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










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












$begingroup$


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










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










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






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

7801730




7801730










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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029585%2fthe-pdf-of-x3-where-x-sim-textnormal0-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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.

25.2k22539




25.2k22539












  • $begingroup$
    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

101k1466117




101k1466117












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029585%2fthe-pdf-of-x3-where-x-sim-textnormal0-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix