That f(x) is of the form of a Gaussian follows from the convolution theorem and the fact that the Fourier transform of a Gaussian is a Gaussian. But this does not guarantee that the product of the densities is in fact a density. As it happens it would be quite supprising if it were.
Originally Posted by vioravis
So we can bust a gut attempting to prove that , but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not so is not a density.
>f1=1/(s1*sqrt(2*pi))* exp( -(x-mu1)^2 / (2*s1^2) );
>f2=1/(s2*sqrt(2*pi))* exp( -(x-mu2)^2/ (2*s2^2) );