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3-D generalization of the Gaussian point spread function

Verification of convolution between gaussian and uniform distributionsExplanation of Approximation for Integral Over Gaussian DistributionIndependence of Gaussian Distribution Function with Different MeansDistinguish Normal Distribution, Gaussian Distribution and Normalised Gaussian Distribution?Deriving the formula for multivariate Gaussian distributionGaussian distribution and proportionalityintegral of 3d gaussian with hollow integral spaceMoment Function of Gaussian Processexpanding the exponential term in the multivariate GaussianCalculating the convolution of an Arcsine law and a Gaussian distribution

6

4

$begingroup$

I would like to extend to 3-D the formulation of the 2-D Gaussian PSF, given by:
$$k_{sigma}(x,y)=frac{1}{sqrt{(2pi)^2}sigma^2}expleft[-frac{x^2+y^2}{2sigma^2}right]$$

Is the following 3-D generalization correct?

$$k_{sigma}(x,y,z)=frac{1}{sqrt{(2pi)^3}sigma^3}expleft[-frac{x^2+y^2+z^2}{2sigma^2}right]$$

normal-distribution 3d

share|cite|improve this question

edited Jul 2 '13 at 16:54

no_name

asked Jul 2 '13 at 16:07

no_nameno_name

205128

$endgroup$

add a comment | 

6

4

$begingroup$

I would like to extend to 3-D the formulation of the 2-D Gaussian PSF, given by:
$$k_{sigma}(x,y)=frac{1}{sqrt{(2pi)^2}sigma^2}expleft[-frac{x^2+y^2}{2sigma^2}right]$$

Is the following 3-D generalization correct?

$$k_{sigma}(x,y,z)=frac{1}{sqrt{(2pi)^3}sigma^3}expleft[-frac{x^2+y^2+z^2}{2sigma^2}right]$$

normal-distribution 3d

share|cite|improve this question

edited Jul 2 '13 at 16:54

no_name

asked Jul 2 '13 at 16:07

no_nameno_name

205128

$endgroup$

add a comment | 

$begingroup$

I would like to extend to 3-D the formulation of the 2-D Gaussian PSF, given by:
$$k_{sigma}(x,y)=frac{1}{sqrt{(2pi)^2}sigma^2}expleft[-frac{x^2+y^2}{2sigma^2}right]$$

Is the following 3-D generalization correct?

$$k_{sigma}(x,y,z)=frac{1}{sqrt{(2pi)^3}sigma^3}expleft[-frac{x^2+y^2+z^2}{2sigma^2}right]$$

normal-distribution 3d

share|cite|improve this question

edited Jul 2 '13 at 16:54

no_name

asked Jul 2 '13 at 16:07

no_nameno_name

205128

$endgroup$

share|cite|improve this question

edited Jul 2 '13 at 16:54

no_name

asked Jul 2 '13 at 16:07

no_nameno_name

205128

share|cite|improve this question

edited Jul 2 '13 at 16:54

no_name

asked Jul 2 '13 at 16:07

no_nameno_name

205128

asked Jul 2 '13 at 16:07

no_nameno_name

205128

asked Jul 2 '13 at 16:07

no_nameno_name

205128

2 Answers
2

active

oldest

votes

8

$begingroup$

Take a look here to see the definition and integration. In short, if you want a Gaussian of the form:
$$Nexpleft(-frac{x^2+y^2+z^2 + ...}{2sigma^2}right) ,,$$
then the constant $N$ depends on the number of variables $n$:
$$N = frac1{sigma^{n}(2pi)^{n/2}} ,.$$
So in your case, $n=3$, and the normalization constant is: $$frac{1}{sigma^3 (2pi)^{3/2}} ,.$$
Note that for the 2D case, this is $1/ 2pi sigma^2$, i.e. you are missing a $sigma$. This is intuitive, since $sigma$ has the dimension of the length of whatever you are trying to measure. Integration over $n$ dimensions adds $n$ powers of that length, and since after integration we get a unit-less quantity (probability), the power of $sigma$ in the Gaussian must always be $-n$.

