This answer draws on the answers to a related question Is joint normality a necessary condition for the sum of normal random variables to be normal? that I asked on stats.SE (and answered too!).
- Let $X_2$ and $Y_2$ denote two independent standard normal random variables. Then, it is well-known that $X_2$ and $Y_2$ are jointly normal random variables and that $X_2+Y_2 \sim N(0,2)$.
- Let $X_1$ and $Y_1$ denote two jointly continuous random variables whose joint density has value $2\phi(x)\phi(y)$ on the shaded regions
shown in the diagram below (borrowed from one of the answers to the question cited above). Here, $\phi(\cdot)$ denotes the standard normal density. Note that $X_1$ and $Y_1$ are dependent random variables.
$\hspace{1.5 cm}$
Since $\phi(\cdot)$ is an even function of its argument, we have that for $x > 0$,
\begin{align}
f_{X_1}(x) &= \int_{-\infty}^{-x} 2\phi(x)\phi(y) \,\mathrm dy
+ \int_{0}^{x} 2\phi(x)\phi(y) \,\mathrm dy\\
&=\phi(x)\left[\int_{-\infty}^{-x} 2\phi(y) \,\mathrm dy
+ \int_{0}^{x} 2\phi(y) \,\mathrm dy\right]\\
&=\phi(x)\left[\int_{-\infty}^{-x} \phi(y) \,\mathrm dy
+ \int_{x}^{\infty} \phi(y) \,\mathrm dy
+ \int_{0}^{x} \phi(y) \,\mathrm dy
+ \int_{-x}^{0} \phi(y) \,\mathrm dy\right]\\
&= \phi(x)\left[\int_{-\infty}^{\infty} \phi(y) \,\mathrm dy
\right]\\
&= \phi(x)
\end{align}
and similarly for $x < 0$. Thus, $X_1 \sim N(0,1)$ and so is $Y_1\sim N(0,1)$ via similar calculations. Note that $X_1$ and $Y_1$ are not jointly
normal random variables. Nonetheless, it is easy to show via the same kinds of calculations exploiting the symmetry of $\phi(\cdot)$ that
$X_1+Y_1 \sim N(0,2)$.
To summarize,
- $X_1, X_2, Y_1, Y_2$ all are standard normal random variables,
- $X_1+Y_1$ and $X_2+Y_2$ are zero-mean normal random variables with variance $2$,
- $X_2$ and $Y_2$ are independent random variables, and
- $X_1$ and $Y_1$ are dependent random variables