The PDF of $X^3$ where $ X sim text{Normal}(0, 1)$












0












$begingroup$


I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)



Here's what I have tried. Let $Y = X^3$, where X is a standard normal distribution with mean 0 and variance 1. Then



$P(Y leq y) = P(X^3 leq y) = P(X leq y^{1/3})$, for both $y$ positive or negative.



The above is the CDF of $X^3$; differentiating w.r.t. to $y$ should give me the PDF, which is



$f_X(y^{1/3})frac{1}{3}y^{-2/3}$, where $f_X$ is the PDF of the standard normal.




But THAT cannot be the answer, because it would take on negative values for $y < 0$. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes $y$ to $|y|$ and results in an integral that converges (in Mathematica).



But where can I introduce the absolute terms!? Much thanks :D






probability probability-theory probability-distributions self-learning







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$endgroup$












  • $begingroup$
    The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
    $endgroup$
    – Did
    Dec 7 '18 at 9:33
















0












$begingroup$


I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)



Here's what I have tried. Let $Y = X^3$, where X is a standard normal distribution with mean 0 and variance 1. Then



$P(Y leq y) = P(X^3 leq y) = P(X leq y^{1/3})$, for both $y$ positive or negative.



The above is the CDF of $X^3$; differentiating w.r.t. to $y$ should give me the PDF, which is



$f_X(y^{1/3})frac{1}{3}y^{-2/3}$, where $f_X$ is the PDF of the standard normal.




But THAT cannot be the answer, because it would take on negative values for $y < 0$. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes $y$ to $|y|$ and results in an integral that converges (in Mathematica).



But where can I introduce the absolute terms!? Much thanks :D






probability probability-theory probability-distributions self-learning







share|cite|improve this question











$endgroup$












  • $begingroup$
    The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
    $endgroup$
    – Did
    Dec 7 '18 at 9:33














0












0








0





$begingroup$


I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)



Here's what I have tried. Let $Y = X^3$, where X is a standard normal distribution with mean 0 and variance 1. Then



$P(Y leq y) = P(X^3 leq y) = P(X leq y^{1/3})$, for both $y$ positive or negative.



The above is the CDF of $X^3$; differentiating w.r.t. to $y$ should give me the PDF, which is



$f_X(y^{1/3})frac{1}{3}y^{-2/3}$, where $f_X$ is the PDF of the standard normal.




But THAT cannot be the answer, because it would take on negative values for $y < 0$. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes $y$ to $|y|$ and results in an integral that converges (in Mathematica).



But where can I introduce the absolute terms!? Much thanks :D






probability probability-theory probability-distributions self-learning







share|cite|improve this question











$endgroup$




I have been stumped for a few days on this...It would be great if anyone can point me to enlightenment :)



Here's what I have tried. Let $Y = X^3$, where X is a standard normal distribution with mean 0 and variance 1. Then



$P(Y leq y) = P(X^3 leq y) = P(X leq y^{1/3})$, for both $y$ positive or negative.



The above is the CDF of $X^3$; differentiating w.r.t. to $y$ should give me the PDF, which is



$f_X(y^{1/3})frac{1}{3}y^{-2/3}$, where $f_X$ is the PDF of the standard normal.




But THAT cannot be the answer, because it would take on negative values for $y < 0$. Whereas a PDF should always be non-negative. I have seen a "correct" answer in a math paper, where it changes $y$ to $|y|$ and results in an integral that converges (in Mathematica).



But where can I introduce the absolute terms!? Much thanks :D






probability probability-theory probability-distributions self-learning




probability probability-theory probability-distributions self-learning






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 7:59









user1101010

7801630




7801630










asked Dec 7 '18 at 6:55









Eve LEve L

125




125












  • $begingroup$
    The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
    $endgroup$
    – Did
    Dec 7 '18 at 9:33


















  • $begingroup$
    The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
    $endgroup$
    – Did
    Dec 7 '18 at 9:33
















$begingroup$
The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
$endgroup$
– Did
Dec 7 '18 at 9:33




$begingroup$
The derivative of $g:ymapsto y^{1/3}$ is positive at every $yne0$, in particular, if $y<0$, $$g'(y)=frac1{3|y|^{2/3}}$$
$endgroup$
– Did
Dec 7 '18 at 9:33










2 Answers
2






active

oldest

votes


















0












$begingroup$

Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $Phi,,phi$ denote the CDF and PDF of $X$. For $yge 0$, $$P(Yle y)=Phi(y^{1/3})implies f_Y(y)=frac{1}{3}y^{-2/3}phi(y^{1/3}).$$Then, for arbitrary $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I can't believe I didn't see this ... Thank you J.G.!
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:44



















0












$begingroup$

If $y ne 0$, the $2$ in the index would make it positive,



$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
    $endgroup$
    – Eve L
    Dec 7 '18 at 7:59












  • $begingroup$
    The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
    $endgroup$
    – Henry
    Dec 7 '18 at 9:28










  • $begingroup$
    thanks for the enlightenment. I interpreted it as always taking the real root previously.
    $endgroup$
    – Siong Thye Goh
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you Henry! That's exactly the problem with using Mathematica :)
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:43











