Joint density of two vectors of multivariate normal random variables
If $bf{X}$ and $bf{Y}$ are dependent multivariate normal random variables, what is the joint density of $bf{X}$ and $bf{Y}$? Is it also multivariate normal?
probability matrices normal-distribution
edited Nov 21 '18 at 10:43
p4sch
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asked Nov 21 '18 at 5:27
shani
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If $bf{X}$ and $bf{Y}$ are dependent multivariate normal random variables, what is the joint density of $bf{X}$ and $bf{Y}$? Is it also multivariate normal?
probability matrices normal-distribution
edited Nov 21 '18 at 10:43
p4sch
4,760217
asked Nov 21 '18 at 5:27
shani
1208
2
On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27
add a comment |
If $bf{X}$ and $bf{Y}$ are dependent multivariate normal random variables, what is the joint density of $bf{X}$ and $bf{Y}$? Is it also multivariate normal?
probability matrices normal-distribution
edited Nov 21 '18 at 10:43
p4sch
4,760217
asked Nov 21 '18 at 5:27
shani
1208
If $bf{X}$ and $bf{Y}$ are dependent multivariate normal random variables, what is the joint density of $bf{X}$ and $bf{Y}$? Is it also multivariate normal?
probability matrices normal-distribution
probability matrices normal-distribution
edited Nov 21 '18 at 10:43
p4sch
4,760217
asked Nov 21 '18 at 5:27
shani
1208
edited Nov 21 '18 at 10:43
p4sch
4,760217
asked Nov 21 '18 at 5:27
shani
1208
edited Nov 21 '18 at 10:43
p4sch
4,760217
edited Nov 21 '18 at 10:43
p4sch
4,760217
edited Nov 21 '18 at 10:43
p4sch
4,760217
4,760217
asked Nov 21 '18 at 5:27
shani
1208
asked Nov 21 '18 at 5:27
shani
1208
asked Nov 21 '18 at 5:27
shani
1208
1208
2
On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27
add a comment |
2
On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27
2
2
On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27
On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27
add a comment |
1 Answer
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As already noted in the comments: In general, it is not true that $X$ and $Y$ have a jointly multivariate normal distribution. Here is one counterexample: Let $X sim mathcal{N}(0,1)$ and $Y$ independent of $X$ with $P(Y=1) =1/2$ and $P(Y=-1)=1/2$. Now set $Z=XY$.
- Show that $Z$ has also normal distribution with mean $0$ and variance $1$, i.e. $Z sim mathcal{N}(0,1)$.
$X$ and $Z$ are uncorrelated.
$X$ and $Z$ are not independent. (For example $P( |X| le t, |Z| >t) =0$, but $P(|X| le t) P(|Z|>t) ne 0$ for $t>0$.)
answered Nov 21 '18 at 10:03
p4sch
4,760217
add a comment |
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1 Answer
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1 Answer
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active
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votes
As already noted in the comments: In general, it is not true that $X$ and $Y$ have a jointly multivariate normal distribution. Here is one counterexample: Let $X sim mathcal{N}(0,1)$ and $Y$ independent of $X$ with $P(Y=1) =1/2$ and $P(Y=-1)=1/2$. Now set $Z=XY$.
- Show that $Z$ has also normal distribution with mean $0$ and variance $1$, i.e. $Z sim mathcal{N}(0,1)$.
$X$ and $Z$ are uncorrelated.
$X$ and $Z$ are not independent. (For example $P( |X| le t, |Z| >t) =0$, but $P(|X| le t) P(|Z|>t) ne 0$ for $t>0$.)
answered Nov 21 '18 at 10:03
p4sch
4,760217
add a comment |
As already noted in the comments: In general, it is not true that $X$ and $Y$ have a jointly multivariate normal distribution. Here is one counterexample: Let $X sim mathcal{N}(0,1)$ and $Y$ independent of $X$ with $P(Y=1) =1/2$ and $P(Y=-1)=1/2$. Now set $Z=XY$.
- Show that $Z$ has also normal distribution with mean $0$ and variance $1$, i.e. $Z sim mathcal{N}(0,1)$.
$X$ and $Z$ are uncorrelated.
$X$ and $Z$ are not independent. (For example $P( |X| le t, |Z| >t) =0$, but $P(|X| le t) P(|Z|>t) ne 0$ for $t>0$.)
answered Nov 21 '18 at 10:03
p4sch
4,760217
add a comment |
As already noted in the comments: In general, it is not true that $X$ and $Y$ have a jointly multivariate normal distribution. Here is one counterexample: Let $X sim mathcal{N}(0,1)$ and $Y$ independent of $X$ with $P(Y=1) =1/2$ and $P(Y=-1)=1/2$. Now set $Z=XY$.
- Show that $Z$ has also normal distribution with mean $0$ and variance $1$, i.e. $Z sim mathcal{N}(0,1)$.
$X$ and $Z$ are uncorrelated.
$X$ and $Z$ are not independent. (For example $P( |X| le t, |Z| >t) =0$, but $P(|X| le t) P(|Z|>t) ne 0$ for $t>0$.)
answered Nov 21 '18 at 10:03
p4sch
4,760217
As already noted in the comments: In general, it is not true that $X$ and $Y$ have a jointly multivariate normal distribution. Here is one counterexample: Let $X sim mathcal{N}(0,1)$ and $Y$ independent of $X$ with $P(Y=1) =1/2$ and $P(Y=-1)=1/2$. Now set $Z=XY$.
- Show that $Z$ has also normal distribution with mean $0$ and variance $1$, i.e. $Z sim mathcal{N}(0,1)$.
$X$ and $Z$ are uncorrelated.
$X$ and $Z$ are not independent. (For example $P( |X| le t, |Z| >t) =0$, but $P(|X| le t) P(|Z|>t) ne 0$ for $t>0$.)
answered Nov 21 '18 at 10:03
p4sch
4,760217
answered Nov 21 '18 at 10:03
p4sch
4,760217
answered Nov 21 '18 at 10:03
p4sch
4,760217
answered Nov 21 '18 at 10:03
p4sch
4,760217
4,760217
add a comment |
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On the other way round: If you know that they are independent, then they are jointly multivariate normal. If you only know they are dependent, then they are not necessary to be jointly multivariate normal in general.
– BGM
Nov 21 '18 at 6:27