I wanted to know if one could write the sum of Gaussian densities with different deviations as a Gaussian density. More specifically,
For $\sigma>0$, let $f_\sigma$ be defined as :$$f_\sigma(x) \triangleq e^{-\frac{x^2}{2\sigma^2}}.$$
Let $\sigma_1,\sigma_2>0$, does it exist a $\sigma>0$ and a coefficient $\alpha>0$ such that: $$\alpha f_{\sigma} = f_{\sigma_1}+f_{\sigma_2}.$$
Any hint, ideas or solutions will be highly appreciated! Thank you very much.
This is not possible unless $\sigma_{1}=\sigma_{2}$ (in which case you can take $\sigma=\sigma_{1}$ and $\alpha=2$).
To see this, proceed assuming $\sigma_{1}\neq\sigma_{2}$. Suppose we can find $\sigma$ and $\alpha$ such that $$ \alpha f_{\sigma}(x)=f_{\sigma_{1}}(x)+f_{\sigma_{2}}(x) $$ for all $x$. Substitute $x=0,1,\sqrt{2}$ to get three equations: \begin{align*} \alpha & =1+1\\ \alpha e^{1/(2\sigma^{2})} & =e^{1/(2\sigma_{1}^{2})}+e^{1/(2\sigma_{2}^{2})}\\ \alpha e^{1/\sigma^{2}} & =e^{1/\sigma_{1}^{2}}+e^{1/\sigma_{2}^{2}} \end{align*} From the first two equations it follows that $\alpha=2$ and $$ \sigma^{2}=\left[2\ln\left(\frac{e^{1/(2\sigma_{1}^{2})}+e^{1/(2\sigma_{2}^{2})}}{2}\right)\right]^{-1}. $$ Substituting these quantities into the third equation, $$ \left(\frac{e^{1/(2\sigma_{1}^{2})}+e^{1/(2\sigma_{2}^{2})}}{2}\right)^{2}=\frac{e^{1/\sigma_{1}^{2}}+e^{1/\sigma_{2}^{2}}}{2}. $$ Defining $u=e^{1/(2\sigma_{1}^{2})}$ and $v=e^{1/(2\sigma_{2}^{2})}$, we can rewrite the above as $$ \left(u - v\right)^2 = 0, $$ a contradiction since $u \neq v$.