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From 1D gaussian to 2D gaussian


Create Fisheye from imageIs there any way to calculate change in derivatives along a vector?normalized Laplacian of GaussianGet 1d kernel from 2d gaussianHow to define the inverse of a vector?Can gaussian low pass filter remove ringing effect from the image?While calculating the arctan (1.01236) do we have to do 2 steps or one step before Taylor series?Derive the Separability of 2D GaussianDoes a summation of two negative Gaussians with different standard deviations have a stationary point?Iterative gaussian convolution and decimation when the decimation/convolution ratios are not divisible by two?













1












$begingroup$


I read this:




The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:



g2D(x,y,$sigma_1^2 + sigma_2^2$) = g1D(x,$sigma_1^2$)g2D(y,$sigma_2^2$)




saying that the product of two 1 dimensional gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.



I tried to deduce this by using:



g1D(x,$sigma1^2$)g2D(y,$sigma2^2$) = $frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



but I fail to obtain



$frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



which is equal to g2D(x,y,$sigma_1^2 + sigma_2^2$).



Someone know how to get there?










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    I read this:




    The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:



    g2D(x,y,$sigma_1^2 + sigma_2^2$) = g1D(x,$sigma_1^2$)g2D(y,$sigma_2^2$)




    saying that the product of two 1 dimensional gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.



    I tried to deduce this by using:



    g1D(x,$sigma1^2$)g2D(y,$sigma2^2$) = $frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



    but I fail to obtain



    $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



    which is equal to g2D(x,y,$sigma_1^2 + sigma_2^2$).



    Someone know how to get there?










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      I read this:




      The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:



      g2D(x,y,$sigma_1^2 + sigma_2^2$) = g1D(x,$sigma_1^2$)g2D(y,$sigma_2^2$)




      saying that the product of two 1 dimensional gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.



      I tried to deduce this by using:



      g1D(x,$sigma1^2$)g2D(y,$sigma2^2$) = $frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



      but I fail to obtain



      $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



      which is equal to g2D(x,y,$sigma_1^2 + sigma_2^2$).



      Someone know how to get there?










      share|cite|improve this question









      $endgroup$




      I read this:




      The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:



      g2D(x,y,$sigma_1^2 + sigma_2^2$) = g1D(x,$sigma_1^2$)g2D(y,$sigma_2^2$)




      saying that the product of two 1 dimensional gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.



      I tried to deduce this by using:



      g1D(x,$sigma1^2$)g2D(y,$sigma2^2$) = $frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



      but I fail to obtain



      $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



      which is equal to g2D(x,y,$sigma_1^2 + sigma_2^2$).



      Someone know how to get there?







      calculus image-processing






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jul 27 '15 at 13:45









      HerwiegHerwieg

      61




      61




















          2 Answers
          2






          active

          oldest

          votes


















          -1












          $begingroup$

          Hint: You see in the exponential that we have
          $$ exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right )=exp left ( frac-x^2 2( sigma_1^2 + sigma_2^2 ) right )exp left ( frac- y^22( sigma_1^2 + sigma_2^2 ) right ) $$
          The normalization hanging around should be
          $$frac12 pi ( sigma_1^2 + sigma_2^2 ) $$
          as you've seen...but under the change of variables
          $$ z_1 = frac sigma_1xsqrtsigma_1^2 + sigma_2^2 quad & quad z_2 = frac sigma_2ysqrtsigma_1^2 + sigma_2^2 $$
          What do you obtain? Also note that
          $$ 2 pi = sqrt2pi sqrt2 pi $$
          Edit: We see that
          $$frac12 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR^2 exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right ) dxdy = $$
          $$= left[ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracx^2 2( sigma_1^2 + sigma_2^2 ) right ) dx right] left [ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracy^2 2( sigma_1^2 + sigma_2^2 ) right )dy right]$$
          $$ =left[ frac1sqrt2 pi sigma_1 int_mathbbR exp left ( - fracz_1^2 2 sigma_1^2 right ) dz_1 right] left [ frac1sqrt2 pi sigma_2 int_mathbbR exp left ( - fracz_2^2 2sigma_2^2 right )dz_2 right] $$






