The PDF of $X^3$ where $ X sim text{Normal}(0, 1)$
$begingroup$
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
edited Dec 7 '18 at 7:59
user1101010
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
$endgroup$
add a comment |
$begingroup$
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
edited Dec 7 '18 at 7:59
user1101010
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 7 '18 at 7:59
user1101010
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
$endgroup$
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
edited Dec 7 '18 at 7:59
user1101010
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
edited Dec 7 '18 at 7:59
user1101010
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
edited Dec 7 '18 at 7:59
user1101010
7391730
edited Dec 7 '18 at 7:59
user1101010
7391730
edited Dec 7 '18 at 7:59
user1101010
7391730
7391730
asked Dec 7 '18 at 6:55
Eve LEve L
125
asked Dec 7 '18 at 6:55
Eve LEve L
125
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
$endgroup$
$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44
add a comment |
$begingroup$
If $y ne 0$, the $2$ in the index would make it positive,
$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
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 $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
$endgroup$
$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44
add a comment |
$begingroup$
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 $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
$endgroup$
$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44
add a comment |
$begingroup$
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 $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
$endgroup$
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 $yinmathbb{R}$,$$f_Y(y)=f_Y(|y|)=frac{1}{3}|y|^{-2/3}phi(|y|^{1/3}).$$
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
answered Dec 7 '18 at 10:14
J.G.J.G.
25.9k22539
25.9k22539
$begingroup$
I can't believe I didn't see this ... Thank you J.G.!
$endgroup$
– Eve L
Dec 7 '18 at 15:44
add a comment |
$begingroup$
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
add a comment |
$begingroup$
If $y ne 0$, the $2$ in the index would make it positive,
$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
If $y ne 0$, the $2$ in the index would make it positive,
$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
If $y ne 0$, the $2$ in the index would make it positive,
$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
If $y ne 0$, the $2$ in the index would make it positive,
$$f_X(y^{1/3})frac{1}{3}y^{-2/3}=f_X(y^{1/3})frac{1}{3}y^{-color{red}2/3} $$
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 7 '18 at 7:11
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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