Linear combination of normal distribution which is not normal












1












$begingroup$


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?










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$endgroup$








  • 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
















1












$begingroup$


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?










share|cite|improve this question









$endgroup$








  • 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














1












1








1





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






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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














  • 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










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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].






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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].






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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].






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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].






        share|cite|improve this answer











        $endgroup$



        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].







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 17 at 23:20

























        answered Jan 17 at 12:22









        Kavi Rama MurthyKavi Rama Murthy

        59.2k42161




        59.2k42161






























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