Mixing
PDF Transformation: What manoeuvres are involved in this case?
Clash Royale CLAN TAG#URR8PPP
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Preamble: Given the illustration in the figure below, the pdf for a random variable $r$ is shown below:
$$f_r(r)= fraclambda pi r exp(-lambda pi r^2/2)1-e^-lambda pi R^2/2$$
where $$0<rleq R$$
Question: In order to transform the above pdf, how was the equation below derived? Are there any special manoeuvres involved? I would appreciate this with steps.
$$f_Z(z)= fracsqrt2lambdae^-lambda pi z^2/2texterfsqrtfracpilambda2(R^2-z^2)1-e^-lambda pi R^2/2$$
where $$0<zleq R$$
With these steps one should be able to transform
$$f_r(r)= frac2rR^2$$ to $$f_Z(z)= frac4sqrtR^2-z^2pi R^2$$
probability probability-distributions
asked Jul 25 at 2:46
Abdulhameed
54111
add a comment |Â
up vote
0
down vote
favorite
Preamble: Given the illustration in the figure below, the pdf for a random variable $r$ is shown below:
$$f_r(r)= fraclambda pi r exp(-lambda pi r^2/2)1-e^-lambda pi R^2/2$$
where $$0<rleq R$$
Question: In order to transform the above pdf, how was the equation below derived? Are there any special manoeuvres involved? I would appreciate this with steps.
$$f_Z(z)= fracsqrt2lambdae^-lambda pi z^2/2texterfsqrtfracpilambda2(R^2-z^2)1-e^-lambda pi R^2/2$$
where $$0<zleq R$$
With these steps one should be able to transform
$$f_r(r)= frac2rR^2$$ to $$f_Z(z)= frac4sqrtR^2-z^2pi R^2$$
probability probability-distributions
asked Jul 25 at 2:46
Abdulhameed
54111
There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Preamble: Given the illustration in the figure below, the pdf for a random variable $r$ is shown below:
$$f_r(r)= fraclambda pi r exp(-lambda pi r^2/2)1-e^-lambda pi R^2/2$$
where $$0<rleq R$$
Question: In order to transform the above pdf, how was the equation below derived? Are there any special manoeuvres involved? I would appreciate this with steps.
$$f_Z(z)= fracsqrt2lambdae^-lambda pi z^2/2texterfsqrtfracpilambda2(R^2-z^2)1-e^-lambda pi R^2/2$$
where $$0<zleq R$$
With these steps one should be able to transform
$$f_r(r)= frac2rR^2$$ to $$f_Z(z)= frac4sqrtR^2-z^2pi R^2$$
probability probability-distributions
asked Jul 25 at 2:46
Abdulhameed
54111
Preamble: Given the illustration in the figure below, the pdf for a random variable $r$ is shown below:
$$f_r(r)= fraclambda pi r exp(-lambda pi r^2/2)1-e^-lambda pi R^2/2$$
where $$0<rleq R$$
Question: In order to transform the above pdf, how was the equation below derived? Are there any special manoeuvres involved? I would appreciate this with steps.
$$f_Z(z)= fracsqrt2lambdae^-lambda pi z^2/2texterfsqrtfracpilambda2(R^2-z^2)1-e^-lambda pi R^2/2$$
where $$0<zleq R$$
With these steps one should be able to transform
$$f_r(r)= frac2rR^2$$ to $$f_Z(z)= frac4sqrtR^2-z^2pi R^2$$
probability probability-distributions
asked Jul 25 at 2:46
Abdulhameed
54111
edited Jul 27 at 6:29
asked Jul 25 at 2:46
Abdulhameed
54111
asked Jul 25 at 2:46
Abdulhameed
54111
asked Jul 25 at 2:46
Abdulhameed
54111
54111
There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56
add a comment |Â
There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56
There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Assume WLOG 2D PDF of the random vector $(hat r,hat theta)$ in the form of
$$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$
then the element of the 2D PDF is
$$dF(r,theta) = Pr((hat rin[r,r+mathrm dr]) wedge(hatthetain[theta,theta+mathrm dtheta])) =dfrac2pi f_r(r),mathrm dr,mathrm dtheta.$$
In according with the illustration,
$$hat z = hat rcoshattheta,quad hat zin(0,hat r],$$
and
$$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$
then
$$f_z(z)=dfrac,mathrm dF_z(z),mathrm dz = dfrac2piintlimits_z^R f_r(r)dfracmathrm drsqrtr^2-z^2.$$
If
$$f_r(r)=dfrac2rR^2,$$
then
$$f_z(z) = dfrac4pi R^2intlimits_z^R dfracrmathrm drsqrtr^2-z^2 = dfrac4pi R^2sqrtr^2-z^2Big|_z^R = mathbfcolorreddfrac4sqrtR^2-z^2pi R^2.$$
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
add a comment |Â
up vote
2
down vote
The density for $r$ corresponds to normal distributions for $z$ and $Y$ that would be independent if the distribution extended over the entire plane; in that case, the marginal distributions for $z$ and $Y$ would just be Gaussians. However, as the distribution is truncated to the circle, $z$ and $Y$ aren't independent. To get the marginal distribution for $z$ you need to integrate out $Y$; this is what leads to the $operatornameerf$ term. To reproduce the result, transform from polar coordinates to Cartesian coordinates ($r^2$ in the exponent becomes $z^2+Y^2$ and $2pi rmathrm drmathrm dphi$ becomes $mathrm dzmathrm dY$) and integrate out $Y$ to get the marginal density for $z$.
