<s>[INST] From 1D gaussian to 2D gaussian 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,$\sigma2^2$) = $\frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-x^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi}\sigma_2}e^{\frac{-y^2}{2\sigma_2^2}}$ = $\frac{1}{2\pi\sigma_1\sigma_2}e^{-(\frac{x^2}{2\sigma_1^2}+\frac{y^2}{2\sigma_2^2})}$
but I fail to obtain 
$\frac{1}{2\pi(\sigma_1^2 + \sigma_2^2)}e^{\frac{-(x^2+y^2)}{2\sigma_1^2 + 2\sigma_2^2}}$
which is equal to g2D(x,y,$\sigma_1^2 + \sigma_2^2$).
Someone know how to get there?
 [/INST] Hint: You see in the exponential that we have
$$ \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right )=\exp \left (  \frac{-x^2 }{2( \sigma_1^2 + \sigma_2^2 ) } \right )\exp \left (  \frac{- y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right )  $$
The normalization hanging around should be
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) } $$
as you've seen...but under the change of variables
$$ z_1 = \frac{ \sigma_1x}{\sqrt{\sigma_1^2 + \sigma_2^2}}  \quad \& \quad z_2 = \frac{ \sigma_2y}{\sqrt{\sigma_1^2 + \sigma_2^2}}  $$
What do you obtain? Also note that
$$ 2 \pi = \sqrt{2\pi} \sqrt{2 \pi} $$
Edit: We see that
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) }  \int_{\mathbb{R}^2} \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) }  \right ) dxdy = $$
$$= \left[ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} }  \int_{\mathbb{R}} \exp \left ( - \frac{x^2 }{2( \sigma_1^2 + \sigma_2^2 ) }  \right ) dx \right] \left [ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} }   \int_{\mathbb{R}} \exp \left ( - \frac{y^2 }{2( \sigma_1^2 + \sigma_2^2 ) }  \right )dy  \right]$$
$$ =\left[ \frac{1}{\sqrt{2 \pi }\sigma_1 }  \int_{\mathbb{R}} \exp \left ( - \frac{z_1^2 }{2 \sigma_1^2   }  \right ) dz_1 \right] \left [ \frac{1}{\sqrt{2 \pi }\sigma_2 }   \int_{\mathbb{R}} \exp \left ( - \frac{z_2^2 }{2\sigma_2^2  }  \right )dz_2  \right] $$
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