share|cite|improve this answer

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the minus sign in the exponential argument is absent? From your words, it seems to me that my 3-D formulation is correct, right? I modified the 2-D formulation.
  $endgroup$
  – no_name
  Jul 2 '13 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @no_name - it's not absent, that the definition of a negative power: $sigma^{-n} = 1/sigma^{n}$. And yes, it is correct.
  $endgroup$
  – nbubis
  Jul 2 '13 at 16:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How would this look if I have three different sigmas and expected values?
  $endgroup$
  – Zloy Smiertniy
  Apr 9 '15 at 15:10
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$Nexpleft(-frac{(x-mx)^2}{sigma_x^2} - frac{(y-my)^2}{sigma_y^2} -frac{(z-mz)^2}{sigma_z^2} right)$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:12
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$ N = frac{1}{sigma_x*sigma_y*sigma_z (2pi)^{3/2}}$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:15
  
  
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

The actual definition of a multivariate normal distribution is:

$$
f_X(x_1,x_2,...,x_n)=f_{X1}(x_1)f_{X2}(x_2)...f_{Xn}(x_n) = prodlimits_{i=1}^nfrac{1}{sigma_isqrt{2pi}}exp[-frac{(x_i-mu_i)^2}{2sigma_i^2}]
$$
Where one could use the more common Cartesian coordinates $x_1=x ,$ $x_2=y ,$ and $x_3=z$

So, in the original case, the suggested solution is true whenever the mean values are null $(mu_i=0)$, and the two/three variables happen to have the same identical standard variation $sigma_x=sigma_y=sigma_zequivsigma$.

share|cite|improve this answer

answered Jul 5 '17 at 16:51

RibbaRibba

1

$endgroup$

add a comment | 

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8

$begingroup$

Take a look here to see the definition and integration. In short, if you want a Gaussian of the form:
$$Nexpleft(-frac{x^2+y^2+z^2 + ...}{2sigma^2}right) ,,$$
then the constant $N$ depends on the number of variables $n$:
$$N = frac1{sigma^{n}(2pi)^{n/2}} ,.$$
So in your case, $n=3$, and the normalization constant is: $$frac{1}{sigma^3 (2pi)^{3/2}} ,.$$
Note that for the 2D case, this is $1/ 2pi sigma^2$, i.e. you are missing a $sigma$. This is intuitive, since $sigma$ has the dimension of the length of whatever you are trying to measure. Integration over $n$ dimensions adds $n$ powers of that length, and since after integration we get a unit-less quantity (probability), the power of $sigma$ in the Gaussian must always be $-n$.

share|cite|improve this answer

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the minus sign in the exponential argument is absent? From your words, it seems to me that my 3-D formulation is correct, right? I modified the 2-D formulation.
  $endgroup$
  – no_name
  Jul 2 '13 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @no_name - it's not absent, that the definition of a negative power: $sigma^{-n} = 1/sigma^{n}$. And yes, it is correct.
  $endgroup$
  – nbubis
  Jul 2 '13 at 16:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How would this look if I have three different sigmas and expected values?
  $endgroup$
  – Zloy Smiertniy
  Apr 9 '15 at 15:10
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$Nexpleft(-frac{(x-mx)^2}{sigma_x^2} - frac{(y-my)^2}{sigma_y^2} -frac{(z-mz)^2}{sigma_z^2} right)$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:12
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$ N = frac{1}{sigma_x*sigma_y*sigma_z (2pi)^{3/2}}$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:15
  
  
  

  
  
  
  

 | 
show 4 more comments

8

$begingroup$

Take a look here to see the definition and integration. In short, if you want a Gaussian of the form:
$$Nexpleft(-frac{x^2+y^2+z^2 + ...}{2sigma^2}right) ,,$$
then the constant $N$ depends on the number of variables $n$:
$$N = frac1{sigma^{n}(2pi)^{n/2}} ,.$$
So in your case, $n=3$, and the normalization constant is: $$frac{1}{sigma^3 (2pi)^{3/2}} ,.$$
Note that for the 2D case, this is $1/ 2pi sigma^2$, i.e. you are missing a $sigma$. This is intuitive, since $sigma$ has the dimension of the length of whatever you are trying to measure. Integration over $n$ dimensions adds $n$ powers of that length, and since after integration we get a unit-less quantity (probability), the power of $sigma$ in the Gaussian must always be $-n$.

