Multivariate normal distribution clarification
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It seems to me that if a random vector $X$ is to have a multivariate normal distribution, then it is necessary and sufficient that $X$ is a vector of independent Gaussians -- is this correct?
In my understanding, a random vector $X$ of length $k$ is said to have a multivariate normal distribution if for any constant vector $a in mathbb{R}^k$, the random variable $Y = a^TX$ has a univariate normal distribution.
Then $e_i^TX$ must be a Gaussian for the elementary basis vectors $e_i$, implying that each $X_i$ must be a Gaussian.
In the other direction, any linear combination of two (or more, by induction) Gaussians should produce another Gaussian.
probability-distributions
asked Dec 3 at 3:51
Elizabeth Han
31119
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up vote
0
down vote
favorite
It seems to me that if a random vector $X$ is to have a multivariate normal distribution, then it is necessary and sufficient that $X$ is a vector of independent Gaussians -- is this correct?
In my understanding, a random vector $X$ of length $k$ is said to have a multivariate normal distribution if for any constant vector $a in mathbb{R}^k$, the random variable $Y = a^TX$ has a univariate normal distribution.
Then $e_i^TX$ must be a Gaussian for the elementary basis vectors $e_i$, implying that each $X_i$ must be a Gaussian.
In the other direction, any linear combination of two (or more, by induction) Gaussians should produce another Gaussian.
probability-distributions
asked Dec 3 at 3:51
Elizabeth Han
31119
1
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It seems to me that if a random vector $X$ is to have a multivariate normal distribution, then it is necessary and sufficient that $X$ is a vector of independent Gaussians -- is this correct?
In my understanding, a random vector $X$ of length $k$ is said to have a multivariate normal distribution if for any constant vector $a in mathbb{R}^k$, the random variable $Y = a^TX$ has a univariate normal distribution.
Then $e_i^TX$ must be a Gaussian for the elementary basis vectors $e_i$, implying that each $X_i$ must be a Gaussian.
In the other direction, any linear combination of two (or more, by induction) Gaussians should produce another Gaussian.
probability-distributions
asked Dec 3 at 3:51
Elizabeth Han
31119
It seems to me that if a random vector $X$ is to have a multivariate normal distribution, then it is necessary and sufficient that $X$ is a vector of independent Gaussians -- is this correct?
In my understanding, a random vector $X$ of length $k$ is said to have a multivariate normal distribution if for any constant vector $a in mathbb{R}^k$, the random variable $Y = a^TX$ has a univariate normal distribution.
Then $e_i^TX$ must be a Gaussian for the elementary basis vectors $e_i$, implying that each $X_i$ must be a Gaussian.
In the other direction, any linear combination of two (or more, by induction) Gaussians should produce another Gaussian.
probability-distributions
probability-distributions
asked Dec 3 at 3:51
Elizabeth Han
31119
asked Dec 3 at 3:51
Elizabeth Han
31119
asked Dec 3 at 3:51
Elizabeth Han
31119
asked Dec 3 at 3:51
Elizabeth Han
31119
asked Dec 3 at 3:51
Elizabeth Han
31119
31119
1
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48
add a comment |
1
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48
1
1
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48
add a comment |
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1
In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$.
– Kavi Rama Murthy
Dec 3 at 6:01
I see, thank you!
– Elizabeth Han
Dec 3 at 15:48