Mixing
Let $Zsim N(0,1)$. Give a clean and rigorous proof that $Z^3$ cannot be normally distributed
1
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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
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
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add a comment |
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
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
$endgroup$
$begingroup$
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.
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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 $tne 0$.
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– kimchi lover
Jan 17 at 21:18
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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
add a comment |
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
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
$endgroup$
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
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
edited Jan 17 at 21:12
Mike Earnest
23.8k12051
23.8k12051
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
asked Jan 17 at 21:06
Mingyang GuoMingyang Guo
61
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
add a comment |
$begingroup$
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
add a comment |
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 $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$
answered Jan 17 at 21:24
Will M.Will M.
2,725315
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
1
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oldest
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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 $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$
answered Jan 17 at 21:24
Will M.Will M.
2,725315
$endgroup$
$begingroup$
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
add a comment |
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 $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$
answered Jan 17 at 21:24
Will M.Will M.
2,725315
$endgroup$
$begingroup$
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
add a comment |
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 $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$
answered Jan 17 at 21:24
Will M.Will M.
2,725315
$endgroup$
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 $mathbf{P}(Z^3 in (-1,1)) = mathbf{P}(Z in (-1,1)).$
answered Jan 17 at 21:24
Will M.Will M.
2,725315
answered Jan 17 at 21:24
Will M.Will M.
2,725315
answered Jan 17 at 21:24
Will M.Will M.
2,725315
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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