share|cite|improve this answer

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the minus sign in the exponential argument is absent? From your words, it seems to me that my 3-D formulation is correct, right? I modified the 2-D formulation.
  $endgroup$
  – no_name
  Jul 2 '13 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @no_name - it's not absent, that the definition of a negative power: $sigma^{-n} = 1/sigma^{n}$. And yes, it is correct.
  $endgroup$
  – nbubis
  Jul 2 '13 at 16:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How would this look if I have three different sigmas and expected values?
  $endgroup$
  – Zloy Smiertniy
  Apr 9 '15 at 15:10
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$Nexpleft(-frac{(x-mx)^2}{sigma_x^2} - frac{(y-my)^2}{sigma_y^2} -frac{(z-mz)^2}{sigma_z^2} right)$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:12
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $$ N = frac{1}{sigma_x*sigma_y*sigma_z (2pi)^{3/2}}$$
  $endgroup$
  – Jumabek Alihanov
  Feb 9 '18 at 1:15
  
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Take a look here to see the definition and integration. In short, if you want a Gaussian of the form:
$$Nexpleft(-frac{x^2+y^2+z^2 + ...}{2sigma^2}right) ,,$$
then the constant $N$ depends on the number of variables $n$:
$$N = frac1{sigma^{n}(2pi)^{n/2}} ,.$$
So in your case, $n=3$, and the normalization constant is: $$frac{1}{sigma^3 (2pi)^{3/2}} ,.$$
Note that for the 2D case, this is $1/ 2pi sigma^2$, i.e. you are missing a $sigma$. This is intuitive, since $sigma$ has the dimension of the length of whatever you are trying to measure. Integration over $n$ dimensions adds $n$ powers of that length, and since after integration we get a unit-less quantity (probability), the power of $sigma$ in the Gaussian must always be $-n$.

share|cite|improve this answer

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

$endgroup$

share|cite|improve this answer

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

edited yesterday

Nanashi No Gombe

537420

edited yesterday

Nanashi No Gombe

537420

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

answered Jul 2 '13 at 16:41

nbubisnbubis

27.3k552110

0

$begingroup$

The actual definition of a multivariate normal distribution is:

$$
f_X(x_1,x_2,...,x_n)=f_{X1}(x_1)f_{X2}(x_2)...f_{Xn}(x_n) = prodlimits_{i=1}^nfrac{1}{sigma_isqrt{2pi}}exp[-frac{(x_i-mu_i)^2}{2sigma_i^2}]
$$
Where one could use the more common Cartesian coordinates $x_1=x ,$ $x_2=y ,$ and $x_3=z$

So, in the original case, the suggested solution is true whenever the mean values are null $(mu_i=0)$, and the two/three variables happen to have the same identical standard variation $sigma_x=sigma_y=sigma_zequivsigma$.

share|cite|improve this answer

answered Jul 5 '17 at 16:51

RibbaRibba

1

$endgroup$

add a comment | 

0

$begingroup$

The actual definition of a multivariate normal distribution is:

$$
f_X(x_1,x_2,...,x_n)=f_{X1}(x_1)f_{X2}(x_2)...f_{Xn}(x_n) = prodlimits_{i=1}^nfrac{1}{sigma_isqrt{2pi}}exp[-frac{(x_i-mu_i)^2}{2sigma_i^2}]
$$
Where one could use the more common Cartesian coordinates $x_1=x ,$ $x_2=y ,$ and $x_3=z$

So, in the original case, the suggested solution is true whenever the mean values are null $(mu_i=0)$, and the two/three variables happen to have the same identical standard variation $sigma_x=sigma_y=sigma_zequivsigma$.

share|cite|improve this answer

answered Jul 5 '17 at 16:51

RibbaRibba

1

$endgroup$

add a comment | 

$begingroup$

The actual definition of a multivariate normal distribution is:

$$
f_X(x_1,x_2,...,x_n)=f_{X1}(x_1)f_{X2}(x_2)...f_{Xn}(x_n) = prodlimits_{i=1}^nfrac{1}{sigma_isqrt{2pi}}exp[-frac{(x_i-mu_i)^2}{2sigma_i^2}]
$$
Where one could use the more common Cartesian coordinates $x_1=x ,$ $x_2=y ,$ and $x_3=z$

So, in the original case, the suggested solution is true whenever the mean values are null $(mu_i=0)$, and the two/three variables happen to have the same identical standard variation $sigma_x=sigma_y=sigma_zequivsigma$.

share|cite|improve this answer

answered Jul 5 '17 at 16:51

RibbaRibba

1

$endgroup$

share|cite|improve this answer

answered Jul 5 '17 at 16:51

RibbaRibba

1

answered Jul 5 '17 at 16:51

RibbaRibba

1

answered Jul 5 '17 at 16:51

RibbaRibba

1