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I am aware of some nice examples of pairs of correlated random variables which are marginally normal but not jointly normal. See this answer by Dilip Sarwate, and this one by Cardinal.

I am also aware of an example of two normal random variables whose sum is not normal. See this answer by Macro. But in this example, the two random variables are uncorrelated.

Is there an example of two normal random variables which have non-zero covariance and whose sum is not normal? Or is it possible to prove that the sum of any two correlated normal random variables, even if they are not bivariate normal, must be normal?

[Context: I have a homework question which asks for the distribution of $aX+bY$ where $X$ and $Y$ are standard normals with correlation $\rho$. I think the question meant to specify that they are bivariate normal. But I am wondering whether anything can be said without this extra assumption for $\rho$ non-zero.]

Thanks!

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    $\begingroup$ Cardinal's answer, which you cite, already contains a solution: see the upper right corner in his panel of examples. $\endgroup$ – whuber Oct 21 '14 at 6:30
  • $\begingroup$ Please can you explain how? He specifies a joint distribution, which yields two normal marginals. It's not clear to me that the sum of the two normal marginals is not normal, which is what I'm after. (See also my comment on Glen_b's answer below.) $\endgroup$ – mww Oct 21 '14 at 18:26
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    $\begingroup$ From the picture alone it is obvious that the density of the sum at zero is zero (because the line $x+y=0$ intersects the plot in a single point, which has measure zero), while the sum itself is just as obviously symmetric about zero, showing that zero is the center of the distribution of the sum. Such a distribution cannot be Normal because Normal distributions have nonzero densities at their centers. $\endgroup$ – whuber Oct 21 '14 at 18:47
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Almost any bivariate copula will produce a pair of normal random variates with some nonzero correlation (some will give zero but they are special cases). Most (nearly all) of them will produce a non-normal sum.

In some copula families any desired (population) Spearman correlation can be produced; the difficulty is only in finding the Pearson correlation for normal margins; it's doable in principle, but the algebra may be fairly complicated in general. [However, if you have the population Spearman correlation, the Pearson correlation - at least for light tailed margins such as the Gaussian - may not be too far from it in many cases.]

All but the first two examples in cardinal's plot should give non-normal sums.


Some examples -- the first two are both from the same copula family as the fifth of cardinal's example bivariate distributions, the third is degenerate.

Example 1:

Clayton copula ($\theta=-0.7$)

histograms of normal margins, non-normal sum & plot of bivariate distribution

Here the sum is very distinctly peaked and fairly strongly right skew

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Example 2:

Clayton copula ($\theta=2$)

histograms of normal margins, non-normal sum & plot of bivariate distribution

Here the sum is mildly left skew. Just in case that's not quite obvious to everyone, here I flipped the distribution (i.e. we have a histogram of $-(x+y)$ in pale purple) and superimposed it so we can see the asymmetry more clearly:

$\hspace{1cm}$superimposed histogram of x+y and -(x+y)

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We could readily interchange the direction of skewness of the sum so that the negative correlation went with the left skew and positive correlation with the right skew (for example, by taking $X^*=-X$ and $Y^*=-Y$ in each of the above cases - the correlation of the new variables would be the same as before, but distribution of the sum would be flipped around 0, reversing the skewness).

On the other hand if we just negate one of them, we would change association between the strength of the skewness with the sign of the correlation (but not the direction of it).

It's worth also playing around with a few different copulas to get a sense of what can happen with the bivariate distribution and normal margins.

The Gaussian margins with a t-copula can be experimented with, without worrying much about details of copulas (generate from correlated bivariate t, which is easy, then transform to uniform margins via the probability integral transform, then transform uniform margins to Gaussian via the inverse normal cdf). It will have a non-normal-but-symmetric sum. So even if you don't have nice copula-packages, you can still do some things fairly readily (e.g. if I was trying to show an example quicly in Excel, I'd probably start with the t-copula).

--

Example 3: (this is more like what I should have started with initially)

Consider a copula based on a standard uniform $U$, and letting $V=U$ for $0\leq U<\frac{1}{2}$ and $V=\frac{3}{2}-U$ for $\frac{1}{2}\leq U\leq 1$. The result has uniform margins for $U$ and $V$, but the bivariate distribution is degenerate. Transforming both margins to normal $X=\Phi^{-1}(U), Y=\Phi^{-1}(V)\,$, we get a distribution for $X+Y$ that looks like this:

enter image description here

In this case the correlation between them is around 0.66.

