Let $Zsim N(0,1)$. Give a clean and rigorous proof that $Z^3$ cannot be normally distributed












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Just as the title suggests. I simply have no idea how to prove that a random variable is not normally distributed. I'd really appreciate some hint.






normal-distribution







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    – dantopa
    Jan 17 at 21:11










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    You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
    $endgroup$
    – kimchi lover
    Jan 17 at 21:18










  • $begingroup$
    As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
    $endgroup$
    – lulu
    Jan 17 at 22:06
















1












$begingroup$


Just as the title suggests. I simply have no idea how to prove that a random variable is not normally distributed. I'd really appreciate some hint.






normal-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
    $endgroup$
    – dantopa
    Jan 17 at 21:11










  • $begingroup$
    You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
    $endgroup$
    – kimchi lover
    Jan 17 at 21:18










  • $begingroup$
    As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
    $endgroup$
    – lulu
    Jan 17 at 22:06














1












1








1





$begingroup$


Just as the title suggests. I simply have no idea how to prove that a random variable is not normally distributed. I'd really appreciate some hint.






normal-distribution







share|cite|improve this question











$endgroup$




Just as the title suggests. I simply have no idea how to prove that a random variable is not normally distributed. I'd really appreciate some hint.






normal-distribution




normal-distribution






share|cite|improve this question















share|cite|improve this question













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edited Jan 17 at 21:12









Mike Earnest

23.8k12051




23.8k12051










asked Jan 17 at 21:06









Mingyang GuoMingyang Guo

61




61












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
    $endgroup$
    – dantopa
    Jan 17 at 21:11










  • $begingroup$
    You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
    $endgroup$
    – kimchi lover
    Jan 17 at 21:18










  • $begingroup$
    As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
    $endgroup$
    – lulu
    Jan 17 at 22:06


















  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
    $endgroup$
    – dantopa
    Jan 17 at 21:11










  • $begingroup$
    You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
    $endgroup$
    – kimchi lover
    Jan 17 at 21:18










  • $begingroup$
    As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
    $endgroup$
    – lulu
    Jan 17 at 22:06
















$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 21:11




$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will hep you understand how best to form questions and answers. The lingua franca for formulation is MathJax.
$endgroup$
– dantopa
Jan 17 at 21:11












$begingroup$
You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
$endgroup$
– kimchi lover
Jan 17 at 21:18




$begingroup$
You can check directly that $Eexp(tZ^3)$ is not finite if $tne 0$.
$endgroup$
– kimchi lover
Jan 17 at 21:18












$begingroup$
As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
$endgroup$
– lulu
Jan 17 at 22:06




$begingroup$
As a hint: Note that $P(Z^3≤1)=P(Z≤1)$ which tells that, were the distribution normal, we'd have $sigma =1 $. But $P(Z^3≤2)neq P(Z≤2)$.
$endgroup$
– lulu
Jan 17 at 22:06










1 Answer
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If $Z^3$ were normally distributed, then it would have variance $mathbf{E}(Z^6) = 15$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) but this contradicts at once the easy fact $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$






share|cite|improve this answer









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  • $begingroup$
    Do you mean the variance or the expectation?
    $endgroup$
    – Mingyang Guo
    Jan 17 at 21:31






  • 1




    $begingroup$
    The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
    $endgroup$
    – Will M.
    Jan 17 at 21:44











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1 Answer
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1 Answer
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active

oldest

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active

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$begingroup$

If $Z^3$ were normally distributed, then it would have variance $mathbf{E}(Z^6) = 15$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) but this contradicts at once the easy fact $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean the variance or the expectation?
    $endgroup$
    – Mingyang Guo
    Jan 17 at 21:31






  • 1




    $begingroup$
    The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
    $endgroup$
    – Will M.
    Jan 17 at 21:44
















2












$begingroup$

If $Z^3$ were normally distributed, then it would have variance $mathbf{E}(Z^6) = 15$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) but this contradicts at once the easy fact $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean the variance or the expectation?
    $endgroup$
    – Mingyang Guo
    Jan 17 at 21:31






  • 1




    $begingroup$
    The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
    $endgroup$
    – Will M.
    Jan 17 at 21:44














2












2








2





$begingroup$

If $Z^3$ were normally distributed, then it would have variance $mathbf{E}(Z^6) = 15$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) but this contradicts at once the easy fact $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$






share|cite|improve this answer









$endgroup$



If $Z^3$ were normally distributed, then it would have variance $mathbf{E}(Z^6) = 15$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) but this contradicts at once the easy fact $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 21:24









Will M.Will M.

2,725315




2,725315












  • $begingroup$
    Do you mean the variance or the expectation?
    $endgroup$
    – Mingyang Guo
    Jan 17 at 21:31






  • 1




    $begingroup$
    The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
    $endgroup$
    – Will M.
    Jan 17 at 21:44


















  • $begingroup$
    Do you mean the variance or the expectation?
    $endgroup$
    – Mingyang Guo
    Jan 17 at 21:31






  • 1




    $begingroup$
    The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
    $endgroup$
    – Will M.
    Jan 17 at 21:44
















$begingroup$
Do you mean the variance or the expectation?
$endgroup$
– Mingyang Guo
Jan 17 at 21:31




$begingroup$
Do you mean the variance or the expectation?
$endgroup$
– Mingyang Guo
Jan 17 at 21:31




1




1




$begingroup$
The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
$endgroup$
– Will M.
Jan 17 at 21:44




$begingroup$
The expectation of $Z^3$ is zero because it is symmetric. Don't make do all the leg work. You should try to do some as well. =)
$endgroup$
– Will M.
Jan 17 at 21:44


















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