Non-normal Bivariate distribution with normal margins

Question

I was working on a question and it stated the following:

Now I am supposed to show that the bivariate distribution is non normal even though the marginal distributions are. I generated 1000 iid variates for X1 and X2 graphed them in the following:

I know that the bivariate distribution is non-normal graphically but someone told me that we can show this analytically which I do not follow. Let $Z=X_1+X_2$, If $(X_1,X_2)$ was bivariate normal then $Z$ would also be normal, but $P(Z=0) \ge P(|X_1| \le 1) \approx 0.68$ i.e. It has positive mass at zero, which cannot be the case with a continuous distribution.

I believe the probabilities come from the way we defined $X_1$ and $X_2$ but I do not really see it.

Why is $P(|X_1| \le 1) \approx 0.68$

And then why is $P(Z=0) \ge P(|X_1| \le 1)$

Answer 1

$X_1 \sim N(0,1)$, hence we are able to check out probability table or find a calculator to compute $$P(|X_1|\le 1)=P(-1 \le X_1 \le 1)=1-2P(X_1 \le -1) \approx 1- 2(0.1587)=0.6826$$

For $Z$, we notice that it is not deterministic. Suppose $Z$ is normally distributed, recall that for a continuous random variable, the probability to take exactly a particular value is $0$. Hence we are supposed to have $P(Z=0)=0$

But $P(Z=0)=P(-1 \le X \le 1)\approx 0.6826$, which is a contradiction since $0$ is not approximately $0.6826$, hence the assumption that $Z$ is a normal random variable is false.

Answer 2

If $Z$ were normal with zero mean, then $P(Z \le 0)=0.5$.

But if $Z$ is normal with any mean and nonzero variance then $P(Z=0)=0$.

If $|X_1| \le 1$ then $X_2 = -X_1$ so $Z=0$. Thus $P(|X_1| \le 1) \le P(Z = 0)$.