So again, $X$ and $Y$ are correlated normals with a (in this case, distinctly) non-normal sum -- because they're not bivariate normal.

[One could generate a range of correlations by flipping the center of $U$ (in $(\frac{1}{2}-c,\frac{1}{2}+c)$, for $c$ in $[0,\frac{1}{2}]$), to obtain $V$. These would have a spike at 0 then a gap either side of that, with normal tails.]


Some code:

library("copula")
par(mfrow=c(2,2))

# Example 1
U <- rCopula(100000, claytonCopula(-.7))
x <- qnorm(U[,1])
y <- qnorm(U[,2])
cor(x,y)
hist(x,n=100)
hist(y,n=100)
xysum <- rowSums(qnorm(U))
hist(xysum,n=100,main="Histogram of x+y")
plot(x,y,cex=.6,
       col=rgb(0,100,0,70,maxColorValue=255),
       main="Bivariate distribution")
text(-3,-1.2,"cor = -0.68")
text(-2.5,-2.8,expression(paste("Clayton: ",theta," = -0.7")))

The second example:

#--
# Example 2:
U <- rCopula(100000, claytonCopula(2))
x <- qnorm(U[,1])
y <- qnorm(U[,2])
cor(x,y)
hist(x,n=100)
hist(y,n=100)
xysum <- rowSums(qnorm(U))
hist(xysum,n=100,main="Histogram of x+y")
plot(x,y,cex=.6,
    col=rgb(0,100,0,70,maxColorValue=255),
    main="Bivariate distribution")
text(3,-2.5,"cor = 0.68")
text(2.5,-3.6,expression(paste("Clayton: ",theta," = 2")))
#
par(mfrow=c(1,1))

Code for the the third example:

#--
# Example 3:
u <- runif(10000)
v <- ifelse(u<.5,u,1.5-u)
x <- qnorm(u)
y <- qnorm(v)
hist(x+y,n=100)
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  • $\begingroup$ Thanks - but if I'm not mistaken, $X+Y=2IZ+(1-I)U+(1-I)V$ is also normal. (When $I=0$, we get $U+V$, and when $I=1$, we get $2Z$. So with probability 1 we get the sum of two independent standard normals, which is normal.) I'm after a case where the sum of two correlated normals is not normal, rather than a case where the joint distribution is not normal. $\endgroup$ – mww Oct 21 '14 at 18:23
  • $\begingroup$ Quite right -- through my various attempts to make a non-bivariate-normal example where I could choose the correlation, somewhere along the line I stopped checking the sum wasn't normal. I'll replace the example with something where I do demonstrate a non-normal sum, but it won't have the ability to directly select $\rho$. Hold on, it might take an hour or so before I can get to it. $\endgroup$ – Glen_b Oct 21 '14 at 20:17
  • $\begingroup$ I've replaced the example with two specific examples using Clayton copulas $\endgroup$ – Glen_b Oct 21 '14 at 22:39
  • $\begingroup$ Fabulous - thanks! Especial thanks for the R code. $\endgroup$ – mww Oct 21 '14 at 23:01
  • $\begingroup$ I added a third example and at the end of that I outline a way to get something like what I was originally attempting - a way to get a tunable correlation between -1 and 1 (aside from special cases at the ends), but for which the sum is non-normal. $\endgroup$ – Glen_b Oct 22 '14 at 6:02
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I came up with one example. X is standard normal variable, and Y=-X. Then X+Y=0, which is constant. Can anyone confirm it's a counterexample?

We know the fact if X,Y are jointly normal, then their sum is also normal. But what if their correlation is -1??

I am a little confused about this. Thx.

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  • $\begingroup$ You get the same thing is true when X=Y and then X-Y =0. These are normal distributions that are not bivariate normal. Hence the property that linear combinations of are normal which applies to the bivariate normal does not need apply. $\endgroup$ – Michael Chernick Apr 4 '18 at 0:12
  • $\begingroup$ @Zirui IMO its a degenerate case of the normal ($\sigma\to 0$) rather than a direct counterexample, though it depends on your definitions. $\endgroup$ – Glen_b Jul 14 '18 at 4:35

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