answered Jul 25 at 6:17
joriki
164k10180328
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assume WLOG 2D PDF of the random vector $(hat r,hat theta)$ in the form of
$$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$
then the element of the 2D PDF is
$$dF(r,theta) = Pr((hat rin[r,r+mathrm dr]) wedge(hatthetain[theta,theta+mathrm dtheta])) =dfrac2pi f_r(r),mathrm dr,mathrm dtheta.$$
In according with the illustration,
$$hat z = hat rcoshattheta,quad hat zin(0,hat r],$$
and
$$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$
then
$$f_z(z)=dfrac,mathrm dF_z(z),mathrm dz = dfrac2piintlimits_z^R f_r(r)dfracmathrm drsqrtr^2-z^2.$$
If
$$f_r(r)=dfrac2rR^2,$$
then
$$f_z(z) = dfrac4pi R^2intlimits_z^R dfracrmathrm drsqrtr^2-z^2 = dfrac4pi R^2sqrtr^2-z^2Big|_z^R = mathbfcolorreddfrac4sqrtR^2-z^2pi R^2.$$
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
add a comment |Â
up vote
1
down vote
accepted
Assume WLOG 2D PDF of the random vector $(hat r,hat theta)$ in the form of
$$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$
then the element of the 2D PDF is
$$dF(r,theta) = Pr((hat rin[r,r+mathrm dr]) wedge(hatthetain[theta,theta+mathrm dtheta])) =dfrac2pi f_r(r),mathrm dr,mathrm dtheta.$$
In according with the illustration,
$$hat z = hat rcoshattheta,quad hat zin(0,hat r],$$
and
$$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$
then
$$f_z(z)=dfrac,mathrm dF_z(z),mathrm dz = dfrac2piintlimits_z^R f_r(r)dfracmathrm drsqrtr^2-z^2.$$
If
$$f_r(r)=dfrac2rR^2,$$
then
$$f_z(z) = dfrac4pi R^2intlimits_z^R dfracrmathrm drsqrtr^2-z^2 = dfrac4pi R^2sqrtr^2-z^2Big|_z^R = mathbfcolorreddfrac4sqrtR^2-z^2pi R^2.$$
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assume WLOG 2D PDF of the random vector $(hat r,hat theta)$ in the form of
$$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$
then the element of the 2D PDF is
$$dF(r,theta) = Pr((hat rin[r,r+mathrm dr]) wedge(hatthetain[theta,theta+mathrm dtheta])) =dfrac2pi f_r(r),mathrm dr,mathrm dtheta.$$
In according with the illustration,
$$hat z = hat rcoshattheta,quad hat zin(0,hat r],$$
and
$$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$
then
$$f_z(z)=dfrac,mathrm dF_z(z),mathrm dz = dfrac2piintlimits_z^R f_r(r)dfracmathrm drsqrtr^2-z^2.$$
If
$$f_r(r)=dfrac2rR^2,$$
then
$$f_z(z) = dfrac4pi R^2intlimits_z^R dfracrmathrm drsqrtr^2-z^2 = dfrac4pi R^2sqrtr^2-z^2Big|_z^R = mathbfcolorreddfrac4sqrtR^2-z^2pi R^2.$$
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
Assume WLOG 2D PDF of the random vector $(hat r,hat theta)$ in the form of
$$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$
then the element of the 2D PDF is
$$dF(r,theta) = Pr((hat rin[r,r+mathrm dr]) wedge(hatthetain[theta,theta+mathrm dtheta])) =dfrac2pi f_r(r),mathrm dr,mathrm dtheta.$$
In according with the illustration,
$$hat z = hat rcoshattheta,quad hat zin(0,hat r],$$
and
$$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$
then
$$f_z(z)=dfrac,mathrm dF_z(z),mathrm dz = dfrac2piintlimits_z^R f_r(r)dfracmathrm drsqrtr^2-z^2.$$
If
$$f_r(r)=dfrac2rR^2,$$
then
$$f_z(z) = dfrac4pi R^2intlimits_z^R dfracrmathrm drsqrtr^2-z^2 = dfrac4pi R^2sqrtr^2-z^2Big|_z^R = mathbfcolorreddfrac4sqrtR^2-z^2pi R^2.$$
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
edited Aug 2 at 5:32
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
answered Aug 2 at 5:16
Yuri Negometyanov
9,2931523
9,2931523
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
add a comment |Â
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
Thank you for this answer. Please, I only need more steps on these three aspects. $$f_r,theta(r,theta)=dfrac2pi f_r(r),quad rin(0,R],quad thetainleft(0,dfracpi2right),$$, $$F_z(z) = Pr(hat z le z) = dfrac2piintlimits_0^Rintlimits_arccosdfrac zr^pi/2 f_r(r),mathrm dr,mathrm dtheta,$$ Specifically, how do I prove the first line. As for the second line how do one handle the lower limit of the integrand of $theta$. The third is that I do not know what does this $wedge$ sign mean in this case?
