Gaussian Functions: Summation of Infinite Variations

[pivoxa15]

Is the summation of an infinite number of different (not just different by constants ) Gaussian functions still a gaussian function?

I think not because you can because you can just pull an e^(ax^2) from the series where a is any value as small as needed.

 

[quasar987]

You're asking if whatever A_n, mu_n and sigma_n,

$$\sum_{n=1}^{\infty}A_n\exp{-\frac{(x-\mu_n)^2}{2\sigma_n^2}}=A\exp{-\frac{(x-\mu)^2}{2\sigma^2}}$$

for some A, mu and sigma. Is that correct?

 

[pivoxa15]

Yeah. Although I think this result doesn't hold?

 

[matt grime]

If you sum two gaussians with mean X and Y (think of them as p.d.f.s) then you get one with mean X+Y. So an infinite sum will, in general, have 'infinite mean'. Similarly, infinite standard deviation. So, no, it is not a Gaussian, and in general won't even exist.

 

[pivoxa15]

But if you sum a finite number of different Gaussian than the sum will still be a gaussian? If so how? how can you get e^a+e^b=e^c where a,b,c are functions that satisfy the exponentials being a gaussian.

 

[matt grime]

So, you're arguing that if property P holds for a finite number of things (terms, whatever), then it holds for an infinite number of things? Please, back off and think about that for a while. And I'm not arguing that e^a+e^b=e^c. What are a,b,c, for a start?

 

[pivoxa15]

What I am saying or rather asking is that I don't even see how it would work for the finite case. That is for any two different Gaussians
http://en.wikipedia.org/wiki/Gaussian_function

How do you add them and produce another gaussian?

or even if you take e^2+e^3. Can you make it into e^a for any number a?

 

[JohnDuck]

What I am saying or rather asking is that I don't even see how it would work for the finite case. That is for any two different Gaussians
http://en.wikipedia.org/wiki/Gaussian_function

How do you add them and produce another gaussian?

or even if you take e^2+e^3. Can you make it into e^a for any number a?

Well, $e^{2} + e^{3} = e^{a} \rightarrow a = ln(e^{2} + e^{3})$.