Originally Posted by

**vioravis** I have the following two densities:

f1(X) = {1/sigma1*sqrt(2*pi}* exp{-(X-mu1^2)/2*sigma1^2}

f2(X) = {1/sigma2*sqrt(2*pi}* exp{-(X-mu2^2)/2* sigma2^2}

So f1 is N(mu1, Sigma1^2) and f2 is N(mu2, Sigma2^2).

Does f(x) = f1(X)*f2(X) follow a normal distribution? Thanks a lot.

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.

So we can bust a gut attempting to prove that $\displaystyle \int_{-\infty}^{\infty} f(x)~dx \ne 1$, but we can just do the experiment and evaluate the thing numerically. The answer is that the integral is not $\displaystyle 1$ so $\displaystyle f$ is not a density.

Code:

>s1=1,mu1=0,s2=2, mu2=5
1
0
2
5
>dx=0.2;
>x=-10+dx/2:dx:20;
>
>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) );
>
>f=f1*f2;
>
>II1=sum(f1)*dx
1
>II2=sum(f2)*dx
1
>II=sum(f)*dx
0.014645
>

RonL