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Is taking $3sigma$ limits for bivariate correlated random variables as the confidence limits an under estimation?
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If the two random variables are uncorrelated, $3sigma$ limit would mean that the random variables lie inside a hyperrectangle with center coordinates being means of random variables. But if two random variables are correlated, is taking a hyperrectangle instead of corresponding hyperellipsoid an under-approximation or altogether not theoretically correct?
probability statistics bivariate-distributions
asked Jul 25 at 14:30
NavE3n Naveen
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If the two random variables are uncorrelated, $3sigma$ limit would mean that the random variables lie inside a hyperrectangle with center coordinates being means of random variables. But if two random variables are correlated, is taking a hyperrectangle instead of corresponding hyperellipsoid an under-approximation or altogether not theoretically correct?
probability statistics bivariate-distributions
asked Jul 25 at 14:30
NavE3n Naveen
61
Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If the two random variables are uncorrelated, $3sigma$ limit would mean that the random variables lie inside a hyperrectangle with center coordinates being means of random variables. But if two random variables are correlated, is taking a hyperrectangle instead of corresponding hyperellipsoid an under-approximation or altogether not theoretically correct?
probability statistics bivariate-distributions
asked Jul 25 at 14:30
NavE3n Naveen
61
If the two random variables are uncorrelated, $3sigma$ limit would mean that the random variables lie inside a hyperrectangle with center coordinates being means of random variables. But if two random variables are correlated, is taking a hyperrectangle instead of corresponding hyperellipsoid an under-approximation or altogether not theoretically correct?
probability statistics bivariate-distributions
asked Jul 25 at 14:30
NavE3n Naveen
61
asked Jul 25 at 14:30
NavE3n Naveen
61
asked Jul 25 at 14:30
NavE3n Naveen
61
asked Jul 25 at 14:30
NavE3n Naveen
61
61
Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45
add a comment |Â
Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45
Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45
Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45
add a comment |Â
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Suppose you are talking about constructing a confidence region for Bivariate normal, then you may refer to, e.g. stats.stackexchange.com/questions/29860/… for using Hotelling's T to construct the region. Note that the region formed by the Cartaesian product of the marginal CI have a lower coverage probability, and it is not optimal in the sense that the volume of the region is larger.
– BGM
Jul 25 at 16:45