Here is a partial solution.
If $(X,Y)$ is jointly normal, with $\sigma^2_x \sigma_y^2>0$ (non-trivial case)
and $\mu_x=0$ then $XY$ cannot be normal.
If it were, then the third cumulant of $XY$
$$\kappa_3= 2\, \sigma_x\, \sigma_y(\sigma_y^2\, \sigma_x^2\, \rho^3\,
+3\, \sigma_y^2\, \sigma_x^2\, \rho
+3\, \sigma_x^2\, \rho\, \mu_y^2
+3\, \sigma_y\, \sigma_x\, \rho^2\, \mu_y\, \mu_x
+3\, \sigma_y\, \sigma_x\, \mu_x\, \mu_y
+3\, \sigma_y^2\rho\, \mu_x^2)$$
would vanish. Since $\mu_x=0$ this leads to
$$0=\rho(\rho^2 \sigma_y^2+3\sigma_y^2+3\mu_y^2).$$
Then $\rho=0$ and we are back in the independent case.
Update:
The more I think about it, the more I realize
that the result is not even plausible.
Generally speaking, the product $XY$ is much too likely to take values near zero
to be normal, or any other nice distribution.
Suppose that $(X,Y)$ has any "nice" bivariate distribution, with
a joint density function that is continuous and non-zero at the origin.
We argue that $XY$ cannot have a "nice" density at zero.
For $0<u<1$ define $$A_u=\{(x,y):u^{1/2}<x<u^{1/4},\, 0<y<u/x\}.$$
These sets shrink towards the origin as $u\to 0$.
We have
$$\mathbb{P}(0<XY\leq u)\geq \mathbb{P}((X,Y)\in A_u),$$
and so letting $\lambda$ denote two-dimensional Lebesgue measure
also
$${\mathbb{P}(0<XY\leq u)\over u}
\geq {\mathbb{P}((X,Y)\in A_u)\over\lambda(A_u)}\cdot{\lambda(A_u)\over u}
= {\mathbb{P}((X,Y)\in A_u)\over\lambda(A_u)}\cdot{\log(1/u)\over 4}.$$
As $u\to 0$, the first factor on the right converges to $f_{(X,Y)}(0,0)$ and the
second one blows up. Therefore, the left hand side also blows up, showing that $XY$
cannot have a continuous density with finite value at $u=0$.
Ok I solved it :)
Since covariance matrix is diagonal we can assume having multiple univariates. And then variance combination is as
$$\hat{\mu} = \frac{n_1\mu_1 + n_2\mu_2}{n_1+n_2}$$
$$\hat{\sigma}^2 = \frac{(\sigma_1^2 + \mu_1^2)n_1 + (\sigma_2^2 + \mu_2^2)n_2}{ (n_1+n_2)} - \hat{\mu}^2$$
Here, I used $\sigma^2 = E[x^2] - E[x]^2$
thanks again
Best Answer