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
            $endgroup$
            – Herwieg
            Jul 28 '15 at 10:08











          • $begingroup$
            What does $dxdy$ become under the change of coordinates I just suggested ?
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:51










          • $begingroup$
            Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:56










          • $begingroup$
            Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
            $endgroup$
            – Herwieg
            Jul 28 '15 at 14:06










          • $begingroup$
            Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
            $endgroup$
            – Jeb
            Jul 28 '15 at 15:08



















          -1












          $begingroup$

          Your first expression,



          $G(x,sigma_1^2) G(y,sigma_2^2) = frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



          is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,



          $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



          is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.



          It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $sigma_1^2 + sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.






          share|cite|improve this answer











          $endgroup$












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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            -1












            $begingroup$

            Hint: You see in the exponential that we have
            $$ exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right )=exp left ( frac-x^2 2( sigma_1^2 + sigma_2^2 ) right )exp left ( frac- y^22( sigma_1^2 + sigma_2^2 ) right ) $$
            The normalization hanging around should be
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) $$
            as you've seen...but under the change of variables
            $$ z_1 = frac sigma_1xsqrtsigma_1^2 + sigma_2^2 quad & quad z_2 = frac sigma_2ysqrtsigma_1^2 + sigma_2^2 $$
            What do you obtain? Also note that
            $$ 2 pi = sqrt2pi sqrt2 pi $$
            Edit: We see that
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR^2 exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right ) dxdy = $$
            $$= left[ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracx^2 2( sigma_1^2 + sigma_2^2 ) right ) dx right] left [ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracy^2 2( sigma_1^2 + sigma_2^2 ) right )dy right]$$
            $$ =left[ frac1sqrt2 pi sigma_1 int_mathbbR exp left ( - fracz_1^2 2 sigma_1^2 right ) dz_1 right] left [ frac1sqrt2 pi sigma_2 int_mathbbR exp left ( - fracz_2^2 2sigma_2^2 right )dz_2 right] $$






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
              $endgroup$
              – Herwieg
              Jul 28 '15 at 10:08











            • $begingroup$
              What does $dxdy$ become under the change of coordinates I just suggested ?
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:51










            • $begingroup$
              Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:56










            • $begingroup$
              Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
              $endgroup$
              – Herwieg
              Jul 28 '15 at 14:06










            • $begingroup$
              Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
              $endgroup$
              – Jeb
              Jul 28 '15 at 15:08
















            -1












            $begingroup$

            Hint: You see in the exponential that we have
            $$ exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right )=exp left ( frac-x^2 2( sigma_1^2 + sigma_2^2 ) right )exp left ( frac- y^22( sigma_1^2 + sigma_2^2 ) right ) $$
            The normalization hanging around should be
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) $$
            as you've seen...but under the change of variables
            $$ z_1 = frac sigma_1xsqrtsigma_1^2 + sigma_2^2 quad & quad z_2 = frac sigma_2ysqrtsigma_1^2 + sigma_2^2 $$
            What do you obtain? Also note that
            $$ 2 pi = sqrt2pi sqrt2 pi $$
            Edit: We see that
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR^2 exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right ) dxdy = $$
            $$= left[ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracx^2 2( sigma_1^2 + sigma_2^2 ) right ) dx right] left [ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracy^2 2( sigma_1^2 + sigma_2^2 ) right )dy right]$$
            $$ =left[ frac1sqrt2 pi sigma_1 int_mathbbR exp left ( - fracz_1^2 2 sigma_1^2 right ) dz_1 right] left [ frac1sqrt2 pi sigma_2 int_mathbbR exp left ( - fracz_2^2 2sigma_2^2 right )dz_2 right] $$






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
              $endgroup$
              – Herwieg
              Jul 28 '15 at 10:08











