A well-known example to show that two random variables whose marginal distributions are normal, do not need necessarily be jointly normal is achieved by letting $X, Y $ have the following joint probability density function: $$ f(x,y):= \frac{1}{\pi}e^{-(x^2+y^2)/2}\cdot 1_{\{(x,y): xy>0\}}. $$
Unless I am doing something wrong, when I compute the characteristic function for $(X, Y), $ I end up with $$ \phi_{X, Y}(s, t) = e^{-(s^2+t^2)/2}. $$ This troubles me, because that is also the ch.f. for a bivariate normal distribution whose components are independent... Not sure if I am making some silly calculus mistake or if this result is correct. I am computing the ch.f. as $$ \phi_{X, Y}(s, t) = \int_0^\infty \int_0^\infty e^{isx+ity}\frac{1}{\pi}e^{-(x^2+y^2)/2}\, dx dy + \int_{-\infty}^0 \int_{-\infty}^0 e^{isx+ity}\frac{1}{\pi}e^{-(x^2+y^2)/2}\, dx dy $$ and completing the square for the exponentials, i.e., for the first of the two integrals above $$ \int_0^\infty \int_0^\infty e^{isx+ity}\frac{1}{\pi}e^{-(x^2+y^2)/2}\, dx dy $$
$$ =\frac{1}{\pi} e^{-t^2/2-s^2/2} \cdot \int_0^\infty e^{-(x^2- 2isx + (is)^2)/2} dx \int_0^\infty e^{-(y^2-2ity + (it)^2)/2} dy + ....$$
Now, both integrals computes at $\pi/2 $ as they are "half a normal" so that we get $$ \int_0^\infty \int_0^\infty e^{isx+ity}\frac{1}{\pi}e^{-(x^2+y^2)/2}\, dx dy = \frac{1}{\pi} e^{-t^2/2-s^2/2} \frac{\pi}{2} = \frac{1}{2} e^{-t^2/2-s^2/2}. $$ Doing the same for the integral $\int_{-\infty}^0 \int_{-\infty}^0 e^{isx+ity}\frac{1}{\pi}e^{-(x^2+y^2)/2}\, dx dy $ gives my result which is strange because this is obviously not the pdf for a bivariate independent standard normal. Hence, I do believe something must have gone wrong at some points, but cannot see where.
Also, if $X $ is a given random variable and we consider the vector ${\bf Y}:= (X, X)^T, $ then its ch.f. is given by $$ \phi_{{\bf Y}}(t_1, t_2) = \phi_X(t_1+t_2) $$ which is interesting as a result in itself. But, if I make the further assumption that $X \sim N(0, 1), $ then we end up with $$ \phi_{{\bf Y}}(t_1, t_2) = e^{-(t_1+t_2)^2/2}. $$ Is this the ch.f. associated with a degenerate bivariate random variable? Technically, such a distribution does not have probability density function, but perhaps one could still get here using generalized inverses?
Thank you for any help,
Maurice