Students of statistics and probability theory sometimes develop misconceptions about the normal distribution, ideas that may seem plausible but are mathematically untrue. For example, it is sometimes mistakenly thought that two linearly uncorrelated, normally distributed random variables must be statistically independent. However, this is untrue, as can be demonstrated by counterexample. Likewise, it is sometimes mistakenly thought that a linear combination of normally distributed random variables will itself be normally distributed, but again, counterexamples prove this wrong. However, it is possible for two random variables <math>X</math> and <math>Y</math> to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.
Examples
A symmetric example
thumb|alt=Two normally distributed, uncorrelated but dependent variables.|Joint range of <math>X</math> and <math>Y</math>. Darker indicates higher value of the density function.
Suppose <math>X</math> has a normal distribution with expected value 0 and variance 1. Let <math>W</math> have the Rademacher distribution, so that <math>W=1</math> or <math>W=-1</math>, each with probability 1/2, and assume <math>W</math> is independent of <math>X</math>. Let <math>Y=WX</math>. Then <math>X</math> and <math>Y</math> are uncorrelated, as can be verified by calculating their covariance. Moreover, both have the same normal distribution. And yet, <math>X</math> and <math>Y</math> are not independent.
To see that <math>X</math> and <math>Y</math> are not independent, observe that <math>|Y|=|X|</math> or that <math>\operatorname{Pr}(Y > 1 \mid -1/2<X<1/2) = \operatorname{Pr}(X > 1 \mid -1/2<X<1/2) = 0</math>.
Finally, the distribution of the simple linear combination <math>X+Y</math> concentrates positive probability at 0: <math>\operatorname{Pr}(X+Y=0) = 1/2</math>. Therefore, the random variable <math>X+Y</math> is not normally distributed, and so also <math>X</math> and <math>Y</math> are not jointly normally distributed (by the definition above).
<math display="block">\begin{align}\Pr(Y \leq x) &= \Pr(\{|X| \leq c\text{ and }X \leq x\}\text{ or }\{|X|>c\text{ and }-X \leq x\})\\
&= \Pr(|X| \leq c\text{ and }X \leq x) + \Pr(|X|>c\text{ and }-X \leq x)\\
&= \Pr(|X| \leq c\text{ and }X \leq x) + \Pr(|X|>c\text{ and }X \leq x) \\
&= \Pr(X \leq x), \end{align}</math>
where the next-to-last equality follows from the symmetry of the distribution of <math>X</math> and the symmetry of the condition that <math>|X| \leq c</math>.
In this example, the difference <math>X-Y</math> is nowhere near being normally distributed, since it has a substantial probability (about 0.88) of it being equal to 0. By contrast, the normal distribution, being a continuous distribution, has no discrete part—that is, it does not concentrate more than zero probability at any single point. Consequently <math>X</math> and <math>Y</math> are not jointly normally distributed, even though they are separately normally distributed.
Examples with support almost everywhere in the plane
Suppose that the coordinates <math>(X,Y)</math> of a random point in the plane are chosen according to the probability density function <math display="block">p(x,y) = \frac{1}{2\pi\sqrt{3 \left[\exp\left(-\frac{2}{3}(x^2 + xy + y^2)\right) + \exp\left(-\frac{2}{3}(x^2 - xy + y^2)\right)\right].</math>Then the random variables <math>X</math> and <math>Y</math> are uncorrelated, and each of them is normally distributed (with mean 0 and variance 1), but they are not independent.
It is well-known that the ratio <math>C</math> of two independent standard normal random deviates <math>X_{i}</math> and <math>Y_{i}</math> has a Cauchy distribution.