Let's Suppose (X,Y) are bivariate random variables with a bivariate normal distribution, and it has a pdf of:
$$f_{XY}(x,y) = \frac{1}{2\pi\sigma_x \sigma_y (1-\rho^2)^{1/2}}\text{exp}\bigg(-\frac{1}{2}q(x,y)\bigg)$$
where:
$$q(x,y) = \frac{1}{1-p^2}\bigg[ \bigg(\frac{x-\mu_x}{\sigma_x} \bigg)^2-2\rho\bigg(\frac{x-\mu_x}{\sigma_x} \bigg)\bigg(\frac{y-\mu_y}{\sigma_y} \bigg) + \bigg(\frac{y-\mu_y}{\sigma_y} \bigg)^2\bigg]$$
Is it always the case, that if I marginalize $f_{XY}(x,y)$ on either x, or y that I always get a Normal distribution?
For instance, is it always true if $f_{XY}(x,y)$ is bivariate normal then:
$f_Y(x) = \int \limits_{-\infty}^{\infty} f_{XY}(x,y)~ dx = \frac{1}{\sigma_y\sqrt{2\pi}}\text{exp}\bigg[\frac{(x-\mu_y)^2}{2\sigma_y^2} \bigg]$
I was just curious if it makes any difference if X & Y are independent ($\rho = 0$), or if X & Y are not independent ($\rho \ne 0$), would the result of the marginalization be the same in either case, that it yields a normal distribution?
I guess that "not independent" means the same as "correlated" for bivariate normal distributions?
