Linear combination of normal distribution which is not normal
$begingroup$
Let $xi_1, xi_2$ be i.i.d N(0,1).
Define
$(X_1,X_2)=begin{cases}
(xi_1, |xi_2|) quad xi_1 geq 0 \
(xi_1,-|xi_2|) quad xi_1 < 0
end{cases}$
This means we can rewrite $X_1=xi_1$ and $X_2=sgn(xi_1)|xi_2|$
I proved that $X_1$ and $X_2$ are uncorrelated standard normal distributions.
Now I have to prove that they are not bivariate gaussian.
I wanted to use the statement that says that $(X_1,X_2)$ are distributed as a bivariate gaussian if each linear combination of them is normally distributed. I want to find a counterexample of a linear combination for which this is not true but I cannot. Can someone suggest me one?
probability normal-distribution bivariate-distributions
asked Jan 17 at 10:40
ank13ank13
255
$endgroup$
add a comment |
$begingroup$
Let $xi_1, xi_2$ be i.i.d N(0,1).
Define
$(X_1,X_2)=begin{cases}
(xi_1, |xi_2|) quad xi_1 geq 0 \
(xi_1,-|xi_2|) quad xi_1 < 0
end{cases}$
This means we can rewrite $X_1=xi_1$ and $X_2=sgn(xi_1)|xi_2|$
I proved that $X_1$ and $X_2$ are uncorrelated standard normal distributions.
Now I have to prove that they are not bivariate gaussian.
I wanted to use the statement that says that $(X_1,X_2)$ are distributed as a bivariate gaussian if each linear combination of them is normally distributed. I want to find a counterexample of a linear combination for which this is not true but I cannot. Can someone suggest me one?
probability normal-distribution bivariate-distributions
asked Jan 17 at 10:40
ank13ank13
255
$endgroup$
1
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17
add a comment |
$begingroup$
Let $xi_1, xi_2$ be i.i.d N(0,1).
Define
$(X_1,X_2)=begin{cases}
(xi_1, |xi_2|) quad xi_1 geq 0 \
(xi_1,-|xi_2|) quad xi_1 < 0
end{cases}$
This means we can rewrite $X_1=xi_1$ and $X_2=sgn(xi_1)|xi_2|$
I proved that $X_1$ and $X_2$ are uncorrelated standard normal distributions.
Now I have to prove that they are not bivariate gaussian.
I wanted to use the statement that says that $(X_1,X_2)$ are distributed as a bivariate gaussian if each linear combination of them is normally distributed. I want to find a counterexample of a linear combination for which this is not true but I cannot. Can someone suggest me one?
probability normal-distribution bivariate-distributions
asked Jan 17 at 10:40
ank13ank13
255
$endgroup$
Let $xi_1, xi_2$ be i.i.d N(0,1).
Define
$(X_1,X_2)=begin{cases}
(xi_1, |xi_2|) quad xi_1 geq 0 \
(xi_1,-|xi_2|) quad xi_1 < 0
end{cases}$
This means we can rewrite $X_1=xi_1$ and $X_2=sgn(xi_1)|xi_2|$
I proved that $X_1$ and $X_2$ are uncorrelated standard normal distributions.
Now I have to prove that they are not bivariate gaussian.
I wanted to use the statement that says that $(X_1,X_2)$ are distributed as a bivariate gaussian if each linear combination of them is normally distributed. I want to find a counterexample of a linear combination for which this is not true but I cannot. Can someone suggest me one?
probability normal-distribution bivariate-distributions
probability normal-distribution bivariate-distributions
asked Jan 17 at 10:40
ank13ank13
255
asked Jan 17 at 10:40
ank13ank13
255
asked Jan 17 at 10:40
ank13ank13
255
asked Jan 17 at 10:40
ank13ank13
255
asked Jan 17 at 10:40
ank13ank13
255
255
1
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17
add a comment |
1
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17
1
1
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First a correction: how did you conclude that $X_1$ and $X_2$ are uncorrelated? $EX_1=0$ and $X_1X_2 geq 0$, $EX_1X_2 >0$ so covariance of $X_1$ and $X_2$ is strictly positive.
If $(X_1,X_2)$ has a bivariate normal distribution then $E(X_2|X_1)=cX_1+d$ for some constants $c$ and $d$. Here $E(X_2|X_1)=E(X_2|xi_1)=I_{xi_1 geq 0} E|xi_2|-I_{xi_1 < 0} E|xi_2|$ which is clearly not of the type $cxi_1+d$. [The left side takes three values].
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
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$begingroup$
First a correction: how did you conclude that $X_1$ and $X_2$ are uncorrelated? $EX_1=0$ and $X_1X_2 geq 0$, $EX_1X_2 >0$ so covariance of $X_1$ and $X_2$ is strictly positive.
If $(X_1,X_2)$ has a bivariate normal distribution then $E(X_2|X_1)=cX_1+d$ for some constants $c$ and $d$. Here $E(X_2|X_1)=E(X_2|xi_1)=I_{xi_1 geq 0} E|xi_2|-I_{xi_1 < 0} E|xi_2|$ which is clearly not of the type $cxi_1+d$. [The left side takes three values].
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
First a correction: how did you conclude that $X_1$ and $X_2$ are uncorrelated? $EX_1=0$ and $X_1X_2 geq 0$, $EX_1X_2 >0$ so covariance of $X_1$ and $X_2$ is strictly positive.
If $(X_1,X_2)$ has a bivariate normal distribution then $E(X_2|X_1)=cX_1+d$ for some constants $c$ and $d$. Here $E(X_2|X_1)=E(X_2|xi_1)=I_{xi_1 geq 0} E|xi_2|-I_{xi_1 < 0} E|xi_2|$ which is clearly not of the type $cxi_1+d$. [The left side takes three values].
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
First a correction: how did you conclude that $X_1$ and $X_2$ are uncorrelated? $EX_1=0$ and $X_1X_2 geq 0$, $EX_1X_2 >0$ so covariance of $X_1$ and $X_2$ is strictly positive.
If $(X_1,X_2)$ has a bivariate normal distribution then $E(X_2|X_1)=cX_1+d$ for some constants $c$ and $d$. Here $E(X_2|X_1)=E(X_2|xi_1)=I_{xi_1 geq 0} E|xi_2|-I_{xi_1 < 0} E|xi_2|$ which is clearly not of the type $cxi_1+d$. [The left side takes three values].
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
First a correction: how did you conclude that $X_1$ and $X_2$ are uncorrelated? $EX_1=0$ and $X_1X_2 geq 0$, $EX_1X_2 >0$ so covariance of $X_1$ and $X_2$ is strictly positive.
If $(X_1,X_2)$ has a bivariate normal distribution then $E(X_2|X_1)=cX_1+d$ for some constants $c$ and $d$. Here $E(X_2|X_1)=E(X_2|xi_1)=I_{xi_1 geq 0} E|xi_2|-I_{xi_1 < 0} E|xi_2|$ which is clearly not of the type $cxi_1+d$. [The left side takes three values].
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
edited Jan 17 at 23:20
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 12:22
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
59.2k42161
add a comment |
add a comment |
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1
$begingroup$
What about using $X_1X_2ge 0$?
$endgroup$
– Song
Jan 17 at 10:51
$begingroup$
What do you mean?
$endgroup$
– ank13
Jan 17 at 11:01
$begingroup$
Note that $X_1X_2=|xi_1||xi_2|ge0$ with probability $1$, so that $(X_1,X_2)$ cannot be bivariate normal.
$endgroup$
– StubbornAtom
Jan 17 at 12:17