionCovariance Quick Questionconditions on covariance of many random variablesCovariance and VarianceCalculate the following covarianceExpression for Covariance- summation over duplicatesStandard normal covariance
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Is covariance transitive?
The 2019 Stack Overflow Developer Survey Results Are InCovariance Questioncovariance and correlation of x and yFinding Variance from CovarianceCovariance QuestionCovariance Quick Questionconditions on covariance of many random variablesCovariance and VarianceCalculate the following covarianceExpression for Covariance- summation over duplicatesStandard normal covariance
$begingroup$
If given:
$$cov(x, y) = 0, cov(x, z) = 0$$
then can we conclude that:
$$cov(y, z) = 0$$
covariance
asked Apr 8 at 0:29
mitmath514mitmath514
213
$endgroup$
add a comment |
$begingroup$
If given:
$$cov(x, y) = 0, cov(x, z) = 0$$
then can we conclude that:
$$cov(y, z) = 0$$
covariance
asked Apr 8 at 0:29
mitmath514mitmath514
213
$endgroup$
add a comment |
$begingroup$
If given:
$$cov(x, y) = 0, cov(x, z) = 0$$
then can we conclude that:
$$cov(y, z) = 0$$
covariance
asked Apr 8 at 0:29
mitmath514mitmath514
213
$endgroup$
If given:
$$cov(x, y) = 0, cov(x, z) = 0$$
then can we conclude that:
$$cov(y, z) = 0$$
covariance
covariance
asked Apr 8 at 0:29
mitmath514mitmath514
213
asked Apr 8 at 0:29
mitmath514mitmath514
213
asked Apr 8 at 0:29
mitmath514mitmath514
213
asked Apr 8 at 0:29
mitmath514mitmath514
213
asked Apr 8 at 0:29
mitmath514mitmath514
213
213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $textcov(x,y)=0 implies textcov(y,y)=0$, but this certainly need not be true!
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
add a comment |
$begingroup$
No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $textcov(x,y)=0 implies textcov(y,y)=0$, but this certainly need not be true!
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
add a comment |
$begingroup$
There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $textcov(x,y)=0 implies textcov(y,y)=0$, but this certainly need not be true!
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
add a comment |
$begingroup$
There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $textcov(x,y)=0 implies textcov(y,y)=0$, but this certainly need not be true!
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
$endgroup$
There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $textcov(x,y)=0 implies textcov(y,y)=0$, but this certainly need not be true!
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
answered Apr 8 at 0:33
Alfred YergerAlfred Yerger
10.6k2249
10.6k2249
add a comment |
add a comment |
$begingroup$
No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
$endgroup$
add a comment |
$begingroup$
No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
$endgroup$
add a comment |
$begingroup$
No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
$endgroup$
No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
answered Apr 8 at 0:44
Holding ArthurHolding Arthur
1,565417
1,565417
add a comment |
add a comment |
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