            • $begingroup$
              What does $dxdy$ become under the change of coordinates I just suggested ?
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:51










            • $begingroup$
              Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:56










            • $begingroup$
              Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
              $endgroup$
              – Herwieg
              Jul 28 '15 at 14:06










            • $begingroup$
              Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
              $endgroup$
              – Jeb
              Jul 28 '15 at 15:08














            -1












            -1








            -1





            $begingroup$

            Hint: You see in the exponential that we have
            $$ exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right )=exp left ( frac-x^2 2( sigma_1^2 + sigma_2^2 ) right )exp left ( frac- y^22( sigma_1^2 + sigma_2^2 ) right ) $$
            The normalization hanging around should be
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) $$
            as you've seen...but under the change of variables
            $$ z_1 = frac sigma_1xsqrtsigma_1^2 + sigma_2^2 quad & quad z_2 = frac sigma_2ysqrtsigma_1^2 + sigma_2^2 $$
            What do you obtain? Also note that
            $$ 2 pi = sqrt2pi sqrt2 pi $$
            Edit: We see that
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR^2 exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right ) dxdy = $$
            $$= left[ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracx^2 2( sigma_1^2 + sigma_2^2 ) right ) dx right] left [ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracy^2 2( sigma_1^2 + sigma_2^2 ) right )dy right]$$
            $$ =left[ frac1sqrt2 pi sigma_1 int_mathbbR exp left ( - fracz_1^2 2 sigma_1^2 right ) dz_1 right] left [ frac1sqrt2 pi sigma_2 int_mathbbR exp left ( - fracz_2^2 2sigma_2^2 right )dz_2 right] $$






            share|cite|improve this answer











            $endgroup$



            Hint: You see in the exponential that we have
            $$ exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right )=exp left ( frac-x^2 2( sigma_1^2 + sigma_2^2 ) right )exp left ( frac- y^22( sigma_1^2 + sigma_2^2 ) right ) $$
            The normalization hanging around should be
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) $$
            as you've seen...but under the change of variables
            $$ z_1 = frac sigma_1xsqrtsigma_1^2 + sigma_2^2 quad & quad z_2 = frac sigma_2ysqrtsigma_1^2 + sigma_2^2 $$
            What do you obtain? Also note that
            $$ 2 pi = sqrt2pi sqrt2 pi $$
            Edit: We see that
            $$frac12 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR^2 exp left ( - fracx^2 + y^22( sigma_1^2 + sigma_2^2 ) right ) dxdy = $$
            $$= left[ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracx^2 2( sigma_1^2 + sigma_2^2 ) right ) dx right] left [ frac1sqrt2 pi ( sigma_1^2 + sigma_2^2 ) int_mathbbR exp left ( - fracy^2 2( sigma_1^2 + sigma_2^2 ) right )dy right]$$
            $$ =left[ frac1sqrt2 pi sigma_1 int_mathbbR exp left ( - fracz_1^2 2 sigma_1^2 right ) dz_1 right] left [ frac1sqrt2 pi sigma_2 int_mathbbR exp left ( - fracz_2^2 2sigma_2^2 right )dz_2 right] $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 28 '15 at 22:52

























            answered Jul 27 '15 at 14:02









            JebJeb

            3,8791813




            3,8791813











            • $begingroup$
              Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
              $endgroup$
              – Herwieg
              Jul 28 '15 at 10:08











            • $begingroup$
              What does $dxdy$ become under the change of coordinates I just suggested ?
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:51










            • $begingroup$
              Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:56










            • $begingroup$
              Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
              $endgroup$
              – Herwieg
              Jul 28 '15 at 14:06










            • $begingroup$
              Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
              $endgroup$
              – Jeb
              Jul 28 '15 at 15:08

















            • $begingroup$
              Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
              $endgroup$
              – Herwieg
              Jul 28 '15 at 10:08











            • $begingroup$
              What does $dxdy$ become under the change of coordinates I just suggested ?
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:51