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $Phi,,phi$ denote the CDF and PDF of $X$. For $yge 0$, $$P(Yle y)=Phi(y^{1/3})implies f_Y(y)=frac{1}{3}y^{-2/3}phi(y^{1/3}).$$Then, for arbitrary $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I can't believe I didn't see this ... Thank you J.G.!
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:44
















0












$begingroup$

Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $Phi,,phi$ denote the CDF and PDF of $X$. For $yge 0$, $$P(Yle y)=Phi(y^{1/3})implies f_Y(y)=frac{1}{3}y^{-2/3}phi(y^{1/3}).$$Then, for arbitrary $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I can't believe I didn't see this ... Thank you J.G.!
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:44














0












0








0





$begingroup$

Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $Phi,,phi$ denote the CDF and PDF of $X$. For $yge 0$, $$P(Yle y)=Phi(y^{1/3})implies f_Y(y)=frac{1}{3}y^{-2/3}phi(y^{1/3}).$$Then, for arbitrary $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$






share|cite|improve this answer









$endgroup$



Here's a way to find the $|y|$-based formula the first time, rather than saying, "Oh gee, it looks like that needs a little something" near the end. Since $X$'s pdf is even, $Y$'s should be even too. Let $Phi,,phi$ denote the CDF and PDF of $X$. For $yge 0$, $$P(Yle y)=Phi(y^{1/3})implies f_Y(y)=frac{1}{3}y^{-2/3}phi(y^{1/3}).$$Then, for arbitrary $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 10:14









J.G.J.G.

24k22539




24k22539












  • $begingroup$
    I can't believe I didn't see this ... Thank you J.G.!
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:44


















  • $begingroup$
    I can't believe I didn't see this ... Thank you J.G.!
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:44
















$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44




$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44











0












$begingroup$

If $y ne 0$, the $2$ in the index would make it positive,



$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
    $endgroup$
    – Eve L
    Dec 7 '18 at 7:59












  • $begingroup$
    The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
    $endgroup$
    – Henry
    Dec 7 '18 at 9:28










  • $begingroup$
    thanks for the enlightenment. I interpreted it as always taking the real root previously.
    $endgroup$
    – Siong Thye Goh
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you Henry! That's exactly the problem with using Mathematica :)
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:43
















0












$begingroup$

If $y ne 0$, the $2$ in the index would make it positive,



$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
    $endgroup$
    – Eve L
    Dec 7 '18 at 7:59












  • $begingroup$
    The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
    $endgroup$
    – Henry
    Dec 7 '18 at 9:28










  • $begingroup$
    thanks for the enlightenment. I interpreted it as always taking the real root previously.
    $endgroup$
    – Siong Thye Goh
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you Henry! That's exactly the problem with using Mathematica :)
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:43














0












0








0





$begingroup$

If $y ne 0$, the $2$ in the index would make it positive,



$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$






share|cite|improve this answer









$endgroup$



If $y ne 0$, the $2$ in the index would make it positive,



$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 7:11









Siong Thye GohSiong Thye Goh

100k1465117




100k1465117












  • $begingroup$
    Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
    $endgroup$
    – Eve L
    Dec 7 '18 at 7:59












  • $begingroup$
    The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
    $endgroup$
    – Henry
    Dec 7 '18 at 9:28










  • $begingroup$
    thanks for the enlightenment. I interpreted it as always taking the real root previously.
    $endgroup$
    – Siong Thye Goh
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you Henry! That's exactly the problem with using Mathematica :)
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:43


















  • $begingroup$
    Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
    $endgroup$
    – Eve L
    Dec 7 '18 at 7:59












  • $begingroup$
    The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
    $endgroup$
    – Henry
    Dec 7 '18 at 9:28










  • $begingroup$
    thanks for the enlightenment. I interpreted it as always taking the real root previously.
    $endgroup$
    – Siong Thye Goh
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you Henry! That's exactly the problem with using Mathematica :)
    $endgroup$
    – Eve L
    Dec 7 '18 at 15:43
















$begingroup$
Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
$endgroup$
– Eve L
Dec 7 '18 at 7:59






$begingroup$
Thanks. That makes sense...which makes me even more confused. Can you explain why in this paper they introduced an absolute sign? Seems superfluous ...projecteuclid.org/download/pdf_1/euclid.aop/1176991795
$endgroup$
– Eve L
Dec 7 '18 at 7:59














$begingroup$
The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
$endgroup$
– Henry
Dec 7 '18 at 9:28




$begingroup$
The problem is that sometimes you might say $(-8)^{-2/3}$ is $-frac14$ and sometimes $-frac18 - frac{sqrt{3}}{8}i$. Wolfram Alpha gives the complex answer
$endgroup$
– Henry
Dec 7 '18 at 9:28












$begingroup$
thanks for the enlightenment. I interpreted it as always taking the real root previously.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 9:46




$begingroup$
thanks for the enlightenment. I interpreted it as always taking the real root previously.
$endgroup$
– Siong Thye Goh
Dec 7 '18 at 9:46












$begingroup$
Thank you Henry! That's exactly the problem with using Mathematica :)
$endgroup$
– Eve L
Dec 7 '18 at 15:43




$begingroup$
Thank you Henry! That's exactly the problem with using Mathematica :)
$endgroup$
– Eve L
Dec 7 '18 at 15:43


















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