I would like to extend to 3-D the formulation of the 2-D Gaussian PSF, given by: $$k_{\sigma}(x,y)=\frac{1}{\sqrt{(2\pi)^2}\sigma^2}\exp\left[-\frac{x^2+y^2}{2\sigma^2}\right]$$
Is the following 3-D generalization correct?
$$k_{\sigma}(x,y,z)=\frac{1}{\sqrt{(2\pi)^3}\sigma^3}\exp\left[-\frac{x^2+y^2+z^2}{2\sigma^2}\right]$$
$\endgroup$2 Answers
$\begingroup$Take a look here to see the definition and integration. In short, if you want a Gaussian of the form:$$N\exp\left(-\frac{x^2+y^2+z^2 + ...}{2\sigma^2}\right) \,,$$then the constant $N$ depends on the number of variables $n$:$$N = \frac1{\sigma^{n}(2\pi)^{n/2}} \,.$$So in your case, $n=3$, and the normalization constant is: $$\frac{1}{\sigma^3 (2\pi)^{3/2}} \,.$$Note that for the 2D case, this is $1/ 2\pi \sigma^2$, 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 $-n$.
$\endgroup$ 9 $\begingroup$The actual definition of a multivariate normal distribution is:
$$ f_X(x_1,x_2,...,x_n)=f_{X1}(x_1)f_{X2}(x_2)...f_{Xn}(x_n) = \prod\limits_{i=1}^n\frac{1}{\sigma_i\sqrt{2\pi}}exp[-\frac{(x_i-\mu_i)^2}{2\sigma_i^2}] $$ 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 $\sigma_x=\sigma_y=\sigma_z\equiv\sigma$.
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