            • $begingroup$
              Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
              $endgroup$
              – Jeb
              Jul 28 '15 at 11:56










            • $begingroup$
              Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
              $endgroup$
              – Herwieg
              Jul 28 '15 at 14:06










            • $begingroup$
              Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
              $endgroup$
              – Jeb
              Jul 28 '15 at 15:08
















            $begingroup$
            Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
            $endgroup$
            – Herwieg
            Jul 28 '15 at 10:08





            $begingroup$
            Using your hint still doesn't give me the correct normalization. Preferably, I would like to go from g1d*g1d to g2d and not the other way around. Could you give me the full derivation?
            $endgroup$
            – Herwieg
            Jul 28 '15 at 10:08













            $begingroup$
            What does $dxdy$ become under the change of coordinates I just suggested ?
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:51




            $begingroup$
            What does $dxdy$ become under the change of coordinates I just suggested ?
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:51












            $begingroup$
            Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:56




            $begingroup$
            Also, if you want to go the other direction, just write the change of variables I gave in terms of $z_1,z_2$
            $endgroup$
            – Jeb
            Jul 28 '15 at 11:56












            $begingroup$
            Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
            $endgroup$
            – Herwieg
            Jul 28 '15 at 14:06




            $begingroup$
            Why do you ask for dxdy, as there is no integration involved? I'm not able to find out the derivation myself with these hints so it would be great if you could give the full derivation.
            $endgroup$
            – Herwieg
            Jul 28 '15 at 14:06












            $begingroup$
            Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
            $endgroup$
            – Jeb
            Jul 28 '15 at 15:08





            $begingroup$
            Oh, you should re-read your notes...there is implicitly an integral hanging around in that statement...it is literally just the change of variables I've written.
            $endgroup$
            – Jeb
            Jul 28 '15 at 15:08












            -1












            $begingroup$

            Your first expression,



            $G(x,sigma_1^2) G(y,sigma_2^2) = frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



            is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,



            $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



            is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.



            It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $sigma_1^2 + sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.






            share|cite|improve this answer











            $endgroup$

















              -1












              $begingroup$

              Your first expression,



              $G(x,sigma_1^2) G(y,sigma_2^2) = frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



              is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,



              $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



              is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.



              It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $sigma_1^2 + sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.






              share|cite|improve this answer











              $endgroup$















                -1












                -1








                -1





                $begingroup$

                Your first expression,



                $G(x,sigma_1^2) G(y,sigma_2^2) = frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



                is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,



                $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



                is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.



                It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $sigma_1^2 + sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.






                share|cite|improve this answer











                $endgroup$



                Your first expression,



                $G(x,sigma_1^2) G(y,sigma_2^2) = frac1sqrt2pisigma_1e^frac-x^22sigma_1^2frac1sqrt2pisigma_2e^frac-y^22sigma_2^2$ = $frac12pisigma_1sigma_2e^-(fracx^22sigma_1^2+fracy^22sigma_2^2)$



                is a correct anisotropic 2D Gaussian. The expression you are trying to obtain,



                $frac12pi(sigma_1^2 + sigma_2^2)e^frac-(x^2+y^2)2sigma_1^2 + 2sigma_2^2$



                is not (it’s an isotropic Gaussian). There is no way you can go from the one to the other. However, one can incorrectly match the two expressions by creating a new coordinate system that is a scaled version of the original one, in such a way that the Gaussian is isotropic in the new coordinate system. The other answer does this. But that is not useful.



                It seems that OP is confused about the combination of the variances. The convolution of two 1-dimensional Gaussian functions with variances $sigma_1^2$ and $sigma_2^2$ is equal to a 1-dimensional Gaussian function with variance $sigma_1^2 + sigma_2^2$. There is no such expectation for the multiplication of Gaussians (in fact, when multiplying them, assuming the same orientation and the same mean, the resulting variance is smaller, not larger.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 17 at 17:04

























                answered Mar 17 at 14:39









                Cris LuengoCris Luengo

                1849




                1849



























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