– Abdulhameed
Aug 2 at 17:09
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
@Abdulhameed From considerations of symmetry, it is sufficient to deal only with the first quarter, and integration over $theta$ must lead to the issue PDF. Besides, $z=costheta$ must be not greater than $r.$ "$wedge$" is the logical AND.
– Yuri Negometyanov
Aug 3 at 2:59
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
Really appreciate your reply. I mean $arccos fraczr$ in the lower limit. Then how was $frac2pif_r(r)$ derived?
– Abdulhameed
Aug 3 at 3:08
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
@Abdulhameed The integration over $theta$ givrmes $dfracpi2,$ and the inverse factor $dfrac2pi$ gives requred norm.
– Yuri Negometyanov
Aug 3 at 3:16
add a comment |Â
up vote
2
down vote
The density for $r$ corresponds to normal distributions for $z$ and $Y$ that would be independent if the distribution extended over the entire plane; in that case, the marginal distributions for $z$ and $Y$ would just be Gaussians. However, as the distribution is truncated to the circle, $z$ and $Y$ aren't independent. To get the marginal distribution for $z$ you need to integrate out $Y$; this is what leads to the $operatornameerf$ term. To reproduce the result, transform from polar coordinates to Cartesian coordinates ($r^2$ in the exponent becomes $z^2+Y^2$ and $2pi rmathrm drmathrm dphi$ becomes $mathrm dzmathrm dY$) and integrate out $Y$ to get the marginal density for $z$.
answered Jul 25 at 6:17
joriki
164k10180328
add a comment |Â
up vote
2
down vote
The density for $r$ corresponds to normal distributions for $z$ and $Y$ that would be independent if the distribution extended over the entire plane; in that case, the marginal distributions for $z$ and $Y$ would just be Gaussians. However, as the distribution is truncated to the circle, $z$ and $Y$ aren't independent. To get the marginal distribution for $z$ you need to integrate out $Y$; this is what leads to the $operatornameerf$ term. To reproduce the result, transform from polar coordinates to Cartesian coordinates ($r^2$ in the exponent becomes $z^2+Y^2$ and $2pi rmathrm drmathrm dphi$ becomes $mathrm dzmathrm dY$) and integrate out $Y$ to get the marginal density for $z$.
answered Jul 25 at 6:17
joriki
164k10180328
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The density for $r$ corresponds to normal distributions for $z$ and $Y$ that would be independent if the distribution extended over the entire plane; in that case, the marginal distributions for $z$ and $Y$ would just be Gaussians. However, as the distribution is truncated to the circle, $z$ and $Y$ aren't independent. To get the marginal distribution for $z$ you need to integrate out $Y$; this is what leads to the $operatornameerf$ term. To reproduce the result, transform from polar coordinates to Cartesian coordinates ($r^2$ in the exponent becomes $z^2+Y^2$ and $2pi rmathrm drmathrm dphi$ becomes $mathrm dzmathrm dY$) and integrate out $Y$ to get the marginal density for $z$.
answered Jul 25 at 6:17
joriki
164k10180328
The density for $r$ corresponds to normal distributions for $z$ and $Y$ that would be independent if the distribution extended over the entire plane; in that case, the marginal distributions for $z$ and $Y$ would just be Gaussians. However, as the distribution is truncated to the circle, $z$ and $Y$ aren't independent. To get the marginal distribution for $z$ you need to integrate out $Y$; this is what leads to the $operatornameerf$ term. To reproduce the result, transform from polar coordinates to Cartesian coordinates ($r^2$ in the exponent becomes $z^2+Y^2$ and $2pi rmathrm drmathrm dphi$ becomes $mathrm dzmathrm dY$) and integrate out $Y$ to get the marginal density for $z$.
answered Jul 25 at 6:17
joriki
164k10180328
answered Jul 25 at 6:17
joriki
164k10180328
answered Jul 25 at 6:17
joriki
164k10180328
answered Jul 25 at 6:17
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
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There's no $r_0$ in the diagram. It's not clear how $r_0$ and $z$ are related.
– joriki
Jul 25 at 5:12
Thanks for that observation. I have edited the question.
– Abdulhameed
Jul 25 at 5:56