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I am using a 3 dimensional gaussian point spread function in the form of

$$\frac{1}{\sqrt{(2\pi)^3}\sigma^3}e^{-\frac{r^2}{2\sigma^2}}$$

being $r^2$ the square of the distances $x^2 + y^2 + z^2$, to distribute a particle with mass on a specific point in space, in a density-probability area

I read somewhere that the $2\pi$ thing is so that the sum of the probabilities is $1$. Could somebody help me, showing how we can go from that $2\pi$ to the $1$?

Thank you very much in advance

Michael
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  • Perhaps you should start with the 1 dimensional gaussian. See eg http://math.stackexchange.com/questions/28558/what-do-pi-and-e-stand-for-in-the-normal-distribution-formula/28564#28564 – leonbloy Oct 04 '15 at 01:00
  • Just what I needed, thank you – Michael Oct 04 '15 at 13:26

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For something to be a density we need $\int_{\Bbb{R}^n} f(r) dr =1$. We have the following identity:

$$\int_{\Bbb{R}^n} e^{-\frac{r^2}{2\sigma^2}}dr=\sqrt{(2\pi)^3}\sigma^3$$

So we must divide to make sure it is normalized! What a density is, is:

$$P(X \in A) = \int_A f(r) dr$$

So if $A$ is the whole space, we should want that probability to be $1$!

  • I understand the first sentence, but I don't know what an identity is, nor what do you mean with "normalized". Still, perhaps leonbloy was right and I should start with simpler questions. I have a long way to go in math. – Michael Oct 04 '15 at 13:29
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    minor oversight, but $n=3$ or else you have to change your exponents to $n$. – Neil G Mar 20 '17 at 04:53