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In comments following this answer of mine to a related question, Users ssdecontrol and Glen_b asked whether joint normality of $X$ and $Y$ is necessary for asserting the normality of the sum $X+Y$? That joint normality is sufficient is, of course, well-known. This supplemental question was not addressed there, and is perhaps worth considering in its own right.

Since joint normality implies marginal normality, I ask

Do there exist normal random variables $X$ and $Y$ such that $X+Y$ is a normal random variable, but $X$ and $Y$ are not jointly normal random variables?

If $X$ and $Y$ are not required to have normal distributions, then it is easy to find such normal random variables. One example can be found in my previous answer (link is given above). I believe that the answer to the highlighted question above is Yes, and have posted (what I think is) an example as an answer to this question.

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    $\begingroup$ How do you want to deal with degenerate distributions? For example, if $X$ is a standard normal and $Y=-2X$, then the joint distribution of $X$ and $Y$ is a degenerate normal distribution and $X+Y$ is a standard normal. $\endgroup$ – Brian Borchers Jan 4 '16 at 15:11
  • $\begingroup$ @BrianBorchers $X$ and $Y = -2X$ are jointly normal random variables even though the distribution is degenerate as you say. The standard definition of joint normality is that $X$ and $Y$ are jointly normal if $aX+bY$ is normal for all choices of $(a,b)$. Here, $(a,b) = (0,0)$ is a degenerate case which is nonetheless called a normal random variable as a courtesy. $\endgroup$ – Dilip Sarwate Jan 4 '16 at 18:02
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Let $U,V$ be iid $N(0,1)$.

Now transform $(U,V) \to (X,Y)$ as follows:

In the first quadrant (i.e. $U>0,V>0$) let $X=\max(U,V)$ and $Y = \min(U,V)$.

For the other quadrants, rotate this mapping about the origin.

The resulting bivariate distribution looks like (seen from above):

$\hspace{1.5 cm}$![enter image description here

-- the purple represents regions with doubled probability and the white regions are ones with no probability. The black circles are contours of constant density (everywhere on the circle for $(U,V)$, but within each colored region for $(X,Y)$).

  1. By symmetry both $X$ and $Y$ are standard normal (looking down a vertical line or along a horizontal line there's a purple point for every white one which we can regard as being flipped across the axis the horizontal or vertical line crosses)

  2. but $(X,Y)$ are clearly not bivariate normal, and

  3. $X+Y = U+V$ which is $\sim N(0,2)$ (equivalently, look along lines of constant $X+Y$ and see that we have symmetry similar to that we discussed in 1., but this time about the $Y=X$ line)

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    $\begingroup$ +1 and a Accept; this construction is much nicer than the one in my own answer! $\endgroup$ – Dilip Sarwate Jan 18 '16 at 6:53
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Consider jointly continuous random variables $U, V, W$ with joint density function \begin{align} f_{U,V,W}(u,v,w) = \begin{cases} 2\phi(u)\phi(v)\phi(w) & ~~~~\text{if}~ u \geq 0, v\geq 0, w \geq 0,\\ & \text{or if}~ u < 0, v < 0, w \geq 0,\\ & \text{or if}~ u < 0, v\geq 0, w < 0,\\ & \text{or if}~ u \geq 0, v< 0, w < 0,\\ & \\ 0 & \text{otherwise} \end{cases}\tag{1} \end{align} where $\phi(\cdot)$ denotes the standard normal density function.

It is clear that $U, V$, and $W$ are dependent random variables. It is also clear that they are not jointly normal random variables. However, all three pairs $(U,V), (U,W), (V,W)$ are pairwise independent random variables: in fact, independent standard normal random variables (and thus pairwise jointly normal random variables). In short, $U,V,W$ are an example of pairwise independent but not mutually independent normal random variables. See this answer of mine for more details.

Notice that the pairwise independence gives us that $U+V, U+W$, and $V-W$ all are zero-mean normal random variables with variance $2$. Now, let us define $$X = U+W, ~Y = V-W \tag{2}$$ and note that $X+Y = U+V$ is also a zero-mean normal random variable with variance $2$. Also, $\operatorname{cov}(X,Y) = -\operatorname{var}(W) = -1$, and so $X$ and $Y$ are dependent and correlated random variables.

$X$ and $Y$ are (correlated) normal random variables that are not jointly normal but have the property that their sum $X+Y$ is a normal random variable.

Put another way, joint normality is a sufficient condition for asserting the normality of a sum of normal random variables, but it is not a necessary condition.

Proof that $X$ and $Y$ are not jointly normal
Since the transformation $(U,V,W) \to (U+W, V-W, W) = (X,Y,W)$ is linear, it is easy to get that $f_{X,Y,W}(x, y, w) = f_{U,V,W}(x-w,y+w,w)$. Therefore we have that $$f_{X,Y}(x,y) = \int_{-\infty}^\infty f_{X,Y,W}(x,y,w)\,\mathrm dw = \int_{-\infty}^\infty f_{U,V,W}(x-w,y+w,w)\,\mathrm dw$$ But $f_{U,V, W}$ has the property that its value is nonzero only when exactly one or all three of its arguments are nonnegative. Now suppose that $x, y > 0$. Then, $f_{U,V,W}(x-w,y+w,w)$ has value $2\phi(x-w)\phi(y+w)\phi(w)$ for $w \in (-\infty,-y) \cup (0,x)$ and is $0$ otherwise. So, for $x, y > 0$, $$f_{X,Y}(x,y) = \int_{-\infty}^{-y} 2\phi(x-w)\phi(y+w)\phi(w)\,\mathrm dw + \int_0^x 2\phi(x-w)\phi(y+w)\phi(w)\,\mathrm dw.\tag{3}$$ Now, \begin{align} (x-w)^2 + (y+w)^2 + w^2 &= 3w^2 -2w(x-y) + x^2 + y^2\\ &= \frac{w^2 - 2w\left(\frac{x-y}{3}\right) + \left(\frac{x-y}{3}\right)^2}{1/3} -\frac 13(x-y)^2 + x^2 + y^2 \end{align} and so by expanding out $2\phi(x-w)\phi(y+w)\phi(w)$ and doing some re-arranging of the integrands in $(3)$, we can write $$f_{X,Y}(x,y) = g(x,y)\big[P\{T \leq -y\} + P\{0 < T \leq x\}\big] \tag{4}$$ where $T$ is a normal random variable with mean $\frac{x-y}{3}$ and variance $\frac 13$. Both terms inside the square brackets involve the standard normal CDF $\Phi(\cdot)$ with arguments that are (different) functions of both $x$ and $y$. Thus, $f_{X,Y}$ is not a bivariate normal density even though both $X$ and $Y$ are normal random variables, and their sum is a normal random variable.


Comment: Joint normality of $X$ and $Y$ suffices for normality of $X+Y$ but it also implies much much more: $aX+bY$ is normal for all choices of $(a,b)$. Here, we need $aX+bY$ to be normal for only three choices of $(a,b)$, viz., $(1,0), (0,1), (1,1)$ where the first two enforce the oft-ignored condition (see e.g. the answer by $Y.H.$) that the (marginal) densities of $X$ and $Y$ must be normal densities, and the third says that the sum must also have a normal density. Thus, we can have normal random variables that are not jointly normal but whose sum is normal because we don't care what happens for other choices of $(a,b)$.

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